M
A
R
T
I
N C. M
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Clarkson College of Technology 
Potsdam, New York 
and 
C
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Rochester Institute of Technology 
Rochester, New York 
E L E M E N T S 
O F 
C L A S S I C A L 
P H Y S I C S 
P
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I N C . 
New York 
· 
Toronto 
Oxford 
· 
Sydney 
· 
Braunschweig 

PERGAMON PRESS INC. 
Maxwell House, Fairview Park, Elmsford, N.Y. 10523 
PERGAMON OF CANADA LTD. 
207 Queen's Quay West, Toronto 117, Ontario 
PERGAMON PRESS LTD. 
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PERGAMON PRESS (AUST.) PTY. LTD. 
Rushcutters Bay, Sydney, N.S.W. 
PERGAMON GmbH 
Burgplatz 1, Braunschweig 
Copyright ©  1975, Pergamon Press Inc. 
Library of Congress Cataloging in Publication Data 
Martin, Martin C 
Elements of classical physics. 
1. 
Physics. 
I. 
Hewett, Charles Á ., joint author. 
Ð. Title. 
QC21.2.M38 
1973 
530.Ã 32 
73-3450 
ISBN 0-08-017098-6 
All Rights Reserved. N o part of this publication may be reproduced, 
stored in a retrieval system or transmitted in any form, or by any 
means, electronic, mechanical, photocopying, recording or otherwise, 
without prior permission of Pergamon Press Inc. 
Printed in the United States of America 

The Authors 
Martin C. Martin (Ph.D. University of Alberta, Edmonton, Alberta, Canada) is 
Associate Professor of Physics at Clarkson College of Technology. His teaching 
efforts have been in the area of classical physics at the undergraduate and graduate 
levels and modem physics at the undergraduate level. H e has carried on research in 
the field of solid state, specifically on the effect of elastic and plastic deformation on 
the physical properties of metals, and the effect of heat treatment on the physical 
properties of metal alloys. 
Charles A. Hewett (Ph.D. University of Missouri-Columbia) is Professor of Physics at 
the Rochester Institute of Technology. His primary teaching efforts have been in the 
areas of classical and modern physics for engineering and science students. Currently 
he is working with Dr. Hrishikesh Banerjee to develop studies in direct energy 
conversion and in the physics of semiconductor electronic devices in response to 
requests of students and faculty members of the College of Engineering. Collabora-
tion with Dr. James Karpick, a local industrial physicist, is expected to result in a 
program of independent study and undergraduate research in 
fluorescence 
and 
photoluminescence of solids that will meet the needs of gifted high school students as 
well as the more usual group of physics majors. 

Preface 
This book is being published at a time when many of the nation's leading physicists already have 
prepared textbooks which implement in an outstanding fashion new approaches developed for 
the teaching of elementary physics to today's students. The appearance of still another textbook 
should therefore be accompanied by a presentation of the ideas behind the structure and content 
of the book. 
It has been our experience that students retain an uneven coverage of the various areas of 
physics from their high school courses. Areas such as elementary heat and light are, in general, 
more easily understood and remembered than mechanics, thermodynamics, sound, and electric-
ity and magnetism, which necessarily involve a greater degree of abstraction in their presentation 
as well as greater mathematical sophistication if quantitative discussions are desired. 
With very few exceptions, the courses in general physics that have been developed for presenta-
tion to entering freshmen begin with mechanics, which represents a relatively serious intellectual 
challenge for the average student, with or without the use of calculus. This is then followed by 
heat, light, and sound, which are in turn followed by electricity and magnetism. As a result, the 
students are confronted by a course that is rather frustrating in the unevenness of its demands 
upon their understanding. If, in addition, calculus is used from the outset, the average student is 
initially discouraged, not only by the relatively foreign quantitative concepts of mechanics, but 
also by the presentation of these concepts in a mathematical "language" that is equally foreign. 
As a result, many students succumb to these demands and terminate their study of physics or 
engineering with the mistaken impression that an understanding of the physical world is beyond 
their grasp. 
With these considerations in mind, we have written a text which: (a) begins with the material 
most readily understood with a minimal mathematical framework, thus providing a smooth 
transition from high school; (b) progresses as uniformly as possible to areas of increasing con-
ceptual difficulty; (c) introduces at appropriate places the necessary mathematical concepts at a 
time when they would already have been presented in a typical concurrent mathematical course, 
and (d) stresses throughout the course the physical concepts and the manner in which these 
concepts can be used to provide quantitative understanding in a wide variety of specific situa-
tions. Many of these are carefully discussed as examples, and the remainder are presented in the 
problem sections as suitable tests of understanding. 
In writing the text, we presume the student has at least a qualitative understanding of the 
meaning of the terms force, pressure, work, and energy. 
After a section devoted to the discussion of dimensions and units, the areas presented are heat, 
light, mechanics, thermodynamics, sound, and electricity and magnetism, respectively. It is our 
opinion that this sequence of topics is best suited for achieving the goals outlined above. How-
ever, those desiring a more conventional course sequence could with very little difficulty begin 
with mechanics (Chapter 14), followed by the remaining topics in their usual order. In addition. 
Chapter 15 on special relativity may be omitted without prejudice to subsequent chapters. N o 
effort has been made to include an extensive presentation of modern physics. There are two 
vii 

vlll 
Preface 
reasons for this: first, the material covered represents a thorough coverage of classical physics 
which is a necessary prerequisite to the proper discussion of modern physics, and second, the 
mathematical preparation required for modern physics in our opinion necessarily makes it a 
second year subject in contrast to the contents of this text. It might also be noted that excellent 
texts covering the subject are available. 
We have emphasized the rationalized MKS units throughout the text. We have also made use of 
some of the various other systems of units in common use today. We introduce the necessary 
conversions for these systems in an appendix. Every system of units has its proponents and 
opponents (who are equally convinced of the correctness of their point of view); our choice 
simply represents a personal preference. 
The authors of this book owe much to many people: our teachers; our colleagues; but, most 
important, our students, whose desire to understand physics has prompted us to try to make then* 
pathway to knowledge as natural as our abilities permit. 

Introduction: 
Dimensions and Units 
In the study of physics, the dimensions and units to be encountered must be understood. A 
student must remember that one can only equate the same kinds of quantities. A test on the 
correctness of an equation can be obtained by checking whether or not the dimensions on one 
side of the equation are the same as those on the other side. 
D I M E N S I O N S 
A dimension may be defined as a name describing certain physical quantities. Therefore, a large 
number of dimensions are possible. This number can be reduced by the fact that certain descrip-
tions can be expressed in terms of other more basic descriptions (dimensions). For example, 
length, area, and volume are dimensions, but area can be measured as a length squared and 
volume as a length cubed. Therefore, the dimensions of area and volume can be stated in terms of 
the more fundamental dimension of length. 
In physics, there are five fundamental dimensions: length, mass, time, temperature, and electric 
charge, which we shall denote by [/], [m], [t], [T], and [<?], respectively. We have already shown 
how area and volume are expressed in terms of length. Now let us consider a few more physical 
quantities which should be familiar from high school physics. 
Velocity = length ^ time = [li 
Acceleration = velocity -j- time = [lt~^] 
Force = mass x acceleration = [mlt~^] 
Pressure = force ^ area = [ml'^i 
Work = force x length = [m/'r'] 
Power = work ^ time = [ml^t~^] 
Density = mass -j- volume = [m/"^] 
Current = charge 
time = [qt~^] 
U N I T S 
A unit may be defined as a particular amount of the dimension or quantity to be measured. Three 
different systems are used today in science and engineering. They are the meter-kilogram-second 
(MKS) system, the centimeter-gram-second (cgs) system, and the foot-slug-second (British) sys-
tem. The rationalized MKS and cgs systems are used universally in scientific work, with the 
MKS actually superseding the two other systems. 
We shall use the rationalized MKS system throughout most of the book. Much of the literature is 
still written in the other two systems of units, so we feel that students should eventually become 
ix 

Introduction 
familiar with the three systems of units. In Table 1 are the MKS units for the physical quantities 
given in Section 1 . For comparison purposes, the equivalent cgs unit is also indicated. 
T a b l e 1 
M K S Units for S o m e Physical Quantities. 
Physical Quantity 
MKS Unit 
cgs Unit 
length 
meter 
centimeter 
mass 
kilogram 
gram 
velocity 
meter/second 
centimeter/second 
acceleration 
meter/second^ 
centimeter/second^ 
force 
newton 
dyne 
pressure 
newton/meter^ 
dyne/centimeter^ 
work or energy 
joule 
erg 
power 
watt 
erg/second 
density 
kilogram/meter^ 
gram/centimeter' 
charge 
coulomb 
statcoulomb 
current 
ampere 
statampere 
The student should be familiar with the above physical quantities and the MKS units associated 
with them, even though a detailed discussion of them will not appear until later in the book. 
Table 2 then gives some prefixes used to represent divisions and multiples of metric quantities. 
T a b l e 2 
Prefixes Used for Divisions and 
Multiples of Metric Quantities. 
deci-
1 0 -
deca-
10· 
centi-
10-^ 
hecto-
10^ 
milli-
1 0 - ' 
kilo-
1 0 ' 
micro-
1 0 " ' 
mega-
1 0 ' 
nano-
1 0 " ' 
giga-
1 0 ' 
pico-
1 0 - " 
tera-
10·^ 

1 
Temperature and 
Thermometry 
1-1 
Thermal Energy, Heat, and 
Temperature 
1 
1-2 
Temperature Measurement 
2 
1-3 
Centigrade and Fahrenheit 
Temperature Scales 
2 
1-4 
General Definition of Temperature 
1-5 
Kelvin and Rankine Temperature 
Scales 
3 
1-6 
Mercury Thermometer 
4 
' 1 -7 
Gas Thermometers 
4 
1-8 
The Platinum Resistance 
Thermometer 
5 
1-9 
Thermocouples 
5 
Consistent with our intent to progress by degrees 
from the easiest to the most difficult topics, we will 
begin by discussing various aspects of thermal 
energy, heat, and temperature. 
1-1 
T H E R M A L E N E R G Y , HEAT, 
A N D T E M P E R A T U R E 
The concept of temperature is fundamental to the 
study of heat. The idea of temperature developed 
from man's sense of hotness and coldness. There-
fore, a body's temperature represents its degree of 
hotness or coldness. In order to define temperature 
in a more concrete way than the above, we will first 
define thermal energy and heat. 
We shall define thermal energy as follows: 
Thermal energy is the energy of the atoms in 
a substance because of their motion on a 
microscopic scale. 
We shall later see in our discussion of kinetic 
theory that the average motional or kinetic energy 
of atoms is directly proportional to the tempera-
ture. 
We shall now define heat as follows: 
Heat is thermal energy transferred between 
two or more material substances or from one 
portion of the substance to another on a 
macroscopic scale. 
This means that a hot body can give up some of 
its thermal energy and therefore affect a neighbor-
ing body. The quantity of thermal energy it gives up 
depends upon the nature and condition of the 
neighboring body and upon the medium separating 
the two bodies. 
Now we are able to give a somewhat better 
definition of temperature; we will define it as fol-
lows: 
The temperature of a body is a measure of its 
ability to transfer heat to other bodies. 
A body is at the same temperature as another 
body if there is no net flow of heat from one to the 
other when they are placed in contact or separated 
only by a conducting wall. We can also say that a 
body X is at a higher temperature than a body Y if 
heat flows from X to Y when they are placed in 
contact or separated only by a conducting wall. 

Temperature ar)d Thermometry 
The above definition of temperature gives us a 
way of defining absolute zero, which is the lower 
limit of temperature. That is, no body can be colder 
or have less thermal energy than a body at absolute 
zero. Therefore, we can define absolute zero as fol-
lows: 
Absolute zero is the temperature of a body 
that is not capable of transferring any thermal 
energy to another body. 
There is a criticism of the above definitions on 
the basis that we do not observe the flow of heat. 
All we can observe is that the cold body gets hotter 
and the hot body gets cooler. However, many phys-
ical processes occur especially in atomic and nu-
clear physics that we cannot observe directly, so 
this is not a critical problem. Other definitions of 
temperature will be given later on in the book which 
remove this criticism. However, we feel that at the 
beginning it is good to have a definition of tempera-
ture even though it may be subject to later slight 
modifications. 
1-2 
T E M P E R A T U R E 
M E A S U R E M E N T 
We shall now discuss thermometry, which is the 
science of measuring temperatures. A number of 
physical properties of substances change with 
temperature. Temperature measuring devices or 
thermometers make use of these properties. Some 
of these properties which change with temperature 
and can be used for thermometric purposes are: 
1. Volume of liquids. 
2. Length of solids. 
3. Volume of gases. 
4. Pressure of gases. 
5. Pressure of saturated vapors. 
6. Electrical resistance of metals. 
7. Thermoelectric currents. 
8. Color of radiated light. 
9. Total radiation intensity. 
10. Magnetic 
susceptibility 
of 
paramagnetic 
salts. 
The construction of a useful thermometer re-
quires a choice of some substance whose ther-
mometric properties indicate changes in tempera-
ture and a choice of the design of the thermometer. 
We then require a thermometer scale, which neces-
sitates a choice of one or more fixed temperature 
points and a selection of numbers to be associated 
with each division. 
1-3 
C E N T I G R A D E A N D F A H R E N H E I T 
T E M P E R A T U R E 
S C A L E S 
The definitions of the Centigrade and Fahrenheit 
scales involve in each case the same two distinct 
fixed thermal points. The lower fixed point is the 
temperature of the melting point of pure ice (ice 
point), which is the temperature at which pure ice 
coexists with air saturated water at one atmosphere 
pressure. The upper fixed point is the temperature 
of the boiling point of pure water (steam point), 
which is the temperature of equilibrium between 
pure water and pure steam at one atmosphere pres-
sure. On the Centigrade scale, the ice point is taken 
as 0 degrees and the steam point as 100 degrees. On 
the Fahrenheit scale, the ice point is taken as 32 
degrees, the steam point as 212 degrees. There are 
100 degrees between the ice and steam points on 
the Centigrade scale, while there are 180 degrees 
between the ice and steam points on the Fahrenheit 
scale. We leave it to the student as an exercise to 
show that the conversion from the Centigrade scale 
to the Fahrenheit scale or vice versa is given by the 
equations 
í°C = | ( í ° F - 3 2 ) , 
í T = |rC + 32. 
(1-1) 
Note that we are using (i) to indicate Centigrade 
and Fahrenheit temperatures. Later in the text 
this same symbol will be used to denote time. The 
double usage of this symbol should not lead to am-
biguity if dimensional analysis is used. 
Example 1 . At what temperatures do the Centi-
grade and Fahrenheit scales coincide? 
SOLUTION 
This means the rC = i^F. Substitute i°C = i'^F in 
Eq. (1-1). 
or 
Therefore, 
í°F = |rF + 32 
^í°F = - 3 2 . 
, o p . 5 ^ 3 2 ) _ 
4 

Kelvin and Rankine Temperature Scales 
since 
1-4 
rC = -40°. 
G E N E R A L DEFINITION OF 
T E M P E R A T U R E 
Let us consider a general definition of tempera-
ture based on the thermometric properties of a 
substance. Let Xo be some thermometric property 
as, for example, length of a solid or height of a 
liquid in a capillary tube measured at 0°C, and Xioo 
and Xt the thermometric property at 100°C and 
some unknown temperature ί Τ . The size of a unit 
interval or degree is Xioo "~ Xo/100, and the size of 
the interval from 0°C to ί Τ  is Xt -Xo. By defini-
tion, i°C is the number of degrees in the interval 
Xt - Xo, and is given by the equation 
X\ ~ XQ 
XlOO — XQ 
100 
or 
\Xioo — X o / 
(1-2) 
Various thermometric properties can be used. 
For example, let us consider a mercury in glass 
thermometer, one with which we are all familiar, 
and a constant volume gas thermometer, which we 
will discuss later. If the length of the mercury col-
umn is substituted in Eq. (1-2) for the thermometric 
property X , we have 
V/ioo-Zo/ 100. 
(1-3) 
If the gas pressure of the gas in the gas thermome-
ter is substituted in Eq. (1-2) for X , we have 
VPioo — Po/ 
(1-4) 
A word of caution is necessary here. Even 
though both the above thermometers read the same 
at OT and lOOT, they do not necessarily read the 
same at in-between temperatures. There is no 
reason to expect them to be the same since the 
temperature as defined by Eq. (1-2) depends upon 
the thermometric property of the substance. Even 
though two substances like mercury and a particu-
lar gas might have the same thermometric property, 
they might be contained in glass which has different 
thermometric properties, so the properties relative 
to the glass would be different. Experiments have 
shown that if we compare scales of temperatures 
based on different liquids or gases, or even on the 
same liquid or gas in different kinds of glass, the 
scale reads slightly differently. However, in the 
range between 0°C and lOOT they generally agree 
to within a few tenths of a degree. 
1-5 
K E L V I N A N D R A N K I N E T E M P E R A T U R E 
S C A L E S 
We shall see later that there is a limit to the low-
est temperature that can ever be attained. This 
lowest temperature is - 273.15°C; it corresponds to 
absolute zero which we previously defined. The ab-
solute temperature scale has absolute zero as its 
zero point. The absolute scale with degree intervals 
equal to those on the Centigrade scale is called the 
Kelvin or absolute scale, and the absolute scale 
with degree intervals equal to those on the 
Fahrenheit scale is called the Rankine scale. The 
temperature of absolute zero on the Fahrenheit 
scale is - 459.67°F. For most temperature measure-
ments, it is only necessary to have three figure ac-
curacy so as to reduce memory work. We shall in 
the calculation in this text relate the Kelvin and 
Centigrade, and the Rankine and Fahrenheit scales 
by the following equations: 
and 
T°K = i°C-h273, 
r°R=i° F + 460. 
(1-5) 
(1-6) 
Both the ice point and the steam point are diffi-
cult to measure with good accuracy. The main diffi-
culty is that when ice melts it becomes surrounded 
with pure water and causes poor contact between 
the ice and the air saturated water. This prevents 
the ice point from being accurately measured. Since 
the boiling point of water is sensitive to small 
changes of pressure, the value of the steam point 
can be in error unless the pressure is kept very con-
stant. 
In 1854, Kelvin proposed a temperature scale 
known as the thermodynamic scale (discussed in 
Chapter 26), which is independent of the properties 
of any substance. He pointed out that the scale 
could be determined by the use of a single fixed 
point. At the Tenth General Conference of Weights 
and Measures in 1954, it was decided to choose this 
fixed point as the temperature and pressure at 

Temperature and Thermometry 
which ice, water, and water vapor coexist in 
equilibrium (the triple point of water), and to assign 
it the value of 273.16°K. This point can be measured 
with greater accuracy than the ice point and the 
steam point. 
1-6 
M E R C U R Y T H E R M O M E T E R 
The mercury thermometer is the most frequently 
used thermometer in laboratories because of its 
simplicity, its fairly large temperature range (- 39°C 
to 357°C), and its quick response to slight tempera-
ture changes. It is made by first blowing a bulb at 
one end of a fine bore glass tube. Mercury is poured 
into the tube, heated to drive out any air, and the 
other end of the tube is sealed. When the bulb is 
brought into contact with a body, the mercury 
either expands or contracts relative to the glass 
tube depending on whether the body is hotter or 
colder than the original surroundings of the bulb. 
The chief errors in a mercury thermometer are 
caused by a non-uniform bore in the tube and im-
perfections of the glass bulb. However, if these and 
other smaller errors are corrected, the mercury 
thermometer is very accurate. A good mercury 
thermometer is only surpassed in accuracy by a 
platinum resistance thermometer in the range 0°C 
to 200T. 
1-7 
G A S T H E R M O M E T E R S 
Gas thermometers are usually not used in 
laboratories because they are somewhat elaborate 
and cumbersome. They are used generally in stan-
dardizing other more simple thermometers. For a 
given temperature rise, gases expand about 20 
times as much as mercury and about 120 times as 
much as the glass in the thermometer. This reduces 
error due to uneven changes in the volume of the 
glass. Gas thermometers have a temperature range 
of - 269°C to 1600°C. 
There are two types of gas thermometers 
—namely, constant volume and constant pressure 
thermometers. With the first type, the temperature 
is measured by the change in pressure of the gas 
with temperature. With the second type, the 
temperature is measured by the change in volume 
of the gas with temperature. Constant volume ther-
mometers are more satisfactory because it is easier 
to measure accurately a change of pressure than a 
change of volume. Figure 1-1 is a simple form of a 
constant volume gas thermometer. 
The volume of the gas in the bulb is kept constant 
by raising or lowering tube C so that the top of the 
mercury is always kept at point B. The pressure is 
measured by measuring the difference between the 
heights h of the mercury surfaces in the tube. The 
pressure of the gas in the bulb is equal to the at-
mospheric pressure plus the pressure due to the 
height h of the mercury in the tube. If we let po be 
the pressure when the bulb is surrounded by melt-
ing ice, ρ 100—the pressure when the bulb is sur-
rounded by steam, and pt—the 
pressure when the 
bulb is surrounded by the substance whose temper-
ature is to be measured, the temperature of the sub-
stance is given by Eq. (1-4), namely. 
i°C 
ÍPt-Po\ 
VP 100-Po/ 100. 
(1-7) 
Hydrogen is the gas usually used, except for ex-
treme low temperatures where helium is used and 
extreme high temperatures where nitrogen is used. 
Figure 1-1 Constant volume gas thermometer as 
described In text. 

Problems 
Helium is used at very low temperatures because it 
does not liquefy until a temperature of about 
- 269°C is reached. Nitrogen is used at high temper-
atures because both hydrogen and helium diffuse 
through the bulb at very high temperatures. 
1-8 
THE PLATINUM R E S I S T A N C E 
T H E R M O M E T E R 
It has been previously stated that the electrical 
resistance of metals changes with temperature. It is 
best to use a noble metal for a thermometer so that 
there is no oxidation. In 1887, Callendar investi-
gated how the resistance of platinum varied with 
temperature. He showed that the change in electri-
cal resistance of platinum could be used for a very 
accurate scale of temperatures over a wide temper-
ature range. Today, it is regarded as the most accu-
rate thermometer in the range -190°C to 660°C. 
If we define the resistance temperature scale by 
applying Eq. (1-2), then 
-(π) 
\lVioo 
-Ko/ 
100, 
(1-8) 
where Ro, Riw, and R , are the resistances of the 
platinum wire at the ice point, steam point, and the 
temperature t, respectively. Callendar also showed 
experimentally that the resistance Rt of a platinum 
wire at temperature t is given by the equation 
Rt = Roil + αί + 
b t \ 
(1-9) 
where Ro is the resistance at the ice point, and a 
and b are constants. In order to determine the con-
stants a and b, three fixed points are necessary. 
These fixed points have been chosen as the ice 
point, the steam point, and the sulphur point which 
is the boiling point of sulphur (444.6°C) at atmos-
pheric pressure. 
Other resistance thermometers of cheaper metals 
(for example, copper) are sometimes used where 
extreme accuracy is not necessary. Also, some 
semi-conductors whose resistance decreases with 
temperature increase are used. They are known as 
thermistors; they have a very large decrease in re-
sistance with increase in temperature and have a 
greater sensitivity than a platinum thermometer, 
but they are less accurate. 
1-9 
T H E R M O C O U P L E S 
Thermocouples are based on the principle that 
electric currents are set up in a closed circuit con-
sisting of two dissimilar metals in contact with one 
another when the junctions are at different temper-
atures (Chapter 34). The reference junction is usu-
ally put in melting ice and the other junction in 
contact with the material whose temperature is to 
be measured. The current (or voltage across a cir-
cuit element) is then related to the temperature by 
appropriate calibrations. 
The most common thermocouples are con-
structed of iron and constantan (an alloy of copper 
and nickel), copper and constantan, and platinum 
and an alloy of platinum-rhodium. Thermocouples 
have the advantage that they can be used for 
measuring temperatures at a particular point in an 
experimental system. 
P R O B L E M S 
1. In a certain city the greatest temperature variation recorded during a 24 hour period was 45°F. This 
corresponds to how many degrees C? 
2. (a) A Centigrade degree is what fraction of a Fahrenheit degree? 
(b) The conversion from the Centigrade scale to the Fahrenheit scale is given by the equation 
í T = | í ° C + 32. 
Justify this relation. 
3. The reading of a thermometer for the temperature of a room is 77° F. What is the reading on the 
Centigrade scale? 
4. (a) What Centigrade temperature corresponds to 233° Kelvin? 
(b) What Fahrenheit temperature corresponds to 77° Kelvin? 
5. At what temperature will the reading of a Fahrenheit thermometer be three times that of a Centigrade 
thermometer? 

Temperature and Thermometry 
6. (a) The temperature of liquid nitrogen is about -1%°C. What is its temperature on the Kelvin scale? on 
the Rankine scale? 
(b) What is the difference between 0°C and 180T in degrees C? in degrees F? 
7. A thermometer was graduated to read 0°C at the boiling point of water, and 155° at the freezing point. 
Find the reading on the Fahrenheit and Centigrade scale corresponding to 80° on this thermometer. 
8. The resistance of a platinum resistance thermometer is found to be 10.000 ohms at the ice point, 
13.855 ohms at the steam point, and 25.261 ohms at the sulphur point. Find the constants a and b in Eq. 
(1-9), and plot R against t in the range 0° to 600°C. 

2 
Quantity of Heat and 
Calorimetry 
2-1 
Quantity of Heat 
7 
2-2 
Specific Heat and Heat Capacity 
7 
2-3 
Calorimetry 
8 
2-4 
Method of Mixtures 
8 
2-5 
Method of Cooling 
9 
2-6 
Latent Heat of Fusion 
10 
2-7 
Latent Heat of Vaporization 
2-8 
Heat of Combustion 
10 
10 
2-1 
QUANTITY O F HEAT 
When a hot and a cold body are brought into con-
tact, there is a transfer of heat from the hot body to 
the cold body until they come to the same tempera-
ture or reach a state that we term thermal equilib-
rium. The transference of heat from a hot body to a 
cold body is analogous to the flow of water from a 
high level to a low level. 
The increase in temperature of a body when 
heated depends on its mass and the material of 
which it is composed. For example, to bring a large 
container of water to the boiling point with a 
laboratory burner requires a long time. The ñnal 
temperature may be relatively low, even though it 
receives a large quantity of heat. If a wire is held in 
a burner flame, it comes to a very high temperature 
in a very short time even though it receives only a 
small quantity of heat. Therefore, the final tempera-
ture reached by two bodies in contact initially at 
different temperatures depends on their masses, the 
materials of which they are composed, and their ini-
tial temperatures. 
The net result of heat transfer to or from bodies 
until thermal equilibrium is reached is given by the 
relation: 
heat gained by cold bodies = heat lost by hot bodies. 
In order to make use of this relation, we must 
define a unit for measuring quantities of heat or 
thermal energy. The unit we use depends on the 
system of units we adopt. In the metric system of 
units, the unit of heat or thermal energy is either the 
calorie or kilocalorie. If we use centimeter-gram-
second (cgs) units, the unit of heat is the calorie; 
while if we use meter-kilogram-second (MKS) 
units, the unit of heat is the kilocalorie. The calorie 
is the quantity of heat required to raise the tempera-
ture of one gram of water through one centigrade 
degree, while the kilocalorie is the quantity of heat 
required to raise the temperature of one kilogram of 
water through one centigrade degree. 
The above definition of the heat unit depends 
slightly on the location of the one degree interval. It 
has been agreed to choose this interval between 
14.5°C and 15.5°C. 
2-2 
S P E C I F I C HEAT A N D HEAT 
C A P A C I T Y 
The specific heat of a body is the heat required to 
raise the temperature of a unit mass of the body one 
degree. 
In the MKS system of units, it is the heat re-
quired to raise the temperature of a one kilogram 

Quantity of Heat and Calorimetry 
mass of the body one degree centigrade. From this 
definition and the definition of the kilocalorie, we 
see that the specific heat of water is 1 kilocalorie 
per kilogram per degree centigrade. Although we 
are concerned here with the MKS system of units, 
it seems noteworthy to mention that the numerical 
value of the specific heats is the same in the cgs 
system of units. 
The quantity of heat necessary to raise the tem-
perature of a body of mass m and specific heat c by 
Δ ί degrees is given by 
Q = mcAt, 
(2-1) 
The specific heat of a substance is roughly con-
stant at ordinary temperatures provided the tem-
perature interval is not too great. As the tempera-
ture is lowered, the specific heats of all substances 
show a decrease. The specific heat of a substance is 
changed by (1) change of state such as from a solid 
to a liquid, liquid to a vapor, and vice versa, (2) 
presence of impurities, and (3) change in tempera-
ture. 
It is convenient, as we shall see in calorimetry, to 
define a quantity called heat capacity. 
The heat capacity of a body is the quantity of 
heat required to raise the temperature of the 
body one degree. It is the product of the mass 
of the body and its specific heat. 
In the MKS system, the unit of heat capacity is 
kilocalories per degree centigrade. 
2-3 
C A L O R I M E T R Y 
Calorimetry refers to the laboratory science of 
making measurements of quantities of heat. In 
making measurements of quantities of heat, a con-
tainer called a calorimeter is used. The ordinary 
calorimeter is a vessel placed within another larger 
vessel, with the two vessels insulated from one 
another. In this way, exchange of heat to the sur-
roundings is minimized. The inner vessel of the 
calorimeter must be provided with a stirrer so as to 
keep its contents at a uniform temperature. 
The fact that when two bodies at different tem-
peratures are placed in contact, heat is transferred 
from the hotter one to the cooler one until their 
temperatures are equal, gives us one of the most 
convenient methods of determining specific heat, 
known as the method of mixtures. 
2-4 
M E T H O D O F M I X T U R E S 
(a) Determination of the Specific Heat of a 
Solid 
Let a solid of mass m, temperature i, of unknown 
specific heat Cx be inunersed in a mass mi of water 
at temperature ti less than t contained in a 
calorimeter of mass mi, known specific heat c, and 
also at temperature ii. Let ti be the final tempera-
ture after thermal equilibrium is reached. 
Heat lost by the solid = mcxit 
- Í2). 
Heat gained by the water = mi(Í2 - íi). 
Heat gained by the calorimeter = m 2 c ( Í 2 - ii). 
Heat lost by the solid = heat gained by the 
water + heat gained by the calorimeter. Therefore, 
mCxit - Í2) = m , ( Í 2 - Í.) + m 2 c ( Í 2 - íi), 
(2-2) 
from which Cx can be determined. 
(b) Determination of the Specific Heat of a 
Liquid 
The specific heat of a liquid may be determined in 
the same way by the method of mixtures. If a solid 
of known mass, specific heat, and temperature is 
immersed in a known mass of the liquid contained 
in a calorimeter at a known temperature, the 
specific heat of the liquid can be calculated. 
Example 1. A 0.10 kg calorimeter of specific 
heat 0.090 kcal/kgT contains 0.15 kg of a liquid at 
20°C. Into this container is placed a 0.20 kg block of 
copper of specific heat 0.095 kcal/kgT and temper-
ature lOO^'C. The final temperature is 40°C. What is 
the specific heat of the liquid? 
SOLUTION 
Heat gained by liquids heat gained by calo-
rimeter = heat lost by copper block. 
(0.15kg)(c,)(40-20)^C 
+ (0.10 kg)(0.090 kcal/kg°C)(40-20)°C 
= (0.20 kg)(0.095 kcal/kg°C)(100-40)^C 
(Cx)(3.0 kg^C) + (0.18 kcal) = 1.14 kcal 
0.%kcal ,Q 32 kcal 
3.0 k g T 
kgT' 
The method of mixtures also can be used to de-
termine the specific heat of gases, but it is very diffi-
cult to perform experimentally. 

Method of Cooling 
9 
2-5 
M E T H O D O F C O O L I N G 
Another way of determining the specific heat of a 
liquid is by the method of cooling. The apparatus 
consists of a calorimeter with a test tube for the 
liquid suspended in the inner calorimeter (Figure 
2-1). A and Β are thermometers and C is a stirrer 
which enables the liquid to be stirred prior to its 
temperature being observed. The space D is filled 
with ice. 
The test tube is first fiUed with hot water, and a 
curve similar to that in Figure 2-2 is plotted. It is 
then filled with an equal volume of the liquid at the 
Figure 2-2 Cooling curve for a liquid. 
same temperature as the hot water, and a cooling 
curve for the liquid is plotted. From the graph, the 
intervals of time are obtained for which the liquid 
and water cool through the same temperature inter-
val Δ ί. Let ti be the time for water, and Í2—the time 
for the other liquid to cool through the same tem-
perature interval. The mass of the water mi and 
that of the liquid m2 is also determined. 
The specific heat of the liquid is calculated as 
follows: 
Heat lost by the liquid ^ m2c(Δ 0 
ti 
Heat lost by the water 
^η ,(1)(Δ ί) 
í,* 
Therefore, 
Figure 2-1 Cooling apparatus for determining the 
specific heat of a liquid. 
c = 
mit2 
rriiti 
(2-3) 
Table 2-1 Specific Heats of Solids, Liquids, and Gases (in kcal/kg°C). 
Metallic solids 
Non-metallic solids 
Liquids 
Gases 
Aluminum 
0.212 
Clay 
0.22 
Alcohol 
0.55-0.60 
Air 
0.24 
Copper 
0.094 
Coal 
0.3 
Mercury 
0.033 
Hydrogen 
3.40 
Brass 
0.090 
Concrete 
0.16 
Oil 
0.5 
Nitrogen 
0.25 
Gold 
0.030 
Glass 
0.12-0.20 
Water 
1.00 
Oxygen 
0.22 
Iron 
0.11 
Granite 
0.19 
Turpentine 
0.41 
Lead 
0.030 
Ice 
0.48 
Nickel 
0.106 
Limestone 
0.22 
Platinum 
0.032 
Marble 
0.21 
Silver 
0.056 
Paraffin 
0.69 
Steel 
0.11 
Rubber 
0.48 
Tin 
0.055 
Wood 
0.3-0.7 
Zinc 
0.092 

10 
Quantity of Heat and Calorimetry 
Other methods of determining the specific heat of 
substances are methods based on change-of-state 
and electrical methods. Table 2-1 is a representative 
listing of the specific heats of various solids, li-
quids, and gases. 
It can be seen from the above table that water has 
a greater specific heat than most substances making 
up the solid surface of the earth. This fact is impor-
tant in explaining why the climate is milder near 
large bodies of water. In the winter, the tem-
perature of the water does not fall very low and 
thus tends to prevent the temperature of the land 
from falling very low, while in the summer the 
water does not rise much in temperature and tends 
to moderate the temperature of the land. 
2-6 
LATENT HEAT OF F U S I O N 
We can add heat to a substance without raising 
its temperature if the substance is undergoing a 
change of state. If, for example, we add heat slowly 
to a mixture of ice and water which is initially at a 
uniform temperature, we will observe that the 
temperature does not rise until all the ice is melted. 
The heat supplied is utilized in changing the ice to 
water. This heat is stored as thermal energy in the 
water, and is given out when the water changes 
back to ice. As this heat cannot be detected by a 
thermometer, it is usually referred to as latent. 
The latent heat of fusion is the quantity of heat 
that must be added to a unit mass of a solid so 
as to melt it, or the quantity of heat that must 
be removed from a unit mass of a liquid so as 
to solidify it, without a change in temperature 
and pressure. 
The latent heat of fusion can be determined by the 
method of mixtures. 
2-7 
LATENT HEAT OF V A P O R I Z A T I O N 
In the last section, we stated that while a solid is 
changing to a liquid or vice versa there is no change 
in temperature. Similarly, when a liquid is changing 
to a vapor or vice versa at constant pressure there 
is no change in temperature. To change a given 
mass of water at the boiling point into vapor or 
steam, a definite amount of heat must be supplied; 
when an equal mass of steam condenses to water, 
an equal amount of heat is given out. The heat 
supplied is stored as thermal energy in the steam, 
and is given out when the steam changes back to 
water; this is called the latent heat of vaporization. 
The latent heat of vaporization is the quantity 
of heat that must be added to a unit mass of a 
liquid so as to vaporize it, or the quantity of 
heat that must be removed from a unit mass of 
a vapor so as to condense it. without a change 
in temperature and pressure. 
The latent heat of vaporization can also be deter-
mined by the method of mixtures. 
Example 2. A 40 gm copper calorimeter contains 
50 gm of ice; both are at 0**C. Ten grams of steam at 
atmospheric pressure are passed into the calorime-
ter. What is the final temperature of the calorimeter 
and its contents? 
SOLUTION 
Heat gained by ice-I-heat gained by calor-
imeter = heat lost by steam. 
[(50 gm)(80 cal/gm)] -f [(50 gm)(i - 0)(1 cal/gm°C)] 
+ [(40 gm)(i - 0)(0.094 cal/gm°C)] 
= (10 gm)(540 cal/gm) 
+ 10 gmdOO^C - 0(1 cal/gm°C) 
4000 cal + 50í cal/°C + 3.76í cal/T 
= 5400 cal + 1000 cal - lOí cal/°C 
63.76í = 2400°C 
í = 37.6°C. 
Table 2-2 gives the melting points, boiling points, 
latent heat of fusion, and vaporization of various 
materials. 
2-8 
HEAT OF C O M B U S T I O N 
The heat given out when substances bum is of 
importance in engineering, chemistry, and physics. 
The heat of combustion of a substance is the 
quantity of heat given out per unit mass or per 
unit volume when the substance is completely 
burned. 
The heats of combustion of solids and liquids are 
usually expressed as the heat per unit mass, while 
the heats of combustion of gases are usually ex-
pressed as the heat per unit volume. The heats of 
combustion of solids and liquids are usually mea-

Problems 
11 
Table 2-2 An Illustrative List of Melting and Boiling 
Points, Latent Heats of Fusion, and Vaporization at 
One Atmosphere Pressure. 
Heat of 
Melting Boiling Heat of 
vapor-
point 
point 
fusion 
ization 
Substance 
°C 
kcal/kg kcal/kg 
Alcohol (ethyl) 
-115 
78 
25 
205 
Alcohol (methyl) 
-98 
65 
22 
267 
Aluminum 
660 
2060 
93 
2000 
Ammonia 
-75 
-34 
108 
327 
Copper 
1080 
2600 
51 
1760 
Gold 
1060 
2970 
16 
446 
Hydrogen 
-259 
-253 
15 
107 
Iron 
1540 
2740 
65 
1620 
Lead 
327 
1744 
6.3 
222 
Mercury 
-39 
357 
2.7 
71 
Methane 
-182 
-161 
14.5 
138 
Nitrogen 
-210 
-196 
6.2 
48 
Oxygen 
-219 
-183 
3.3 
51 
Platinum 
1774 
4407 
27 
640 
Silver 
960 
2212 
24 
552 
Tin 
232 
2270 
14 
650 
Tungsten 
3400 
5930 
44 
1180 
Water 
0 
100 
80 
540 
Zinc 
419 
907 
24 
362 
sured with a bomb calorimeter (Figure 2-3). A cer-
tain mass of fuel is put in the bomb (a steel cylinder 
fitted with a gas tight cover) with pure oxygen 
under pressure so as to insure complete combus-
tion. The bomb is surrounded by water in a 
calorimeter. The fuel is ignited by sending an elec-
tric current through a heater wire within the bomb. 
The heat of combustion is calculated from the rise 
in temperature of the water, bomb, and calorimeter. 
The heat of combustion of gases is usually meas-
ured by using a continuous flow calorimeter. Here, 
water is let flow continuously through a tube heated 
Leads for 
electric current 
Oxygen-
Bomb 
Stirrer 
Insulating 
"container 
Figure 2-3 Bomb calorimeter for measuring the heats 
of combustion of solids and liquids. 
with the gas flame. From the change in temperature 
of the water, the mass of water that flows through 
the tube, and the volume of gas consumed by the 
flame, all in the same interval of time, the heat of 
combustion of the gas can be calculated. 
Table 2-3 contains an illustrative list of the heats 
of combustion of a number of solids, liquids, and 
gases. 
Table 2-3 Heats of Combustion. 
Substance 
kcal/kg 
Substance 
kcal/m' 
Solids: 
Gases: 
Coal 
6000-7800 
Acetylene 
12,900 
Wood 
4500-5000 
Coal gas 
4300 
Liquids: 
Natural gas 8000-18,000 
Alcohol 
6400 
Propane 
20,600 
Gasoline 
11,400 
Diesel oil 
10,600 
Kerosene 
11,000 
P R O B L E M S 
1. Find the equilibrium temperature when 80 gm of iron at 100°C are dropped into 200 gm of water con-
tained in a 50 gm iron vessel which are both at 20°C. 
2. A 50 gm sample of a solid material, at a temperature of lOOX, is dropped into a 100 gm copper calorime-
ter containing 200 gm of water initially at 20°C. The final temperature of the calorimeter and its contents 
is 22°C. What is the specific heat of the sample? 

12 
Quantity of Heat and Calorimetry 
3. A piece of solid metal of mass 0.150 kg was heated at 659°C and then plunged into 0.100 kg of water at a 
temperature of 25T. When the water and metal reached equilibrium in temperature, their combined 
mass was found to be 0.247 kg. 
(a) What became of the mass that was lost? 
(b) What was the final equilibrium temperature? 
4. Fifty grams of aluminum at 98°C is dropped into 37.2 gm of alcohol at 10°C in a 50 gm copper calorime-
ter. The final temperature of the mixture is 35°C. What is the specific heat of the alcohol? 
5. To find the specific heat of methyl alcohol, a 40 gm mass of aluminum at 200°C is added to 120 gm of the 
alcohol, which is initially at 20°C. The final temperature of the mixture is 40°C. What does this give for 
the specific heat of the alcohol? 
6. A 0.300 kg aluminum vessel contains 0.600 kg of water at 20*'C. How much steam at 100°C must be 
added to raise the temperature of the water to lOOT? 
7. How much heat is required to melt 10 gm of ice (already at 0°C), heat the resulting water to lOOT, and 
boil it all away? 
8. Ice at - 10°C and steam at 110°C come into thermal equilibrium as water at 50°C in a calorimeter. Take 
the specific heat of steam at 0.50 kcal/kgT. What is the ratio of the mass of ice to the mass of steam? 
9. Twenty grams of steam at 100°C were condensed in 362 gm of water at lOT. The heat capacity of the 
container is 40cairC, and the final temperature is 40°C. Calculate the heat of vaporization. 
10. Two kilograms of molten tin at its melting point are dropped into a 1 kg copper vessel containing 3 kg of 
water at 10°C. What is the final equilibrium temperature? 
11. If 200 gm of iron at 120T, 20 gm of ice at OT, and 100 gm of water at 20°C are aU mixed, and it is found 
that there is no change in the temperature of the water, what must be the specific heat of the iron? 
12. In an insulated container are 50 gm of ice and 200 gm of water at 0°C. Ten grams of steam at lOO^'C and 
atmospheric pressure is allowed to flow into the mixture, condensing therein. Neglecting the heat 
absorbed by the container, compute the final steady state temperature of the mixture. 
13. Calculate the equilibrium temperature 
(a) when we mix 100 gm of ice at - 20*C with 400 gm of water at 30°C, 
(b) when we mix 100 gm of ice at -20°C with 200 gm of water at 30°C. 
14. A mixing faucet is supplied with cold water at a rate of 0.6 gallons per minute and hot water at a rate of 1 
gallon per minute. If the temperature of the cold water is 50°F and the hot water is 150°F, what is the 
temperature of the water from the faucet? 
15. If 10 gm of gasoline are burned in a bomb calorimeter containing 3 kg of water, what is the heat of 
combustion of the gasoline if the heat capacity of the calorimeter is 1 kcal^C and the rise in temperature 
is 25°C? 
16. Twenty grams of water contained in a 30 gm copper calorimeter was found to cool from 70°F to 50°F in 
10 minutes. An equal volume of another liquid with a mass of 15 gm cools from 70**F to 50T in 6 minutes 
in the same calorimeter. Find the specific heat of the liquid. 

3 
Expansion of Solids, 
Liquids, and Gases 
3-1 
Expansion of Solids 
13 
3-2 
Coefficient of Linear Expansion of 
Solids 
13 
3-3 
Coefficient of Area Expansion 
14 
3-4 
Coefficient of Volume Expansion 
14 
3-5 
Importance and Applications of 
the Expansion of Solids 
14 
3-6 
Expansion of Liquids 
15 
3-7 
Expansion of Gases 
16 
3-8 
Boyle's Law 
16 
3-9 
Charles' Law 
16 
3-10 
The General Gas Law 
17 
3-1 
E X P A N S I O N O F S O L I D S 
Most substances expand when heated and con-
tract when cooled. Solids have the property of re-
taining their shape when heated without the support 
of a container. All three dimensions change with 
temperature. If the body expands isotropically, that 
is, if the body expands equally in all directions, we 
may then take any dimension as the length of the 
body. Solid bodies of different materials, having dif-
ferent space arrangements of their atoms, expand 
different amounts for equal temperature changes. 
To express numerically the expansion of a particu-
lar substance, we need a quantity characteristic of 
that substance. This quantity is called the coeffi-
cient of expansion. 
3-2 
COEFFICIENT O F L I N E A R 
E X P A N S I O N 
OF S O L I D S 
Suppose a rod of length lo is heated through a 
temperature interval At until its length is L The 
difference /, - /o = Δ / is the amount the rod has ex-
panded on heating. It is found experimentally that 
for moderate temperature intervals the quantity 
Δ / / / ο Δ ί equals a constant which is characteristic of 
the material in the rod. This constant a is called the 
coefficient of linear expansion. That is. 
A/ 
/ο Δ ί* 
(3-1) 
The coefficient of linear expansion may thus be 
defined as follows: 
The coefficient of linear expansion is the 
increase in length per unit length per degree 
rise in temperature. 
Equation (3-1) can be written in a different form by 
replacing Δ / by /, - /o, namely. 
t =/ο (1 + αΔ ί). 
(3-2) 
This is sometimes more convenient than Eq. (3-1). 
The units of a are °C"' or °F~\ Since the 
Fahrenheit degree is only i as large as the Centi-
grade degree, a (per °F) is equal to la (per °C). 
Typical values of the coefficient of linear expansion 
of various solids are given in Table 3-1. 
13 

14 
Expansion of Soiids, Liquids, and Gases 
Table 3-1 Coefficients of Linear Expansion of Solids. 
a in 
a in 
units of 
units of 
Substance 
Substance 
10-* T -
Aluminum 
24 
Magnesium 
26 
Brass 
19 
Nickel 
13 
Copper 
14 
Platinum 
9 
Glass 
Quartz 
(ordinary) 
9 
(fused) 
0.4 
Glass 
(pyrex) 
3 
Silver 
19 
Gold 
14 
Steel 
12 
Iron 
11 
Tin 
20 
Invar 
0.9 
Tungsten 
4.5 
Lead 
30 
Zinc 
30 
Example 1. An aluminum rod is 200 cm long at 
OT. What is its length at lOOT? 
SOLUTION 
Length at 100°C = /,oo = /o(l + aAt) 
= 200cm[l + (24xlO-'°C-') 
x(lOOT)] 
= 200 cm + (200 cm)(24 x 10"') 
= 200.48 cm. 
3-3 
C O E F F I C I E N T OF A R E A E X P A N S I O N 
Consider a square sheet of length lo heated 
through a temperature interval Δ ί until its length 
is 
If a is the coefficient of linear expansion of the 
material, its new area is given by 
Λ  = I' = [lod + αΔ Ο ΐ' = loW + 2αΔ ί + α'Δ ί') 
= Λ ο (1 + 2αΔ ί + α ' Δ ί ^ 
Since αΔ ί is a very small quantity for all solids, 
α^Δί^ may be neglected in comparison with 2αΔ ί. 
Therefore, 
Λ = Α ο ( 1 + 2αΔ ί), 
(3-3) 
and the coefficient of area expansion is twice the 
coefficient of linear expansion. Although the above 
result was derived for a square sheet, it holds for a 
sheet of any shape. 
The coefficient of area expansion of a solid 
is the increase in area per unit area per degree 
rise in temperature, and is twice the coeffi-
cient of linear expansion for an isotropic 
solid. 
It is worth noting that Eq. (3-3) is applicable to 
the increase in area of a hole in a sheet of material 
as well as the increase in area of the material itself. 
Can you explain this? 
3-4 
C O E F F I C I E N T O F V O L U M E 
E X P A N S I O N 
Consider a cube of material of edge length k 
heated through a temperature interval Δ ί until its 
length is /,. If a is the coefficient of linear expansion 
of the material, its new volume is given by 
ν , = /,' = [/ο (1 + α Δ ί)^ 
= loM + 3αΔ ί + 3α'Δ ί' + α'Δ ί') 
= ν ο (1 + 3 α Δ ί + 3 α ' Δ ί ' + α'Δ ί'). 
Since we can neglect the terms containing α^Δί^ 
and α'Δ ί', 
Vt = Vo(l + 3 a A i ) 
(3-4) 
is an equation which holds for an isotropic sub-
stance of any shape. 
The coefficient of volume expansion of a solid 
is the increase in volume per unit volume per 
degree rise in temperature, and is three times 
the coefficient of linear expansion for an 
isotropic solid. 
Equation (3-4) is applicable to the increase in 
volume of a cavity in a solid as well as the increase 
in volume of the solid itself. Can you explain this? 
3-5 
I M P O R T A N C E A N D A P P L I C A T I O N S 
O F THE E X P A N S I O N O F S O L I D S 
In pipelines, bridges, and non-welded rails, allow-
ance must be made for temperature changes, other-
wise they will bend with a rise in temperature. This 
is taken care of in pipelines by expansion joints and 
in bridges and rails by leaving a small space be-
tween the bridge sections or rails. The joint is usu-
ally made by means of two plates, one on each side 
of the bridge sections or rails. Bolts are put through 
the plates and through slotted holes in the bridge 
section or rail so that they are free to slide. Clock 
pendulums must be compensated for expansion in 
order for them to keep time correctly. 
The fact that different metals have different 
coefficients of expansion has been applied to the 
construction of thermometers and thermostats. If 
two thin strips of different metals welded or riveted 

Expansion of Liquids 
15 
1 
/ 
Zinc^\\ 
-^Iron 
/ i f 
\\V--lron 
m 
i 
T<To 
T= \ 
m 
-To 
T>To 
Figure 3-1 
element. 
Bimetallic thermometer or thermostat 
together are heated, the resultant strip will bend into 
a curve (Figure 3-1). The metal with the larger 
coefficient of expansion will be on the convex side of 
the curve. When used as a thermostat, one end is 
fixed and the other end opens or closes an electrical 
circuit. That is, it operates as a switch which is 
governed by the temperature. The temperature 
control of a cooking oven operates on this principle. 
3-6 
E X P A N S I O N OF LIQUIDS 
Since a liquid must be held in a solid container, 
the true expansion of the liquid can only be ob-
tained if we also know the expansion of the con-
tainer. 
The apparent volume expansion of the liquid is 
equal to the difference between the true volume ex-
pansion of the liquid and the volume expansion of 
the container. The new volume of the container is 
given by 
cVr = Vo(l + 3 a e A i ) , 
(3-5) 
where ac is the linear expansion coefficient of the 
container. The new true volume of the liquid is 
given by 
/V, = Vo(l + ß , A i ) , 
(3-6) 
where ß i is the volume expansion coefficient of the 
liquid. 
The apparent volume expansion of the liquid is 
, V . - c V . = V o ( / 3 , - 3 a c ) A i . 
(3-7) 
Hence, the apparent volume expansion coeffi-
cient of the liquid is 
VoAi' 
where we have replaced /V, - c V r by AV. 
(3-8) 
ε  
I 
I i 
Ο -
Ι.010 
1.008 
1.006 
1.004 
1.002 
1.000 
5 1015 20 25 30 35 40 45 50 
Temperature in **C 
Figure 3-2 The anomalous expansion of water. 
In the above experiment, we could measure the 
value of β / if we knew ac. If we had filled the 
container with water both enclosed in a bath at 0°C 
and raised the temperature of the bath to 4 T , we 
would have found that the water decreased in vol-
ume. The reason is that water has a peculiar expan-
sion property. It has a negative expansion coeffi-
cient between O'^C and 4 T or, in other words, when 
water is heated from OT it contracts with tempera-
ture rise until it reaches 4^C and then begins to 
expand. The volume per gram of water at tempera-
tures between 0°C and 100**C is shown graphically 
in Figure 3-2. 
Water has its greatest density at 4**C. This is 
important in nature. It is the reason why the surface 
of lakes and rivers freeze first. The water at the 
bottom stays at 4**C, enabling the fish to enjoy a 
warmer winter than some land dwellers. There are a 
number of accurate methods for determining the 
expansion coefficient of liquids. The interested 
student may read about these methods in books on 
Table 3-2 Coefficients of Volume Ex-
pansion of Liquids. 
Substance 
/310-'°C-
Alcohol (ethyl) 
112 
Alcohol (methyl) 
120 
Benzene 
124 
Carbon tetrachloride 
124 
Ether (ethyl) 
166 
Glycerin 
50 
Mercury 
18 
Petroleum 
95 
Turpentine 
97 
Water 
21 

16 
Expansion of Solids, Liquids, and Gases 
heat that are available in the library. Typical values 
of the coefficient of volume expansion of various 
liquids are given in Table 3-2. 
Example 2. A pyrex glass bottle is ffiled with 
mercury at 20° C. The volume of the bottle is 
100 cm\ How many cm^ spill over when the bottle 
is placed in steam at 100° C? 
SOLUTION 
Volume that spills over = Δ V = (/3„ - 3ac) VoAi 
= [(180 - 3 X 3) X 10"' ° C-'](100 cm')(80T) 
= (171 X 10"'° C-*)(100 cm')(80° C) 
= 1.37 cm'. 
3-7 
E X P A N S I O N O F G A S E S 
Gases do not have a definite volume. If a gas is 
put into a container, it fills the complete volume of 
the container. It also exerts a definite pressure upon 
the walls of the container. 
Suppose we have a container of definite volume, 
containing a definite mass of gas, which exerts a 
definite pressure on its walls. If the container is a 
cylinder fitted with a piston and we change the vol-
ume by pushing the piston in or out of the cylinder, 
the pressure changes. If we keep the piston fixed 
and change the temperature of the gas, the pressure 
changes. If we allow the piston to move freely and 
change either the pressure or the temperature, the 
volume changes. 
3-8 
B O Y L E ' S LAW 
In 1660, Robert Boyle found a relation between 
the pressure and volume of a gas if the temperature 
of the gas is held constant. The relation is known as 
Boyle's law, and is stated as follows: 
The product of the pressure and the volume of 
a given mass of gas at constant temperature is 
a constant. 
Mathematically, it is written 
pV = constant 
(for constant i ) , 
(3-9) 
or, alternately. 
P i V, = Po Vo 
(for constant 0, 
(3-10) 
where p i is the gas pressure when its volume is V i , 
and Po is the initial pressure when its initial volume 
is Vo, both states being at the same temperature. 
Ρ  
( I P s . 
r, 
V 
Figure 3-3 Graphic illustration of p\/=constant for 
various temperatures for one amount of gas. 
The constant in Eq. (3-9) depends upon the temper-
ature of the gas. Figure 3-3 is a graphical illustration 
of Eq. (3-9). Curve (1) represents Boyle's law for a 
given mass of gas at absolute temperature T i , curve 
(2) for an absolute temperature Ti, etc. They are 
called isothermal curves (curves of constant tem-
perature). 
3-9 
C H A R L E S ' L A W 
The volume of gases, like most solids and liquids, 
increases with an increase in temperature. The vol-
ume expansion coefficient of a gas is defined in the 
same way as that of a liquid, namely 
ß  = 
A V 
VoAi* 
(3-11) 
Gases, however, differ from liquids as they all have 
very nearly the same value of β  at constant pres-
sure. At 0°C, the volume expansion coefficient of all 
permanent gases is approximately m''C~\ 
Consider a gas whose volume is Vo at atmos-
pheric pressure po and temperature 0°C. We heat 
the gas, keeping its pressure constant, to a tempera-
ture t Its new volume is given by 
or 
or 
V = Vo(l + /3i) 
V _ V _ V o i - . . ( r - 2 7 3 ) _ V o T 
273 
V, = 
273 
VoT 
2 7 3 ' 
273 
(3-12) 
where Τ  is the absolute temperature in "K. 

The General Gas Law 
17 
We see that the volume of the gas is directly 
proportional to the absolute temperature at con-
stant pressure. This is known as Charles' law, and is 
stated as follows: 
At constant pressure, the volume of a given 
mass of a gas varies directly as the absolute 
temperature. 
Symbolically, it is stated as 
V 
γ = constant 
(at constant pressure). (3-13) 
Alternately, it is written 
or 
V , ^ Vo 
T, 
To 
Yi = L 
Vo 
To* 
(3-14) 
(3-15) 
Similarly, at constant volume, the pressure of a 
given mass of gas varies directly as the absolute 
temperature; an alternative form of Charles' law is 
Pl = I l 
Po 
To 
(3-16) 
We may well ask what happens to the pressure 
and volume of a gas when Τ  equals zero. Actually, 
all gases liquefy at temperatures above absolute 
zero, so the question has no physical significance. 
Figure 3-4 is a graph of a volume of a particular 
mass of gas at constant pressure versus its tempera-
ture in both Centigrade and Kelvin degrees. We 
see that the volume of the gas is directly propor-
tional to its absolute temperature. 
\ 
-273 
0 
°C ' 
0 
273 
Figure 3-4 Variation of volume of a gas with tempera-
ture at constant pressure. 
3-10 
THE G E N E R A L G A S L A W 
Boyle's law and Charles' law provide the basis 
for a more general relation between pressure, vol-
ume, and temperature known as the general gas 
law. Combining Eqs. (3-10) and (3-15), we get 
P i V . ^ P o V o 
T, 
To · 
(3-17) 
Since pi, Vi, and Ti can be any set of pressure, vol-
ume, and temperature values, we can remove the 
subscripts. The value of Vo, if taken as the gas vol-
ume at atmospheric pressure po and temperature To 
(0°C), depends on the mass and kind of gas used. 
However, 
one 
kilomole 
(kilogram 
molecular 
weight) of any gas has a volume of 22.4 m' at at-
mospheric pressure and OT. Hence, if the gas con-
tains η  kilomoles, we can write Eq. (3-17) as 
where 
pV = nRZ 
_ P o V o 
To · 
(3-18) 
The value of R has the same value for all gases, 
and is known as the universal gas constant. 
1? = 8.31 X 10' joules/(kü omole °C). 
Equation (3-18) is called the general gas law. 
If we substitute To = 273 and T, = 273 + ί in Eq. 
(3-17), we obtain 
P . V , = p o V o ( ? ^ ^ ) 
(3-19) 
p . V . = p o V o ( l + 2 ^ ) 
(3-20) 
or 
or 
P i V , = poVo(l + j3i), 
(3-21) 
where /3 = zfe °C"'. 
It should be pointed out that these equations do 
not hold exactly at all pressures and temperatures. 
However, it is quite remarkable that these simple 
equations specify the behavior of all gases without 
too much error except at high pressures and very 
low temperatures. 
Example 3. The volume of a gas at 27°C and at-
mospheric pressure is 1 m'. What is the volume 
when the pressure is equal to four times the at-
mospheric pressure and the temperature is 327°C. 

18 
Expansion of Solids, Liquids, and Gases 
SOLUTION 
PiVi^PiVi 
T2 
T, 
4 p i V 2 ^ ( p . ) ( l m') 
600°K 
300°K 
V2 = 600 
1200 = 0.50 m'. 
P R O B L E M S 
1. At 25°C an iron rod is 2 cm in diameter. A brass ring has an interior diameter 1.995 cm at the same 
temperature. At what temperature, both the rod and ring being heated, will the ring just slide onto the 
rod? 
2. A square brass plate with a circular hole through its center is heated from 5°C to 605T. When cold, the 
plate is 32 cm on an edge and the diameter of the hole is 8 cm. What is the diameter of the hole when the 
plate is hot? 
3. The density of a liquid is defined by the expression ρ = m/V, where m is the mass and V is the volume 
of the liquid. Show that 
Δ ρ = - β ρ Δ Τ , 
where Δ Γ is the change in temperature. 
4. A copper container has a capacity of 500 cm' at 5 T . If 470 cm' of methyl alcohol also at 5°C are placed 
in the container, at what temperature will the container be just filled with alcohol? 
5. A thin aluminum wire is bent into the form of a rectangle 50 cm long and 40 cm wide. When the 
temperature of the wire is changed from 10°C to lOOT, what is the change in area enclosed by the wire? 
6. A glass flask holds exactly 1000 cm' of water at 5°C. When the flask and the water are heated to 95°C, 
15 cm' of water overflow. What is the linear coefficient of expansion of the glass? 
7. We have two square parallel plates in contact, one of steel and the other of aluminum. At 0°C, the 
aluminum plate is 1 m on a side, and the steel plate is 1.001 m on a side. It is arranged to have an etching 
process begin when the plates have precisely the same area. Find the temperature at which the etching 
process begins. 
8. A pyrex glass flask in a dark cupboard contains 1000 cm' of ethyl alcohol at 20*'C. The flask is removed 
from the cupboard and set in sunlight, so that in 20 minutes the temperature rises to 60**C. What volume 
of alcohol is present at the end of 20 minutes? 
9. A narrow-necked ordinary glass bottle is just filled with 100 cm' of a liquid when the bottle and the liquid 
are at - 20*'C. The coefficient of volume expansion of the liquid is 12.4 x 10"^ per "^C. What volume of the 
liquid has spilled over by the time enough heat has been added to raise the temperature of the bottle and 
the liquid to 10°C? 
10. A certain medical prescription requires 10 cm' of ethyl alcohol at 20**C. A freshly sterilized graduated 
pyrex flask is used for measuring out the alcohol. The flask is at a temperature of %.3°C. 
(a) What volume of alcohol at %.3°C must be 
used? 
(b) What volume on the graduated pyrex flask 
will give the required amount of alcohol if 
the flask was calibrated at 20**C? 
11. Two rods of metal A and one rod of metal Β are 
connected as shown in Figure 3-5. The overall 
length is L and the length of Β is d. If the 
coefficients of linear expansion are «Λ  and aa, 
determine d in terms of L, «A, and «Β , if L does 
not change with temperature. 
Figure 3-5 

Problems 
19 
12. If 1 m'of a certain gas at atmospheric pressure and 0°C has a mass of 2 gm, what would be the mass of 
the same volume of the same gas at 0.50 atmospheric pressure and 100°C? 
13. A steel rod and a brass rod are fastened to the 
same end plate as shown in Figure 3-6. If the 
distance between the ends of the rods d is to 
remain 8 cm over a range of temperatures, what 
must be the length of each rod? 
Figure 3-6 
14. A hand pump with a cylinder 0.50 m long is used to pump air into a tire. Assuming the temperature of the 
air to remain constant, find how far the piston must be moved before air can enter the tire if the pressure 
of air in the tire is twice the atmospheric pressure at the beginning of the stroke. 
15. A certain mass of gas occupies a volume of 380 cm' at 1T*C and pressure 1.053 atmospheres. What will 
its volume be at - lOT and pressure 1 atmosphere? 
16. An air bubble of 10 cm' volume is at the bottom of a lake, where the pressure is 4 times that at the 
surface and the temperature is 5°C. The bubble rises to the surface, which is at a temperature of 20°C. If 
the temperature of the air in the bubble is the same as the surrounding water, what is its volume just 
before it reaches the surface? 
17. An aluminum container holds 400 cm' of a liquid when completely filled at 10°C. When the temperature 
of the container and liquid is raised to 80T, 25 cm' of the liquid spill out of the container. Determine the 
coefficient of volume expansion of this liquid. 
18. (a) If 1 m' of air at 1 atmosphere pressure and 27T temperature is compressed to 0.50 m' at 3 
atmospheres, what will be the resulting temperature? 
(b) If the gas is now allowed to cool down to its original temperature at constant volume, what will be 
the final pressure? 
19. The cross-sectional area of the mercury colunm in a barometer is 1.20 cm^, the length of the vacuum at 
the top is 8 cm when the barometer reads 76.4 cm. Calculate the volume of external air that must be 
inserted into the tube in order to lower the mercury colunm to 38.2 cm. Assume the atmospheric pres-
sure and the temperature remain constant. 
20. Owing to air above the mercury in a barometer, it reads 74.5 cm when the actual pressure is 75.2 cm, and 
73.3 cm when the actual pressure is 73.7 cm of mercury. Calculate the reading of the barometer when the 
true pressure is 76 cm. 
21. A metal container has an internal volume of 1000 cm' when its temperature is O^'C, but when its 
temperature is 100°C the container holds 1008 cm'. 
(a) Determine the coefficient of linear expansion for this metal. 
(b) If a liquid just fills the container at O'^C, and 10 cm' spill out at lOO^'C, determine the coefficient of 
volume expansion of the liquid. 
22. A thermometer is made of a capillary tube of ordinary glass of 0.020 mm^ cross section at 0°C. At 0°C 
there are 2 cm' of mercury in the thermometer. How far does the mercury move up the tube at 30°C? 

4 
Heat 
Transfer 
4-1 
Methods of Heat Transfer 
21 
4-2 
Coefficient of Thermal Conductivity 
21 
4-3 
Conductivity of Liquids and Gases 
22 
4-4 
Convection 
23 
4-5 
Radiation 
24 
4-6 
Prevost's Theory of Exchanges 
24 
4-1 
M E T H O D S O F HEAT T R A N S F E R 
There are three ways in which heat is transferred 
from one place to another. The three ways are by 
conduction, convection, and radiation. 
Conduction is the transfer of heat in which 
thermal energy is transferred from molecule 
to molecule in a material with no perceptible 
motion of the material. 
Convection is the transfer of heat by mass 
motion of the heated material. 
Radiation is the transfer of thermal energy by 
electromagnetic waves. 
In the fírst two methods, a material medium is 
required. In radiation, no material is needed, and 
the heat is transferred with the speed of light. 
4-2 
COEFFICIENT OF T H E R M A L 
CONDUCTIVITY 
twist them together at one end as illustrated in Fig-
ure 4-1. At equal intervals along the wires we attach 
small metal balls by means of wax. We then heat 
the ends which were twisted together. As the heat is 
conducted along the wires, the balls drop off when 
the temperature of the wire is sufficient to melt the 
wax. After a particular time interval, the balls will 
stop dropping off, and the temperature is said to be 
in a steady state, which means that the heat passing 
along the wire is equal to the heat that escapes from 
it. When this state is reached, it is found that the 
copper has lost the largest number, the aluminum 
the next largest number, and the steel the least 
number of balls. This means that the copper wire is 
the best conductor, the aluminum is next best, and 
the steel is poorest. 
Consider the conduction of heat through a slab of 
Burner 
lyCopper 
fV-Aluminum 
^ ~ S t e e l 
Materials differ widely in their ability to conduct 
heat. Metals are good conductors while non-
metallic materials are poor conductors. However, 
there is a fair difference in the conductivity of 
different metals as well as of different non-metals. 
These differences may be illustrated in a simple 
way for metals. Suppose we take three equal 
diameter wires of aluminum, steel, and copper, and 
Figure 4-1 Comparison of conductivity of metals. 
Wax 
balls 
21 

22 
Heat Transfer 
Figure 4-2 Conductivity of heat through a slab of 
material. 
material of face area A and thickness Δ Χ , with a 
difference of temperature between the faces of Δ ί 
after a steady state is reached, as illustrated in Fig-
ure 4-2. H, the heat per unit time, passing through 
the slab is found experimentally to be proportional 
to the temperature gradient At I AX and the area Λ . 
Then 
Δ ί 
H o c 
A Δ Χ  
or 
H = KA AX' 
(4-1) 
where the proportionality constant Κ  is called the 
coefficient of thermal conductivity of the material, 
and is defined as follows: 
The coefficient of thermal conductivity of a 
material is the time rate of heat flow by 
conduction per unit area per unit temperature 
gradient. 
Various systems of units are used in specifying 
K. In MKS units, Κ  is expressed in kcal/sec m°C. 
Table 4-1 gives typical values of Κ  for a number of 
substances. 
Example 1 . An aluminum pan has a 0.10 m^ heat-
ing surface, 0.20 cm thick. How much water 
evaporates per minute if the outer surface is kept at 
HOT? 
SOLUTION 
From Eq. (4-1), 
Η  = KAj^ 
= (480 X 10"' kcal/sec m°C) 
X (0.10 
m ) 
( 2 Χ 1 0 - Π1 ) 
= 24 kcal/sec. 
Therefore, the heat flow in 1 minute 
= (24 kcal/sec)(60 sec/nun) 
= 1.44x10' kcal/min. 
L„, the heat of vaporization of water, is 540 kcal/kg. 
Therefore, the mass of water vaporized in 1 minute 
is given by 
(1.44 X 10' k c a l / m i n ) ( ^ ¿ ^ ) = 2.67 kg/min. 
4-3 
CONDUCTIVITY 
O F LIQUIDS 
A N D G A S E S 
We can see from Table 4-1 that all liquids have 
very low conductivities. As a demonstration, we 
can partially fill a test tube with water and boil the 
upper portion of it without appreciably heating the 
bottom, as illustrated in Figure 4-3. 
Table 4-1 Typical Values of the Coefficient of Ther-
mal Conductivity (K) for Various Substances. 
Substance 
Κ  
10-* kcal/sec m**C 
Metals: 
Aluminum 
Copper 
Brass 
Iron and Steel 
Silver 
Non-Metals: 
Brick (fire) 
Brick (insulating) 
Concrete 
Cork 
Glass 
Ice 
Wood 
Gases: 
Air 
Hydrogen 
Oxygen 
480 
920 
260 
110 
1000 
2.5 
0.4 
2.0 
0.1 
2.0 
4.0 
0.2 
5.7 X 10" 
33.0x10" 
5.6x10-

Convection 
23 
Figure 4-3 Demonstration that water is a poor heat 
conductor. 
The conductivity of gases is extremely small as 
shown in Table 4-1. Many solid materials are good 
insulators because they are porous and contain 
much air. Storm windows on a house are effective 
chiefly because of the enclosed air between them 
and the regular windows. The most 
effective 
insulator of all is a vacuum, because heat can only 
be transferred through a vacuum by radiation. 
4-4 
C O N V E C T I O N 
Convection occurs only in liquids and gases. As 
previously stated, it is the transfer of heat by the 
mass motion of the heated material. The material 
motion is due to the difference in density of a hot 
and cold portion of the material. 
When a container of water is heated on a stove, 
the heat passes by conduction through the bottom 
of the container to the water. The lowest layer of 
water is heated and expands, thus becoming less 
dense than the colder water above. It is forced up-
ward by the colder water, which sinks. This circula-
tion continues until all the water is heated to the 
boiling point. The process is an example of convec-
tion. Hot water and hot air heating of a building are 
also due to convection. 
Unequal heating of land and water gives rise to 
winds. Local land and sea breezes, as illustrated in 
Figure 4-4, are accounted for by convection cur-
rents owing to the higher specific heat of water. 
Also, the unequal heating of large bodies of water 
gives rise to ocean currents. 
In determining the quantity of heat conducted 
through a wall or window, it is not correct to use the 
difference between the outdoor and indoor temper-
ature as the difference in temperature between the 
outer and inner surfaces of the wall or window. Be-
cause of convection effects, the outer surface tem-
perature may be well above the outdoor tempera-
ture, and the inner surface temperature well below 
the indoor temperature. Only part of the tempera-
ture drop occurs in the wall or window; the rest 
occurs in the layers of air in contact with them. In 
the case of an ordinary window pane in a home with 
an indoor temperature of 70**F and outdoor temper-
SEA BREEZE 
DAY 
Land Warmer than Water 
LAND BREEZE 
NIGHT 
Land Cooler than Water 
Figure 4-4 Illustration of sea and land breezes. 

24 
Heat Transfer 
ature of 0°F, the temperature of the glass will be 
about 35*T. The temperature difference between 
the outside and inside of the glass will be 0.2T. 
The mathematical theory of convection is quite 
complex. The heat per unit time (H) transferred by 
convection to or from a surface can be calculated 
from the following equation: 
H = hAAt, 
(4-2) 
where h is the convection coefficient, A is the area 
of the surface. At is the temperature difference be-
tween the surface and the main body of the fluid. 
The value of h depends on many circumstances, 
such as the shape and orientation of the surfaces; 
the density, viscosity, specific heat, thermal con-
ductivity, and velocity of the fluid; and whether or 
not evaporation or condensation takes place. A 
large amount of research has been done in the de-
termination of h. As a result, there are in existence 
many formulae, graphs, and tables from which h 
can be obtained for various conditions. 
A law discovered by Newton relates the rate of 
cooling or heating of a given substance to the differ-
ence in temperature between it and its surround-
ings. The law is stated as follows: 
The time rate of heat flow from a body to the 
surroundings or vice versa is directly pro-
portional to the difference in temperature 
between the body and the surroundings pro-
vided the difference in temperature is small. 
Mathematically, it is stated as 
Η  = CAÍ, 
(4-3) 
where C is a constant. The law agrees well with ex-
periment when the heat transfer is chiefly by con-
vection. 
Example 2. The temperature of a room is 20T, 
and the inside temperature of a window pane 0.2 cm 
thick is 0**C. The value of the convection coefficient 
for the pane is 9x 10"^ kcal/sec m^°C. What is the 
heat transferred per unit area? What is the outside 
temperature of the window and the outdoor tem-
perature? 
SOLUTION 
H = hAAt=(9x 
10' kcal/sec m ' O d m')(20°C) 
= 18x10"' kcal/sec m'. 
Hence, 18x 10'kcal/secm^ must be conducted 
through the window pane itself. Using Eq. (4-1), 
18x10"' kcal/sec m' 
= (2 X 10-^ kcal/sec m X ) ( ^ ^ ) 
Δ ί =0.18T. 
Therefore, the outside temperature of the pane = 
- 0 . 1 8 T . 
Since the same quantity of heat must be trans-
ferred to the outdoors from the outside of the win-
dow as from the room to the inside of the window, 
the difference in temperature in both cases must be 
equal. Therefore, the outdoor temperature will be 
-20.18°C. 
4-5 
RADIATION 
All bodies not at absolute zero temperature emit 
electromagnetic radiation. Electromagnetic radia-
tion includes thermal radiation, light, and X-rays. 
The principal difference in these types of radiation 
is wavelength (see Section 10-5). The quantity of 
radiation emitted by a body depends on its tempera-
ture and the nature of its surface. When elec-
tromagnetic radiation of any kind falls on a material 
surface, some may be absorbed, some reflected, 
and some transmitted. The part absorbed is trans-
formed into thermal energy or other forms of 
energy within the absorbing material. 
Heat transfer by radiation does not require a 
material medium and the heat in transit travels with 
the speed of light. An example of this is the transfer 
of heat from the Sun to the Earth. It travels through 
ninety million miles of space in which there is no 
material substance in about 8 minutes. 
4-6 
P R E V O S T ' S T H E O R Y O F E X C H A N G E S 
If a number of bodies at different temperatures 
are placed in an evacuated container with insulated 
walls, they transfer thermal radiation between 
themselves and the container walls until they finally 
all reach the same temperature. 
After they all reach the same temperature, the 
radiation does not stop. A thermal equilibrium is 
reached and each body absorbs as much radiation 
as it emits per unit time. This observation was first 
made by Prevost in 1792. From his observations, he 

Prevost's Theory of Exchanges 
25 
proposed a theory which can be stated as follows: 
In a state of equilibrium, the amount of energy 
radiated per unit time from an object is equal 
to the energy absorbed by it in the form of 
radiations from surrounding objects. 
The amount of radiation absorbed or emitted by a 
body is dependent on the nature of its surface. A 
polished surface is a poor absorber and a poor emit-
ter, whereas a black surface is a good absorber and 
a good emitter. This can be demonstrated with the 
apparatus shown in Figure 4-5. The two bulbs A 
and Β are connected to an air thermometer C. If a 
can half painted black and the other half left shiny 
is filled with hot water and placed between the 
bulbs, it can be demonstrated that the black surface 
is the better radiator. Next, if we blacken one of the 
bulbs and use a totally black can filled with hot 
water, we can demonstrate that a black surface is a 
better absorber than a shiny one. 
Figure 4-5 Demonstration that a black surface is a 
better radiator than a polished surface. 
If radiation falls on an opaque body, some is 
absorbed and some is reflected. If a denotes the 
fraction that is absorbed and ρ the fraction that is 
reflected, then 
α + p = 1, 
(4-4) 
where a is called the absorption factor and ρ the 
reflection factor. 
We previously gave a demonstration that a good 
radiator is a good absorber. This means that the 
radiation rate of a surface must be proportional to 
the absorption factor of the surface. The following 
law, called Kirchhoff's 
Law of Radiation, 
states 
this fact: 
The ratio of the rates of radiation of any two 
surfaces Is equal to the ratio of the absorption 
factors of the two surfaces. 
Symbolically, the law is written as follows: 
W2 
«2' 
(4-5) 
where Wi and W2 are the radiation rates or radia-
tion powers of the two surfaces and are expressed 
in joules/sec m^ or watts/m\ 
(Note: 
1 watt = 
1 joule/sec.) 
A perfect radiator is a body that is a perfect 
absorber, and is called a black body. 
No surface is a perfect absorber with α = 1. An 
approximation to a perfect absorber may be ob-
tained by a hollow sphere having a small opening 
with the inside walls having a rough, dull surface as 
shown in Figure 4-6. The radiation enters or leaves 
the cavity through a small hole. Part of the radia-
tion entering the cavity will be absorbed by its walls 
and part reflected. Only a very small part of the 
reflected radiations escape through the hole, so that 
after many internal reflections nearly all the radia-
tion is absorbed and the body approximates a black 
body. 
Figure 4-6 An approximation of a perfect absorber. 
It was mentioned at the beginning that the quan-
tity of radiation emitted by a body depends on its 
temperature. In fact, the total radiation emitted by a 
body increases very rapidly as the temperature is 
raised. The quantitative relation between the rate of 

26 
Heat Transfer 
radiation from a body and its surface temperature 
is given by the following law, known as the 
Stefan-Bolt zmann Law: 
The rate at which a radiator emits radiant 
energy is directly proportional to the fourth 
power of its absolute temperatqre. 
Symbolically, the law is written as follows: 
W = ασ Τ \ 
(4-6) 
where W is the radiation power per unit area in 
joules/sec m^ or watts/m^ a is the absorption factor 
which varies between zero and unity, and σ  is the 
Stefan-Boltzmann constant with a numerical value 
of 5.669 xlO-Vatts/m'°K\ 
Example 3. If the radiation from a small opening 
in a coal stove approximates black body radiation, 
and it radiates 2.84 x 10^ watts/m\ calculate its tem-
perature in degrees Centigrade. 
2.84 X 10' watts/m' 
SOLUTION 
W= ασ Τ ' 
= (1)(5.67 X 10"" watts/m'^K')(r) 
2.84xl0'watts/m' 
L5.67xlO-"watts/m'°K'J 
= 14%°K 
i= 1223°C. 
P R O B L E M S 
1. Water in a glass beaker is boiling away at a rate of 30 gm/min. The bottom of the beaker has an area of 
300 cm^ and it is 0.2 cm thick. Calculate the temperature at the underside of the bottom of the beaker. 
2. A boiler made of steel plate 1.5 cm thick has a surface area of 8 m^ The boiler contains water at 100°C, 
and its outer surface is 80°C. How much heat does it lose by conduction? 
3. A certain object radiates energy at the rate of 10 kcal/sec m^ when it is at 27'*C. What is its rate of 
radiation at 127T? 
4. If 2.80 kcal/sec is the rate of conductive heat transfer through a flat 4m^ slab of glass, find the 
temperature gradient in the glass. 
5. We have determined that a very hot blue-white star has a surface temperature of 23,000°K. Measure-
ments and simple calculation show that this star radiates at the rate of 1.4 x 10'° watts per square meter 
of its surface. Find the reflection factor for the star. 
6. The filament in a light bulb has a diameter of 0.20 mm and an absorption factor of 0.30. What will its 
temperature be when it radiates 5 watts per cm length? 
7. The rear wall of a fireplace has an effective area of 0.50 ml It can be considered a black body surface, 
and has a temperature of 327°C. How many watts does it radiate to the room? 
8. Why is a thermos bottle 
(a) evacuated? 
(b) double-walled? 
(c) silvered on the inside? 
9. By measurement and simple calculations, it has been found that the Sun radiates at the rate of 
6.25 X 10^ watts per square meter of its surface. A certain physical model of a hot gas predicts that the 
Sun should have a reflection factor of 0.10. Find the Sun's surface temperature if this model holds. 
10. Consider two spherical bodies A and Β of the same size in a black enclosure. Each is free to radiate 
thermal energy to the other. Each body is in thermal equilibrium with its immediate surroundings and 
with the other. The temperature of the bodies and their immediate surroundings has a fixed value T. A 
has a reflection factor of 0.70 and Β has an absorption factor of 0.60. 
(a) Which body, A or B, radiates energy at the greater rate? 
(b) Explain your answer. 

Problems 
27 
11. A beer cooler made of oak contains 8 kg of ice at O'^C. Its dimensions are 0.60 x 0.40 x 0.40 m. The wood 
is 5 cm thick, and has a thermal conductivity of 6x 10"'kcal/m sec°C. If the outside temperature is 
20T, 
(a) what is the rate of heat flow into the ice box? 
(b) how long will it take to melt the ice? 
12. A long rod, insulated to prevent heat losses, has one end immersed in boiling water (at atmospheric pres-
sure) and the other end in a water-ice mixture. The rod consists of 1 m of copper (one end in steam) and 
a length L2 of steel (one end in ice). Both rods are of cross-sectional area 5 cm\ The temperature of the 
copper iron junction is 60T, after a steady state has been set up. 
(a) How much heat flows per second from the steam bath to the ice water mixture? 
(b) How long is L2? 
13. Heat, sufficient to vaporize water at lOOT at a rate of 1.80 kg per hour, passes through the bottom of an 
aluminum pan, 2 mm thick and 250 cm^ in area, with a certain flame under the pan. Calculate the 
temperature of the underside of the pan next to the flame. 
14. The operating temperature of a tungsten ñlament in an incandescent lamp is 2227°C, and its absorption 
factor is 0.30. Find the surface area of the filament of a 100 watt lamp. 
15. In a brick house, the air in a room has a temperature of 20X when the outside temperature is - 18°C. 
The convection coefficient is 10"' kcal/sec m^°C inside and 2 x 10"' kcal/sec m^°C outside. The rate of 
heat transfer per unit area through the walls is 2.25 x 10"^ kcal/sec ml Find: 
(a) the temperature of the inside surface of the wall and 
(b) the temperature of the outside wall. 
16. An uninsulated steam pipe 10 cm in diameter with an outside temperature of 95°C passes through a 
room. The absorption factor is 0.70, and the convection coefficient when the room is 25T is 2x 
10"' kcal/sec m'°C. 
(a) What is the heat loss per meter of pipe by radiation? 
(b) What is the heat loss per meter of pipe by natural convection? 

5 
Thermal Properties 
of Substances 
5-1 
Explanation of Change in State 
29 
5-2 
Pressure-Temperature Diagram for 
a Pure Substance 
31 
5-3 
Real Gases 
31 
5-4 
Expansion of Gases 
32 
5-5 
Liquefaction of Gases 
32 
5-6 
Atmospheric Humidity 
33 
5-1 
E X P L A N A T I O N OF C H A N G E IN S T A T E 
We will now consider the various states of matter 
and the interconversions among them. If we apply 
heat to a crystalline solid in which the atoms or 
molecules are arranged in a regular pattern, the 
temperature, energy, and amplitude of atomic vi-
brations increase until the atoms can no longer hold 
together in their regular position. The atoms be-
come free to move around and to slide over each 
other, and the solid changes to a liquid. 
In the case of pure crystalline substances such as 
ice, the fusion or melting point is sharply marked. 
For a specific pressure, there is a definite tempera-
ture above which the substance is wholly liquid and 
below which it is solid. In the case of amorphous 
substances, such as glass, there is no definite melt-
ing point. Such substances are pliable over a range 
of temperatures. Figure 5-1 is a temperature-time 
graph for both types of substances when heat is 
being added to them. The flat portion of curve (b) in 
Figure 5-1 represents the time interval during which 
the change in state occurs. The constant tempera-
ture indicated by this flat portion is the temperature 
corresponding to the change in state, for example, 
melting point. Curve (a) of Figure 5-1 does not 
have a flat portion, consistent with the fact that 
there is no definite melting point. If a substance ex-
a—Amorphous 
b—Crystalline 
TIME 
Figure 5-1 Temperature-time graph for a crystalline 
and amorphous substance. 
pands on solidifying, an increase of pressure will 
lower its melting point; if it contracts on solidifying, 
an increase in pressure will raise its melting point. 
This is an example of Le Chatelier's principle 
which says that a system will react to applied forces 
by assuming a configuration that seeks to minimize 
the forces applied. Thus, applying a pressure to 
a solid causes a decrease in volume. Since the 
29 

30 
Thermal Properties of Substances 
volume decreases upon melting for the first case, 
an increase in pressure lowers the melting point 
and vice versa for the second case. The pressure 
required is quite large for any significant change in 
the melting point. Consider a wire with a weight on 
it hung around a block of ice, as illustrated in Figure 
5-2. The ice directly below the wire melts because 
its melting point is lowered below the temperature 
of the surrounding ice. The water resulting from the 
melted ice flows to the top of the wire and freezes 
again into ice because of the reduced pressure. 
Figure 5-2 Wire passing through block of ice. 
Eventually, the wire will pass through the block of 
ice leaving it intact. The process of melting under 
pressure and freezing again, as in the above exam-
ple, is called regelation. Another example of regela-
tion is ice skating. The ice beneath the skates melts 
because of the pressure due to the weight of the 
skater. The skater therefore glides on a thin film of 
water which freezes again behind him. 
In the change in state from liquid to vapor, the 
process is different than from solid to liquid. In a 
liquid, not all the molecules have the same energy. 
There are always a few which have energies and 
velocities greater than the average, just as there are 
a few with energies and velocities that are lower. 
When a liquid is exposed to the open ah-, some of 
these molecules with high energy escape from the 
liquid surface and may be carried away by air cur-
rents. Any liquid exposed to the air evaporates. 
Evaporation occurs at all temperatures. If, how-
ever, a liquid is put in a closed container, mole-
cules leaving the liquid accumulate in the space 
above the liquid. Some of these vapor mole-
cules will settle on the liquid surface and recon-
dense. Soon, a steady or equilibrium state is 
reached, where the number of molecules leaving 
the liquid surface becomes equal to the number 
reentering it. When this equilibrium is reached, the 
space above the liquid is said to be saturated with 
vapor; the pressure of such a vapor is called the 
saturated vapor pressure. The rate of evaporation 
depends on the temperature, and the rate of recon-
densation depends on the vapor pressure. For this 
reason, there is a direct connection between the li-
quid temperature and the vapor pressure. The satu-
rated vapor pressure of any substance is found to 
depend only on the temperature. It does not depend 
on the amount of vapor present. The saturated 
vapor pressures for water at various temperatures 
is given in Table 5-1. Figure 5-3 is a typical satu-
rated vapor pressure versus temperature graph for 
a substance. The end of the vaporization curve is 
called the critical point. Above the critical point, no 
distinct liquid-to-vapor phase transition can be de-
tected. The temperature and pressure at which this 
happens are called the critical temperature (tc) and 
critical pressure ( p c ) , respectively. Table 5-2 gives 
the critical temperatures and pressures for some 
common gases. 
Since the molecules of higher energy escape 
from the liquid, the average molecular energy of the 
remaining liquid-is decreased. Therefore, the liquid 
is cooled by evaporation unless heat is supplied to 
compensate for this energy loss. 
A liquid boils in the open atmosphere at a tem-
perature for which the vapor pressure just above 
the liquid surface is equal to the atmospheric pres-
Table 5-1 Pressure of Saturated Water Vapor. 
Temperature 
Pressure 
Temperature 
Pressure 
r c ) 
(mm Hg) 
C^C) 
(atm) 
0 
4.58 
100 
1.0 
5 
6.51 
n o 
1.41 
10 
8.94 
120 
1.96 
15 
12.67 
140 
3.57 
20 
17.5 
160 
6.10 
30 
31.8 
180 
9.90 
40 
55.1 
200 
15.4 
50 
92.5 
220 
22.8 
60 
149.0 
250 
39.3 
70 
234.0 
300 
84.8 
80 
355,0 
350 
163,2 
90 
526.0 
374 
218.4 

Real Gases 
31 
Critical point 
TEMPERATURE 
Figure 5-3 Vapor pressure versus temperature. 
Tabie 5-2 Critical Temperatures and Pressures for 
Various Gases. 
Critical 
Critical 
temperature 
pressure 
Substance 
(in °C) 
(in atmospheres) 
Air 
- 140.7 
37.2 
Ammonia 
132.4 
111.5 
Argon 
- 122.0 
48.0 
Carbon dioxide 
31.0 
73.0 
Helium 
- 267.9 
2.26 
Hydrogen 
-240.0 
12.8 
Nitrogen 
- 147.1 
33.5 
Oxygen 
- 118.8 
49.7 
Water 
374.0 
218.4 
sure. Figure 5-3 is, therefore, also a graph of the 
variation of the boiling point with external pres-
sure. The higher the external pressure, the higher 
the boiling point must be. This is why water boils at 
a higher temperature in a pressure cooker. The boil-
ing points of various substances at standard at-
mospheric pressure were given in Table 2-2 of 
Chapter 2. 
5-2 
P R E S S U R E - T E M P E R A T U R E 
D I A G R A M 
FOR A P U R E S U B S T A N C E 
Figure 5-4 Triple point diagram for water. 
one another. Similar curves representing the pres-
sure and temperature at which the solid and liquid 
and the solid and vapor are in equilibrium can be 
drawn. For a suitable range of temperature and 
pressure, a single pressure-temperature diagram 
will show all three curves. These three curves in-
tersect at one point—called the triple point. As a 
result, the diagram is usually given the name triple 
point diagram. Figure 5-4 is such a diagram for 
water. The curves representing the pressure and 
temperature at which the solid and liquid phase are 
in equilibrium is called the fusion or melting point 
curve, the solid and vapor in equilibrium—the sub-
limation 
curve, and the liquid and vapor in 
equilibrium—the vaporization 
or boiling 
point 
curve. The point at which all three phases can coex-
ist in equilibrium is, of course, the triple point. 
While the pressure corresponding to the triple 
point of water lies well below atmospheric pres-
sure, this pressure for some substances, such as 
CO2, lies above atmospheric pressure. At atmos-
pheric pressure, CO2 sublimes or, in other words, 
passes directly from the solid to the vapor at 
- 78°C. Solid CO2 is commonly known as dry ice. It 
is used to keep things cool; it is more desirable than 
ordinary ice because it is less messy since it does 
not go through the liquid state. 
5-3 
R E A L G A S E S 
We have seen that the vapor pressure curve rep-
resents the pressure and temperature at which the 
liquid and vapor phases exist in equilibrium with 
Real gases do not obey Boyle's law or the general 
gas law perfectly. This is best illustrated by a ρ - V 
diagram (Figure 5-5) for a real gas. Suppose the gas 

32 
Thermal Properties of Substances 
Figure 5-5 Pressure-volume diagram for a real gas. 
is compressed at a temperature ii along the curve 
ABCD in the diagram. If the pressure is low, AB in 
the diagram, the curve follows closely that of an 
ideal gas, and can be fitted approximately by 
Boyle's law. When point Β  is reached, however, a 
sharp change occurs in the nature of the curve and 
the volume continues to decrease without further 
increase of pressure (EC in the diagram). This is 
because, at point B, the gas begins to condense and 
gradually changes to a liquid along EC. At point C, 
all the gas has completely changed to liquid and a 
large increase in pressure is required for a small de-
crease in volume (CD in the diagram). 
At a higher temperature, ti, condensation begins 
at a higher pressure and at a smaller volume. This is 
because the effect of the attractive forces between 
the molecules is smaller since the molecules are 
moving faster at a higher temperature. Finally, if 
the temperature is increased sufficiently, a tempera-
ture will be reached where the thermal agitation 
will be so rapid that condensation into the liquid 
state is prevented. This is, as we stated in the pre-
vious section, the critical temperature (tc on the 
diagram). 
The dashed line and the critical temperature 
curve divide the ρ - V diagram of Figure 5-5 into 
four regions. Below the dashed curve, the sub-
stance is a mixture of liquid and vapor in equilib-
rium. Below the critical temperature to the right of 
the dashed curve, the substance is a vapor or gas; to 
the left of the dashed curve, it is a liquid. Above the 
critical temperature curve, the substance is a gas. 
The general gas law, which applies to an ideal 
gas, does not take into account the forces between 
the molecules of the gas. These forces vary with 
pressure, volume, and temperature of the confined 
gas. There are two types of intermolecular forces, 
repulsive and attractive. The repulsive forces are 
short-range forces which become strong when the 
molecules come very close together. The attractive 
forces are of longer range and are weaker than the 
repulsive forces. They are usually called van der 
Waal's forces. Van der Waal studied the character 
of intermolecular forces and developed an equation 
of state for a real gas. This equation takes into ac-
count the volume of, and the forces between, the 
molecules. The equation is 
b) = RT. 
(5-1) 
The quantities a and b are constants for a particu-
lar gas. The term a/V^ takes account of the inter-
molecular forces, and the term b is proportional to 
the volume occupied by the molecules themselves. 
5-4 
E X P A N S I O N OF G A S E S 
If a gas is compressed in a cylinder fitted with a 
piston, work is done on the gas, and the gas be-
comes heated. If, on the other hand, a gas is al-
lowed to expand, the gas does work in pushing the 
piston out against atmospheric pressure, and the 
gas is cooled. 
Consider the following question: If the gas were 
to expand without doing work against atmospheric 
pressure, that is, expand freely into a vacuum, 
would the gas cool? In other words, does the ther-
mal energy of a gas change on free expansion? The 
answer is no for an ideal gas. The thermal energy 
of an ideal gas is independent of volume and pres-
sure changes, provided the temperature does not 
change. However, there is no ideal gas. The thermal 
energy of a real gas depends on pressure as well as 
on temperature. At temperatures not too far below 
the critical temperature, a free expansion causes a 
decrease in temperature of the gas. 
5-5 
LIQUEFACTION 
OF G A S E S 
All gases can be liquefied. Air has a critical tem-
perature of - 140°C, and cannot be liquefied by 
merely compressing it at room temperature. The 
following is a brief description of one conunercial 
method used for the liquefaction of air. It is known 

Atmospheric Humidity 
33 
as the Linde process, and a simple drawing of the 
apparatus used in this process is shown in Figure 
5-6. Air is compressed in the compressor, circulated 
through the coils in the cooler, and allowed to ex-
pand through a small opening at O. As a result of 
the expansion, the air cools further. This cooled air 
is again compressed and circulated through the 
coils in the cooler so that the approaching air be-
comes progressively cooler before expansion. By 
the continuous operation of this cycle, the expand-
ing air is finally cooled to the point where it liquefies 
and is collected at the bottom of the container 
below the expansion tube. 
Cooler 
Compressor 
Liquid air 
Figure 5-6 Linde apparatus for liquid air. 
In 1908, Kammerlingh Onnes liquefied helium, 
which is the most difficult gas to liquefy. He was 
able to do this by passing pure compressed helium 
gas through liquid hydrogen boiling at a reduced 
pressure. The helium was then allowed to expand 
and liquefied at a temperature of 4.2°K. Tempera-
tures as low as 0.7°K can be obtained by the rapid 
evaporation of liquid helium. Temperatures ap-
proaching absolute zero have been reached by de-
magnetizing magnetic materials which were initially 
at liquid helium temperatures. 
All gases, except helium, which have first been li-
quefied can subsequently be solidified by pumping 
away the vapor above the liquid. The solid state of 
helium only exists under high pressure. 
5-6 
ATI\AOSPHERIC HUlVilDITY 
Water vapor is present in the air. Water vapor is 
lighter than air. When water is evaporated into the 
air, it displaces a volume of air equal to its own 
volume. The term humidity is used to describe the 
water vapor content of the atmosphere. The actual 
vapor pressure at any time or place cannot exceed 
the saturation pressure for the existing temperature 
(Figure 5-3), for condensation begins to take place 
as soon as this value is reached. The higher the 
temperature, the more water vapor the air is capa-
ble of holding. When warm air is cooled, some of 
the water vapor condenses and reappears as water. 
The temperature at which the water vapor in the air 
is sufficient to saturate it is called the dewpoint. 
The measurement of humidity is called 
hy-
grometry. 
In hygrometry, either the absolute 
humidity or relative humidity is measured. 
The absolute humidity is the mass of water vapor 
per unit volume in the atmosphere at a given tem-
perature. The relative humidity is the ratio of the 
amount of water vapor present in the atmosphere to 
the amount required to saturate it at the same tem-
perature. 
Since the vapor pressure is proportional to the 
mass of vapor present, the relative humidity is 
equal to the ratio of the actual pressure of the vapor 
present to the saturation vapor pressure at the same 
temperature. It is usually expressed in percent: 
relative humidity (%) = { ^ { 100), 
(5-2) 
where ρ equals the vapor pressure of the water 
vapor present in the air, and p„ equals the vapor 
pressure of the water vapor if the air were saturated 
at the same temperature. 
Example 1. Find the relative humidity if the 
vapor pressure of the water vapor in the air is 5 nun 
of mercury at a temperature of 10°C. 
S O L U T I O N 
From Table 5-1, the saturated water vapor pres-
sure at 10°C is 8.94 mm of mercury. Therefore, 
relative humidity = ( ^ | | ^ ^ ) ( , 0 0 ) = 56%. 
There are various ways of measuring the relative 
humidity. One of the most accurate ways is by de-
termining the dewpoint. This temperature may be 
found by partly filling a brightly polished container 
with water and dropping pieces of ice into it, while 
stirring it with a thermometer. The dewpoint is the 
temperature indicated by the thermometer when 
moisture first appears on the polished surface. The 

34 
Thermal Properties of Substances 
saturated vapor pressure of water at the dewpoint 
is a measure of the water vapor present in the air. 
Therefore, if we know the dewpoint and the tem-
perature of the air, we can calculate the relative 
humidity. 
Example 2. If the room temperature is 20°C and 
the dewpoint is 15°C, find the relative humidity. 
S O L U T I O N 
From Table 5-1, the saturated vapor pressures at 
15T and 20°C are 12.67 and 17.5 nun of mercury, 
respectively. Therefore, 
relative humidity = (12-67 mm of HgV 
^ ^ 3 % . 
\ 17.5 nun of Hg / 
Another way of measuring the relative humidity 
is with the wet and dry bulb hygrometer, which 
makes use of the principle of cooling by evapora-
tion. It consists of two mercury thermometers 
placed side by side. The bulb of one thermometer is 
kept dry, while that of the other is kept continually 
wet by a piece of porous cloth attached to a wick 
which dips into a container of water as illustrated in 
Figure 5-7. The temperature of the wet bulb ther-
mometer will read lower than that of the dry bulb 
thermometer because of the evaporation taking 
place at its surface. If the air is already saturated, 
no evaporation takes place and both thermometers 
give the same reading. Thus, the drier the air, the 
Figure 5-7 Wet and dry bulb hygrometer. 
greater the difference in the readings of the two 
thermometers. Tables have been constructed per-
mitting the relative humidity and dewpoint to be 
determined from the readings of the two thermome-
ters. 
P R O B L E M S 
1. For a given substance, it is observed that the melting point decreases with an increase in pressure. 
(a) Can the substance be liquefied? 
(b) Can the substance be solidified? 
(c) Is the heat of fusion positive or negative? 
Give reasons for your answers. 
2. An air conditioning unit works in the sunmier time by cooling the air below the desired temperature to 
condense out excess water vapor, and then reheating the air. If the outside air is at 86°F and 70% relative 
humidity, and the delivered air is at 68°F, to what temperature must it be cooled to lower the water 
content in the delivered air to 50% relative humidity? 
3. The measured volume of a quantity of hydrogen collected over water is 780 cm\ the temperature being 
16°C and the barometer reading 740 mm. The volume is measured with the water level the same inside 
and outside the bottle. If the quantity of hydrogen collected has a volume of 705 cm^ at O^'C and 1 
atmosphere pressure when dried, calculate the vapor pressure of saturated water vapor at 16T. 
4. The temperature in a room is 20*'C. A can containing water is gradually cooled by adding ice to it. At 10°C, 
the surface of the can clouds over. What is the relative humidity in the room? 

Problems 
35 
5. In the morning, a hygrometer in a room at a temperature of 20T shows the relative humidity to be 37%. 
What is the dewpoint? Later on in the day, the hygrometer is broken, making it necessary to determine 
the relative humidity by an alternative method. The temperature in the room remains at 20°C. The 
dewpoint is measured by the method given in Problem 4 and found to be 10°C. What is the relative 
humidity? 
6. A small amount of liquid is put in a pyrex glass tube. The tube is evacuated and sealed off. Describe the 
behavior of the liquid surface when the temperature is raised 
(a) if the volume of the tube is much less than the critical volume, 
(b) if the volume of the tube is much greater than the critical volume. 
7. The pressure of water vapor in equilibrium with liquid water is given approximately by 
ρ (atmospheres) = \0~U^ 
for t between lOT and 100°C. The volume of a cylinder completely filled with water at 20°C is doubled 
by partly withdrawing a tight fitting piston. After equilibrium is restored, the temperature is lOT. What is 
the pressure of the water vapor which now fills one-half of the cylinder? 
8. What pressure must be maintained in a pressure cooker to boil water at 110°C, when the ordinary boiling 
point is lOOT? 
9. What is the maximum value that could be obtained for the melting point of ice? 

6 
Reflection and Refraction 
at a Plane Surface 
6-1 
Introduction 
37 
6-2 
Rectilinear Propagation 
37 
6-3 
Reflection 
38 
6-4 
The Formation of Images by a Plane 
Mirror 
38 
6-5 
Refraction 
39 
6-6 
The Critical Angle and Total Internal 
Reflection 
40 
6-7 
Dispersion 
40 
6-1 
INTRODUCTION 
In the next eight chapters we will discuss optical 
phenomena. A truly complete understanding of this 
subject becomes possible only after three basic 
questions have been answered: 
What is the nature of light? 
How does it propagate from point to point? 
How does the presence of matter affect its prop-
agation? 
Full answers for these questions would necessarily 
involve a discussion of a major part of physics as 
we understand it today. We are not prepared for 
such detail at present, so what follows in the next 
several chapters must be regarded as incomplete. 
On the other hand, much of our present knowl-
edge of light and its behavior was already known to 
early civilizations (although our present under-
standing is due mainly to the work of scientists 
from the seventeenth to the nineteenth century). It 
is therefore meaningful to consider first the more 
readily observed characteristics of light and some 
practical applications. 
6-2 
R E C T I L I N E A R P R O P A G A T I O N 
In this chapter and the two that follow, we shall 
be concerned with phenomena that can be ex-
plained by noting that a narrow beam of light travel-
ing in an isotropic medium (one whose properties 
are the same in all directions) will travel in a 
straight line—a property called rectilinear propaga-
tion. This is consistent with our experience that ob-
jects illuminated by point sources of light appar-
ently cast sharply defined shadows, and that we 
cannot see around corners. 
Most common sources of light are extended 
sources, not point sources. By using appropriate 
apertures, such a source can be used to define a 
narrow pencil of light.t If the cross section of the 
pencil is reduced to the ideal limit of a point, the 
beam which results is called a ray. The concept of a 
light ray is very useful, for it allows us to use 
straight lines and geometrical constructions to dis-
cuss on paper the behavior of light as it reaches the 
boundary of an isotropic medium or passes from 
one isotropic medium to another of different optical 
character. This method of discussing 
optical 
phenomena is called geometrical optics; we will use 
it in the following pages to discuss mirrors, lenses, 
and optical instruments. 
tThis statement does not apply to laser light, which is 
discussed in chapter 11. 
37 

38 
Reflection and Refraction at a Plane Surface 
6-3 
REFLECTION 
When a ray of light traveling in an isotropic me-
dium arrives at a perfectly smooth boundary sur-
face of a second medium, two different situations 
can occur. The ray may be reflected from the 
boundary surface, thus remaining in the ñrst 
medium, or it may enter the second medium, in 
which case it is said to be refracted. The extent to 
which either of the two situations predominates de-
pends upon several factors to be discussed later. 
Reflection from a smooth, uniform surface is 
termed regular or specular reflection. The laws gov-
erning the situation are easily stated, and can be 
readily veriñed with a small hand mirror or similar 
reflector. 
The laws of specular reflection are as follows: 
1. The ray incident upon the smooth boundary, 
the ray reflected from the boundary, and a 
perpendicular line (the normal) drawn to the 
plane containing the boundary (reflecting sur-
face) at the point of incidence all lie in a single 
plane. 
2. The angle between the incident ray and the 
normal (the angle of incidence, /) is equal in 
magnitude to the angle between the reflected 
ray and the normal (the angle of reflection, r). 
Figure 6-1 illustrates these laws and the quantities 
introduced in them. Thus, Law 1 makes it possible 
to draw a two-dimensional diagram, while Law 2 is 
a statement of the symmetry of the reflection pro-
cess. 
If the pencil of light is large in cross section, and 
it strikes a surface which is very irregular (for 
example, a crumpled sheet of aluminum foil), the 
laws of reflection will not seem to be successful in 
predicting the paths of the reflected light rays, 
which will be visible in all directions. This type 
lncident^,^^^\ 
ray 
Medium 1 
\ 
^^^,^^Normal 
^¿^-^Reflected 
/^j^rm////m/////// 
of reflection is called diffuse reflection. To under-
stand the apparently random directions of the re-
flected rays we note the following: If a normal to 
the irregular reflecting surface is constructed at the 
point of incidence of each (parallel) ray of a small 
pencil of light, it will be observed that the direction 
of each reflected ray will satisfy the laws of reflec-
tion. Therefore, we can see that it is sufficient to 
study specular reflection in detail, since diffuse re-
flection can be reduced to the specular case by suit-
ably limiting either the cross section of the light 
pencil or the extent of the reflecting surface. 
6-4 
THE F O R M A T I O N O F I M A G E S 
BY A P L A N E M I R R O R 
The image formed by a plane mirror of an ex-
tended illuminated object can be constructed by ap-
plying the laws of reflection to individual points on 
the object. A point source of light, in the absence of 
pin holes, will send out light rays in all directions. 
As seen in Figure 6-2, of the many rays striking a 
Observer 
////////////ί?Μ ////////////// 
Figure 6-1 Reflection at a plane boundary. 
Figure 6-2 Image of a point source. 
reflecting surface, only those intercepted by the 
observer's eye need be drawn. The reflected rays, 
extended back to their point of intersection, form 
an image of the source at /. To the observer, the 
light rays seem to come by rectilinear propagation 
dh-ectly from this image. Because the image is be-
hind the reflecting surface where no real light rays 
exist, it is only an apparent or virtual image. Thus, it 
would not appear on a screen placed at /, although 
it certainly seems real to an observer. Since ί = r 

Refraction 
39 
and i' = r', triangles SOO' and lOO' are congruent 
(why?), which establishes the fact that S and / are 
symmetrically located with respect to the mirror. It 
is left as a problem to show that an extended object 
in front of a plane mirror will have a virtual image 
equal in size to the object and symmetrically lo-
cated with respect to the mirror. 
6-5 
R E F R A C T I O N 
A monochromatic (single color) light ray that is 
refracted at a plane boundary between two differ-
ent isotropic media obeys the laws of refraction. 
These are as follows: 
1. The incident ray, the refracted ray, and the 
normal to the plane boundary at the point of 
incidence lie in a plane. 
2. The ratio of the sine of the angle of incidence 
to the sine of the angle of refraction is a 
constant, which is independent of the angle of 
incidence and depends only upon the two me-
dia and the color of the light used. 
These laws are illustrated in Figure 6-3. 
Ν  
r' —  
Medium 1 
O 
Medium 2 / 
////// '//, Ψ /// 
ψ/ 
Figure 6-3 Refraction at a plane boundary. 
The second law is called Snell's law after Wille-
brord Snell, who discovered it in 1621. Expressed in 
equation form, Snell's law reads 
sm I 
sin r 
where «21 is a color dependent constant called the 
relative index of refraction of medium 2 with re-
spect to medium 1. If the roles of the incident and 
refracted rays are reversed, it is found experimen-
tally that the diagram remains the same, and only 
the directions of the rays are reversed.! In this 
case, Snell's law states that 
sm Γ 
sm f 
sm I 
sm Γ 
7= 
Πη , 
and we conclude that 
ni2 = 
— . 
rill 
(6-2) 
(6-3) 
We can show that the relative index of refraction 
of one medium relative to another can be deter-
mined at once, providing the indices of refraction 
of the two media relative to a third are both known. 
If a ray of light enters a slab of material with paral-
lel plane sides, the ray emergent from the slab is 
directed parallel to the incident ray but displaced 
laterally by an amount depending upon the thick-
ness of the slab, the angle of incidence, and the 
relative index of refraction. By successively adding 
other plane parallel slabs, the ray emergent from 
the final slab will still be parallel to the incident ray, 
and the total lateral displacement is simply the sum 
of the individual displacements due to each slab. 
Figure 6-4 shows this situation (where φ ι = Φ Α ). 
Medium 3 
Medium 1 
Lateral displacement 
of the incident beam 
= «21, 
(6-1) 
Figure 6-4 Multiple lateral displacement of a light 
ray. 
tThe reflected rays are not the same when the incident 
and refracted rays are reversed. See Section 6-6. 

40 
Reflection and Refraction at a Plane Surface 
Repeated application of Snell's law yields 
^sin φ ι 
sin Φ 2 
^ sin φ ^ ^ sin ^3 
sin^2' 
sinks' 
sin</>4 
sin^r 
Multiplication of these equations yields riiin^iriis = 
1, from which it follows that 
«21 = 
1 
^ W23 
«32«13 
«13 
(6-4) 
because of Eq. (6-3). 
It is customary, when quoting indices of refrac-
tion, to compare a given material with a vacuum, 
which by convention has been adopted as a stan-
dard or reference medium. The index of refraction 
of a medium with respect to a vacuum is then iden-
tified as the absolute index of refraction, or simply 
the index of refraction of the medium. When the 
term index of refraction is used in this book, it is 
implicit that the comparison medium is the vacuum 
and that, hence, a second subscript is not neces-
sary. Table 6-1 lists some approximate values for 
indices of refraction of various materials. It can be 
seen that the index of refraction is qualitatively re-
lated to the density of the material. Thus, for exam-
ple, there is little difference between the various 
gases and a vacuum. One might, therefore, specu-
late that the index of refraction indicates the extent 
to which a given material interferes with or impedes 
the passage of light. More dense materials would be 
expected to give more effective interference than 
those of lesser density since there are more indi-
vidual atoms or molecules to act. We shall return to 
this point in a later chapter. 
Table 6-1 Typical Indices of Refraction. 
Substance 
Index of refraction 
Dry air (STP) 
1.0003 
C02(STP) 
1.0004 
Water 
1.33 
Alcohol 
1.36 
Glass 
1.50-1.70 
Quartz (fused) 
1.46 
Diamond 
2.42 
A more synunetric form of Snell's law is obtained 
if we use absolute indices of refraction as exhibited 
in Eq. (6-4). Thus, since 
sin φ ι 
«2 vacuum « 2 
—
- 
= « 2 1 = 
= — -
sin Φ
2 
« 1 vacuum « 1 
we can write 
« 1 sin φ ι = « 2 sin Φ
2 . 
(6-5) 
It should be noted that the index of refraction is a 
function of the color of the light ray involved. 
6-6 
T H E C R I T I C A L A N G L E A N D TOTAL 
I N T E R N A L R E F L E C T I O N 
A medium is said to have an optical density great-
er or less than that of a second medium if its index 
of refraction is greater or less than that of the sec-
ond medium. If a ray of light goes from an optical-
ly less dense medium to an optically more dense 
medium ( « 2 > «0, we see from Snell's law that the 
refracted ray will be bent toward the normal. Since 
the angle of incidence in medium 1 cannot be great-
er than 90**, it follows that the angle of refraction 
can take on values between 0** (for ί = 0°) and a 
maximum value θο  (for i = 90°). 
This latter angle, called the critical angle, has the 
following significance. Consider a series of light 
rays in medium 1 with incidence angles between 0° 
and 90°. The refraction angles in medium 2 will be 
as shown in Figure 6-5(a). On the other hand, if we 
now consider, as in Figure 6-5(b), a series of light 
rays originating in medium 2 with angles between 0° 
and 90°, it is clear that Snell's law cannot be 
satisfied when the angle exceeds the critical angle 
since it would imply |sin ö | ^ 1. The physical expla-
nation of this situation is that the "refracted" ray 
for i > ö c is not refracted at all. Instead, it is re-
flected back into medium 2 from the interface and, 
therefore, will obey the laws of reflection. It can be 
shown experimentally that rays originating in the 
optically denser medium will experience both re-
flection and refraction at an interface with a less 
dense medium. The fraction of the incident light 
that is refracted is greatest at i = 0°, and will be 
equal to 0 for angles equal to or greater than ft. This 
latter situation is called total internal reflection. 
6-7 
D I S P E R S I O N 
In Section 6-5, it was stated that the index of 
refraction of a medium is a function of the color of 
light in the medium. A schematic illustration of this 
relationship is shown in the graph of Figure 6-6. 
Table 6-2 gives a set of typical values of the index 
of refraction for various colors. In Chapter 11, a 
more quantitative method of distinguishing colors 

Dispersion 
41 
Medium 1 
Medium 2 
Medium 1 
^ 
Medium 2 
5-
Figure 6-5 The critical angle and total internal reflection. 
Figure 6-6 Variation of index of refraction with color. 
Table 6-2 Color Dependence of the Index of Refrac-
tion for a Typical Flint Glass. 
Red 
Yellow 
1.622 
1.627 
Blue 
Violet 
1.639 
1.662 
will be presented. For the present, we merely note 
that red light is refracted less than blue or violet 
light. We note also that the values for the index of 
refraction in Table 6-1 must be regarded as approxi-
mate or average values since a white light source is 
assumed. 
White light is a combination of all the various col-
ors of light. The percentages of the various colors 
in a beam of white light depends markedly upon the 
nature of the source. Now let us consider what 
effect the variation of index of refraction with color 
will have upon the appearance of a ray of white 
light passing through a transparent substance with 
non-paraUel sides—^for example, a prism. The situa-
tion is illustrated by Figure 6-7. 
Figure 6-7 Dispersion of white light by a prism. 
At the first surface, the rays of the various colors 
will be bent by an amount that increases from red to 
violet. In addition, the non parallel sides of the 
substance accentuate the separation of the various 
colors by providing different path lengths to be tra-
versed by the separate rays. Finally, when these 
separated rays leave the transparent medium, the 
separation is further enhanced by a bending away 
from the normal that is greatest for violet and least 
for red light. Why? (It is implied here that the 

42 
Reflection and Refraction at a Plane Surface 
medium surrounding the prism of transparent mater-
ial is a vacuum or, at any rate, a medium with an 
index of refraction lower than that of the prism.) It is 
clear from the preceding discussion that the devia-
tion angles for the various colors depend not only 
upon the index of refraction but also upon the apex 
angle A of the prism. This "spreading-out" of the 
several colors in the ray of light is known as 
dispersion. If the constituent colors of a light ray are 
known, they can be used to deduce the refractive 
properties of a material. Conversely, if the refrac-
tive properties of the material are known, they can 
be used to analyze a beam of light whose chromatic 
makeup or color composition is unknown. 
P R O B L E M S 
1. Show that a light ray reflected from a plane mirror will be rotated through an angle 2Θ  when the mirror is 
rotated through an angle Θ . Hint: Consider the light ray originally at normal incidence and rotate the 
mirror. 
2. Use the laws of specular reflection to show that an extended object in front of a plane mirror will have a 
virtual image equal in size to the object and synunetrically located with respect to the mirror. 
3. A ray of light is incident at grazing incidence onto one face of a glass cube. The ray emerges from an 
adjacent face of the cube at an angle θ relative to the normal to that face. Show that sin Ö = cot ft, where 
ft is the critical angle. 
4. A circular disc floats on the surface of a liquid for which η  = 1.4. What must its diameter be if a point 
object located 10 cm below the surface is not to be visible from above? 
5. A ray of light strikes a plane sheet of glass (n = 1.60) that is 3 cm thick. The angle of incidence is 45°. 
Determine the lateral displacement D of the ray as it emerges from the glass. (See Figure 6-4.) 
6. If a ray of light makes an angle of incidence of 53° at an air-diamond interface, what are the angles of 
reflection and refraction for the light? 
7. When an object is under water, it appears to be closer to the surface than it actually is. Show that the 
apparent depth is equal to the actual depth divided by the index of refraction for the water, η  = 1.33. 
8. At what angle should a fish look up toward the surface of the water in order to see objects on the shore? 
9. Two plane mirrors meet at an angle of 60°. A ray of light enters parallel to one mirror and strikes the 
other. By making a ray diagram, trace the ray through the system and tell how it leaves. 
10. By means of ray diagrams, show the number of images that a person will see by standing in front of a 
pair of vertical plane mirrors joined along one edge and making an angle of 90° with each other. Consider 
both single and multiple reflections. 
11. A ray of white light is incident on the face of a 
60° prism so that the violet light traverses the 
prism parallel to the base of the prism as shown 
in Figure 6-8. The index of refraction for the 
violet light is 1.662. Calculate the deviation 
angle for violet light. 
Figure 6-8 

7 
Reflection and Refraction 
at a Curved Surface 
7-1 
Terminology and Sign Convention 
7-2 
Reflection from a Spherical 
Surface—Ray Tracing Method 
45 
7-3 
Reflection from a Spherical 
Surface—Analytical Method 
46 
43 
7-4 
Refraction at a Spherical Surface 
7-5 
Combinations of Reflecting and 
Refracting Surfaces 
48 
47 
7-1 
T E R M I N O L O G Y A N D S I G N 
C O N V E N T I O N 
In the last chapter, the laws of reflection and re-
fraction were applied to situations involving plane 
surfaces. The plane surface is a special case of a 
curved surface (a spherical surface with an infinite 
radius of curvature), so it is appropriate to consider 
next reflection and refraction for general curved 
surfaces. Experimental studies show that the laws 
of reflection and refraction as stated are sufficient, 
when applied to individual light rays, to provide a 
satisfactory analysis in all situations for which geo-
metrical optics is appropriate. 
It is found that the specific behavior of a given 
light ray depends upon the radius of curvature of 
the optical surface (the interface between two 
media of differing indices of refraction). Thus, a 
concave mirror with its center of curvature in front 
of the reflecting surface as in Figure 7-1(a) will tend 
to focus or bring into convergence a beam of light 
rays parallel to the axis of synunetry of the mirror. 
On the other hand, a convex mirror, with its center 
of curvature behind the reflecting surface as in 
Figure 7-1(b), will tend to defocus or cause a 
divergence of a similar light beam. Even though the 
magnitude of the two radii of curvature may be the 
same, the effects on a beam of light are markedly 
different. 
The location of the image of an object providing 
the oncoming light beam is also affected by the 
radius of curvature. For a concave mirror the light 
rays converge to a real image at a point in front of 
the mirror. The image formed by a convex mirror is 
virtual since the reflected rays appear to be diver-
ging from an image point behind the mirror where 
no light rays actually exist. To distinguish between 
these two cases, it is reasonable to identify object 
and image distances and radii of curvature as either 
positive or negative. (We shall see that the same 
designations are appropriate for all other distances 
relevant to reflection and refraction situations.) 
Therefore, it is desirable to introduce a choice of 
signs for each of the several distances or dimen-
sions to be encountered in any situation before be-
ginning an analysis of any specific case. Such a set 
of choices is called a sign convention. The reader 
can, by a rapid survey of physics textbooks, dis-
cover that there are a number of different sign 
conventions in use. The effect of a given sign 
convention is to label some distances positive, 
others negative. Another sign convention may pro-
duce different signs for some or all of the distances 
involved, but the magnitudes of all distances will 
43 

44 
Reflection and Refraction at a Curved Surface 
CONCAVE MIRROR 
CONVEX MIRROR 
C 
I 
C 
\ 
(a) 
\ 
(b) 
Figure 7-1 
remain unchanged. The interpretation of algebraic 
results, however, can be extremely confusing for 
the reader who turns to one or more additional texts 
while studying geometrical optics. It is, therefore, 
mandatory that in consulting any text on geometri-
cal optics one must first determine the sign conven-
tion adopted if confusion is to be minimized. 
In this text we provide a simple sign convention 
and apply it to the case of reflection by way of 
illustration. We shall see, however, that it can be 
applied consistently to cases of refraction as well, 
with similar analytical results and interpretation. 
The sign convention for geometrical optics is as 
follows: 
1. All distances are to be measured from the op-
tical surface to the point in question along the 
axis of the optical surface. 
2. The distance from the surface to the object 
(the object distance p) is positive if the direc-
tion traveled is opposite to that of the ap-
proaching light. 
3. The distance from the surface to the image 
(the image distance q) is positive if the direc-
tion traveled is the same as that of the depart-
ing light. 
4. Any radius of curvature (R) is positive if the 
direction from the surface to the center of cur-
vature is the same as that of the departing 
light. 
5. Any object or image dimension above the axis 
of the optical surface is positive; below the 
axis—negative. 
As used in geometrical optics, the term axis re-
fers to the axis of synunetry of the optical surface. 
As an example, consider the heavy horizontal line 
in either Figure 7-l(a) or (b). An axis of synunetry 
implies a conic surface, and one commonly encoun-
ters discussions of spherical and paraboloidal opti-
cal surfaces. The reflection or refraction of light is 
physically less complicated for the latter case than 
for the former case. To see this, refer to Figure 7-2, 
where the laws of reflection are applied to the indi-
vidual light rays striking a spherical [Fig. 7-2(a)] and 
a paraboloidal [Fig. 7-2(b)] reflecting surface. In the 
latter case, all the reflected rays cross the axis at a 
common point, called the focal point. For the 
spherical surface, the reflected rays cross the axis 
at a point which is dependent upon the vertical dis-
tance from the axis of the incoming horizontal ray. 
This is known as spherical aberration. 
Thus, there is no unique focal point for a spheri-
cal reflecting surface, unless one restricts the reflec-
tion from the spherical surface to a small central 
portion. When this is true, the central part of the 
spherical surface approximates the exact curvature 
of the paraboloidal surface. It is desirable to con-
sider spherical surfaces in preference to the para-
boloidal surface because of the mathematical sim-
plification thus afforded. In the following discus-
sions, therefore, references to spherical surfaces 
must be understood to refer to centrally limited 
spherical surfaces. 
In many situations, one does not have a beam of 
light rays parallel to the axis of a surface upon 
which the beam is incident. For non-parallel rays, 
the analysis is complicated by the appearance of 
trigonometric functions. To avoid this difficulty, 
one generally considers situations for which inci-
dent and reflected (or refracted) rays all make 

Reflection from a Splierical Surface— Ray Tracing Method 
45 
SPHERICAL MIRROR 
PARABOLIC MIRROR 
» 
1 
\ 
o 2 —^ 
3 
—  
* 
/ f 1 / 
3 
/ 
3 
2 / 
2 
1 
/ 
(a) 
(b) 
Figure 7-2 Reflections of parallel light rays. 
angles relative to the axis of the surface which are 
small enough to permit use of the approximate 
relationst 
sinÖ«ö  
tanÖ«ö , 
where Ö is measured in radians. For angles less 
than 0.2 radians (« 12°), these relations are in error 
by less than 1%; thus, relatively large angles (:s 12°) 
can be considered. Rays satisfying these relations 
are called paraxial rays since they are only approx-
imately parallel to the axis of the surface. In the 
analytical solutions that follow, we shall see that 
the use of paraxial rays makes possible greatly sim-
plified discussions of reflection and refraction from 
curved surfaces. 
7-2 
R E F L E C T I O N F R O M A S P H E R I C A L 
S U R F A C E — R A Y T R A C I N G 
M E T H O D 
It was stated in the last chapter that to analyze 
any reflection it is sufficient to break up the beam of 
light into individual rays and apply the laws of re-
flection to each of them. The ray tracing method 
does this for a set of rays that have properties lend-
ing themselves to easy graphical construction. The 
paths of three such rays are shown in Figure 7-3. 
tSee the tables of Natural Trigonometric Functions in 
the Appendix. (Note that 2ir radians = 360^) 
Constructing normals to the spherical surface 
(radii) at the point of intersection of the light rays 
with the surface (neglecting spherical aberration), 
and using the laws of reflection, establishes the fact 
that these rays have the following properties: 
1. A ray parallel to the axis upon reflection from 
a converging surface crosses the axis at the fo-
cal point, which is at a point halfway between 
the center of curvature and the center of the 
surface, along the axis, or appears to come 
from the focal point for a diverging surface. 
2. A ray passing through the center of curvature 
(or directed toward the center of curvature for 
a diverging surface) is reflected back upon it-
self since it is incident normally upon the sur-
face. 
3. A ray passing through the focal point (or di-
rected toward the.focal point for a diverging 
surface) is reflected parallel to the axis. 
4. A ray striking the center of the surface will be 
reflected at an equal angle on the opposite side 
of the optic axis. 
Notice that this construction yields not only the lo-
cation of the image but also its size, orientation, and 
(by virtue of its location) whether it is real or vir-
tual. As the figures indicate, two of the rays above 
are sufficient to determine the characteristics of the 
image. Use of the third ray or fourth ray, however, 
provides a desirable check on the accuracy of the 
construction. 

46 
Reflection and Refraction at a Curved Surface 
CONCAVE MIRROR 
CONVEX MIRROR 
(b) 
Figure 7-3 Graphical constructions for reflection. 
7-3 
R E F L E C T I O N F R O M A S P H E R I C A L 
S U R F A C E — A N A L Y T I C A L 
M E T H O D 
Consider a point source on the axis of a concave 
(converging) mirror as in Figure 7-4. The center of 
curvature is at C, the object and image points are at 
Ρ  and 0, respectively, and O is called the vertex of 
the optical surface. 
From the figure and the fact that the sum of the 
angles of a triangle is 180°, we see that 
from which 
/3 = α + δ  
γ + α = 2j3. 
(7-1) 
Applying our sign convention to the figure, we ob-
tain 
h 
, 
^ 
h 
, 
h 
tan a = — t a n β  = ^ 
tan y = 
p-€  
q-€  
As asserted earlier, these equations do not admit a 
simple solution unless the angles are small (paraxial 
rays). For such a case, we can equate the tangents 
to the angles and neglect the smaU distance € , so 
that Eq. (7-1) becomes (for paraxial rays) 
Ρ  
Q 
R 
or 
1+1 = 1 = 1 
ρ q 
R 
r 
(7-2) 
Figure 7-4 Geometry for the law of reflection. 
Notice that the definition of the focal point (the 
value of <j f or ρ = oo) leads by means of Eq. (7-2) to 
the result that the focal length has the value RH as 
asserted earlier. Equation (7-2) also includes the 
special case of the plane mirror. In this case, Ä = oo, 
^ = - p, as in Problem 6-2. 
Next we investigate reflection for an extended 
object. Figure 7-5 shows that unless the object 
height y is small, the distance from the top of the 
object to the surface will differ from the distance 
from the bottom of the object to the surface. This 
will produce a curved image, and this effect is an-
other aberration of spherical surfaces, called cur-
vature of field. We assume here that the object is 
limited in height so that this curvature is effectively 

Refraction at a Spherical Surface 
47 
y 
ρ 
- 
- 
-ì  
-^-^ 
Figure 7-5 Reflection of an extended object. 
absent. Since the angles of incidence and reflection 
are equal, we can write 
tanö  = ^ = 
^ , 
where the image height is negative by virtue of our 
sign convention. Defining the magnification m as 
the ratio of image to object height, we obtain 
y 
Ñ  
(7-3) 
Notice that for this case (a real image) the mag-
nification is negative since the object and image 
distances are positive, as is the object height, while 
the image height is negative. For a virtual image, 
the image distance would be negative and the mag-
nification would therefore be positive. In this way, 
we see that the magnification gives information not 
only as to the increase or decrease in size of the 
image relative to the object, but also indicates its 
orientation. The point here is that the use of the 
sign convention together with Eqs. (7-2) and (7-3) 
are sufficient to analyze any reflection problem 
without recourse to a graphical construction. 
7-4 
R E F R A C T I O N AT A S P H E R I C A L 
S U R F A C E 
Let us now consider the case of refraction at a 
spherical surface, illustrated in Figure 7-6. An ob-
ject at point Ñ  in a medium of refractive index η  
has an image at point Q in a medium of refractive 
index n'. The radius of curvature of the optical sur-
face is R. From the triangles PIC and C/Q, we have 
Figure 7-6 Refraction at a spherical surface. 
Snell's law in this case is 
ç  sin ö  = n' sin φ \ 
In addition, we can write 
tan a = —;—, 
tan /3 = -= 
, tan y = 
. 
p + e ' 
^ 
R-e' 
^ 
q-€  
Again, restricting the discussion to paraxial rays 
permits use of the relations 
sin Ö =^ tan Ö « Ö, 
and the small distance e can be neglected. Then 
Snell's law becomes 
η φ  = 
η 'φ \ 
and the tangent relations become (approximately) 
h 
^ 
h 
h 
^=p' 
y = q' 
Using these results in Eqs. (7-4), 
or 
and, finally. 
na 
=(η '-η )â  
η  
n' 
^(n'-n) 
Ñ  
q 
R 
(7-5) 
or . 
(7-4) 
1 + Í L = ( ! L : _ I ) 1 
(7.6) 
ρ 
q 
\n 
IR 
For a given object distance, therefore, we see 
that the image distance depends not only on the 
radius of curvature of the optical surface, as in 
reflection, but also upon the indices of refraction of 

48 
Reflection and Refraction at a Curved Surface 
the two media involved. We can also investigate the 
question of a focal point and a related focal length 
for a refracting surface. In this case, one can con-
sider an infinitely distant object with an image at a 
focal point or an infinitely distant image with the 
object at a focal point. The magnitudes of the focal 
lengths in each case follow from Eq. (7-5), and it is 
likewise clear that they are on opposite sides of the 
surface. 
To determine the magnification of an extended 
object (assuming small angles as before), we refer 
to Figure 7-7. Here we have 
t a n ^ = A 
t a n ^ ' = — ^ 
Ρ  
Q 
and 
η  sin φ  = n' %\η φ \ 
I " ' 
y 
y 
Figure 7-7 Refraction for an extended object by a 
spherical surface. 
For small angles. 
or 
Therefore, 
n'p 
q 
^ 
_y^-nq 
tn —  —  —
 — ;— . 
y 
n'p 
(7-7) 
Once again, Eqs. (7-5) and (7-7) with our sign 
convention are sufficient for a complete analysis of 
single surface refraction (for paraxial rays). Notice 
that Eq. (7-6) has the same form as Eq. (7-2) for 
reflection. This is in part due to the choice of sign 
convention adopted. 
Although we do not do so, the reader may de-
monstrate that if the two focal lengths are known 
from experiment—or by calculation from Eq. 
(7-6)—one can apply the ray tracing method of 
Section 7-2 without change to situations involving 
refraction. 
7-5 
C O M B I N A T I O N S O F R E F L E C T I N G A N D 
R E F R A C T I N G 
S U R F A C E S 
In this section, we consider situations involving 
more than one optical surface, either reflecting or 
refracting. As the examples below illustrate, no 
new concepts are required. Instead, one simply 
solves the problem serially. That is, the image pro-
duced by the first surface becomes the object for 
the second surface, the new image is then used as 
the object for the thh-d surface, and so forth. There-
fore, adherence to the sign convention and applica-
tion of Eqs. (7-2), (7-3), (7-6), and (7-7) to the vari-
ous surfaces for any number of surfaces is suffi-
cient. It must be stressed, however, that application 
must be made in the order that they affect the light 
from the original object. 
Example 1. Let us describe the image formed for 
an object placed a distance of 5R to the left of the 
center of a glass sphere (index n' = 1.50) of 
radius R, as shown in Figure 7-8. The height of 
the object is RI3. 
S O L U T I O N 
Before we analyze the problem quantitatively, it 
can be inferred from Snell's law that the final image 
will be nearer the center of the sphere than is the 
object, and it will be reduced in size. This is true 
because the front (left) surface causes a conver-
gence of light rays from the object (n' >n). Then, 
the already convergent rays are converged even 
more by the second (right) surface since the light 
rays are now entering a less dense medium and 
therefore bend away from the normal at the sur-
face. This produces rays that cross the axis nearer 
the center of the sphere as asserted. 
From Eq. (7-5), we have 
1 
4R 
for the first surface, or 
2qi 
\2 
')R 
qi = eR, 
Thus, with respect to the second surface, p 2 = 
- (61? - 2 Ä ) . Why? Equation (7-6) now yields 
or 

Combinations of Reflecting and Refracting Surfaces 
49 
n=1.00 
y=R/3 
-5R-
15. 
Figure 7-8 Refraction by a sphere. 
Therefore, the final image is located at a distance 
TR to the right of the center of the sphere. Since 
it is outside the sphere, it will be a real image. The 
total magnification will be the product of that pro-
duced by the first surface and that produced by the 
second surface. Therefore, using Eq. (7-7) twice 
gives 
MTOTAI —  
m\m2 
The real image is therefore inverted and reduced 
from a height y = Ä/3 to a height 
-3^R_-R 
y 
= rritotai — -ψ- 
3" ~ 
~7~· 
In the figure, a ray from the object through the 
center of the sphere passes through undeviated, 
whereas a ray parallel to the axis from the original 
object has been properly located by means of 
Snell's law at each surface to yield the result ob-
tained above analytically. 
Example 2. Repeat the analysis of Example 1 for 
an object of height y = Ä/3 located a distance 5R 
to the left of the center of a hemisphere (index η ' = 
1.50) of radius R if the plane surface of the hemi-
sphere is silvered (totally reflecting). 
S O L U T I O N 
The situation is illustrated in Figure 7-9. As the 
rays sketched indicate, the image will be real, in-
Silvered plane surface 
= R/3 
n=1.00 ^.^y^ 
\ 
\ 
r^^^ 
' 
^ 
\1.50i 
7" 
5R 
Figure 7-9 Refraction and reflection by a silvered 
hemisphere. 
verted, and near the front (left) surface. Using Eq. 
(7-5), 
R^lqi 
\2 
')R 
AR 
yields qx = 6R 
so that p2 = -{6R-R) 
= 
-5R. 
Equation (7-2) applies next for the reflection at the 
plane surface, giving 
or 
q2 = 5R. 
This requires that ρζ  = - {5R - R) = - AR for the 
second refraction at the curved surface. Now Eq. 
(7-5) gives 
© (-ά)^έ=(-^)(4)· 

50 
Reflection and Refraction at a Curved Surface 
and 
= iR as shown. Here, the magnification is 
In the next chapter, we shall apply the considera-
tions of this chapter to situations involving two 
7-Τ αΛ  /-Ä"\ Γ-η 'αΛ  
refracting surfaces separated either by finite or neg-
mtotai = mim2m3 
= (—7-7) ^ ( 
Μ  x (——^ I 
ligible distances compared to the radii of curvature 
\n ρ J 
\ η  P2 / 
\ npi J 
γ . 
^ 
. 
0 1 . · . . · 
1 
of the two surfaces. Such situations are known as 
^ _ (cjxQiQÄ  
^ _ 
(6Jg)(5J?)(fj?) 
thick and thin lenses, respectively, and their appli-
\p 1P2P3/ 
(AR)(- 
5R)(- 
AR) 
14* 
cation, either singly or in combination, lead to such 
^ ^ , . 
. 
, . 
. 
, . 
J 
optical instruments as the microscope, the tele-
Thus, the final image m this case is real, inverted, 
, 
J - 
, . , 
scope, and the camera, 
and has a height 
6 R 
R 
y = m t o t a i y = -
3
- = - ^ . 
P R O B L E M S ! 
1. It was asserted in Section 7-1 that there is no unique focal point for a spherical surface. Demonstrate 
this for reflection by making a scale diagram of a concave spherical mirror with a radius of curvature of 
10 cm. With a straight edge, draw incident rays parallel to the axis at distances of 1 , 2 , 3 , 4 , and 5 cm from 
the axis. Using a protractor, construct the reflected rays and determine the points at which they cross 
the axis. 
2. Show by ray tracing methods that the focal length / of a concave mirror with radius of curvature R is 
given by the relation f = RI2. 
3. Construct by ray tracing methods the image produced by an object located in front of a convex mirror at 
a distance greater than the focal length. Also determine the magnification and whether the image is real 
or virtual. Is the image erect or inverted? 
4. A student finds a spherical metal bowl which is shiny inside and out. Looking into the concave side, he 
sees his inverted image 4 cm from the bottom of the bowl. Turning the bowl over (and keeping the 
distance from himself to the bowl surface the same), he sees his erect image at 3 cm from the bowl. What 
is the radius of curvature of the bowl? 
5. A concave mirror has a focal length of 1.5 cm. Where should an object be placed if its image is to be 
erect and twice the object size? 
6. A spherical shaving mirror (concave; why?) has a radius of curvature of 30 cm. What is the magnifica-
tion produced when the face is 10 cm from the mirror? 
7. To determine the height of a tree, a resourceful physics student notes that the image of the tree just 
covers the length of a 5 cm plane mirror held 30 cm from his eye. If the tree is 90 m from the mirror, what 
is its height? 
8. A spherical fish bowl has a radius R . Neglecting the effect of the glass wall, where would a goldfish 
actually at the center of the bowl appear to be when viewed from the outside? If the goldfish is 3 cm 
long, what will be its apparent length? For water η  = 1.33. 
9. A goldfish swimming inside a spherical fish bowl of radius R appears to be at a distance 2R from the 
front of the bowl. Neglecting the effect of the glass wall, where is the goldfish actually located relative to 
the front of the bowl? If the goldfish is 3 cm long what will be its apparent length? For water η  = 1.33. 
10. When refraction occurs at a spherical surface, Eq. (7-5) can be used to define two focal lengths / i , ¡2. 
When the object is at infinity, q=fu 
and when the image is at infinity, ρ = / 2 . Show that 
η  —  η  
η  —  η  
tExcept for Problem 1, it is assumed that all light rays are paraxial. 

Problems 
51 
11. An object is placed in front of a glass hemi-
sphere of radius R, index of refraction π  = 1.50 
as shown in Figure 7-10. If the object is at a 
distance 3R from the spherical surface and has 
a height y = Ä/3, find the location, orientation, 
and size of the final image, and indicate whether 
it is real or virtual. 
Figure 7-10 
12. For the hemisphere of Problem 11, find the final image position when the object is an infinite distance 
from the spherical surface. Then repeat the calculation when the hemisphere is reversed (so that rays 
from the object now strike the plane surface first). 

8 
Lenses and Optical 
Instruments 
8-1 
Thick and Thin Lenses 
53 
8-5 
The Compound Microscope 
57 
8-2 
Lens Combinations 
55 
8-6 
Telescopes 
57 
8-3 
The Eye 
56 
8-7 
The Camera 
58 
8-4 
The Simple Microscope or 
Magnifier 
57 
8-1 
THICK A N D THIN L E N S E S 
A lens may be described as a material of given in-
dex of refraction which is bounded by two spheri-
cal surfaces. If the separation of the two surfaces at 
the optical axis is not negligible compared to focal 
lengths and/or object and image distances, the lens 
is said to be a thick lens. Conversely, if the separa-
tion is negligible, it is said to be a thin lens. In either 
case, the method of Section 7-5 can be applied to 
situations involving one or more lenses, as was as-
serted there. (In fact Example 1 of Section 7-5 is a 
particular case of a thick lens whose radii of curva-
ture are equal in magnitude.) 
For thin lenses, it is possible to simplify the 
mathematical analysis of a given situation and also 
to develop a simple ray tracing technique analogous 
to the method of Section 7-2 for reflection. The ray 
tracing technique is also applicable to thick lenses 
provided the concept of principal planes is properly 
developed. Since such a discussion is more appro-
priate to an advanced study of optical systems, we 
will restrict ourselves to the case of thin lenses 
only. 
Consider a double convex (converging) lens with 
radii of curvature Äi - Ä2 as in Figure 8-1, with an 
object located at a distance pi in front of it. From 
A 
j 
c 
Τ  
Figure 8-1 Geometry of an object and a thin lens of 
thickness t. 
Section 7-5, we can write 
η  ^ n' 
_(n'-n) 
Pi 
Qi 
Ri 
(8-1) 
for the ñrst surface. Thus, the object distance p2 for 
the second surface will be given by 
P2 = - (^1 - 1 ) 
« - 
(since t is assumed negligible 
for thin lenses). 
Therefore, 
n' . η  
. 
,x/ 
1 \ 
n' , η  
— + — = (m - 
w ) ( — 5 - ) = 
+ 
(8-2) 
P2 φ  
V RiJ 
qx 
qi 
53 

54 
Lenses and Optical Instruments 
Combining Eqs. (8-1) and (8-2), 
^ + ^ = ( „ ' - n ) ( i - + i - ) . 
(8-3) 
Pi 
<?2 
\Rl 
RlJ 
We have previously defined focal lengths for both 
reflection and refraction. If that concept is applied 
here we see that when g = oo, 
and when ρ = oo, 
, = / , = [ ( „ - _ „ ) ( ¿ ^ ¿ . ) ] - ' . 
(8-4) 
Thus, the magnitudes of the two focal lengths of a 
thin lens are the same, as could be checked experi-
mentally by reversing the lens and noting that this 
has no effect on the image produced. Equation (8-4) 
is known as the lens maker's equation. The net 
effect of the lens on the light rays from an object 
can be written simply as 
case for reflection. The three rays chosen in this 
case are illustrated in Figure 8-2: 
n_ η  _ j_ 
(8-5) 
where the subscripts 1 and 2 are no longer neces-
sary if we understand ρ to be the original object 
distance, q to be the final image distance, and note 
that the two focal lengths are identical in mag-
nitude. If the lens is in air (n = l)t then Eq. (8-5) 
becomes 
Ρ  
q 
f 
which is identical in form to Eq. (7-2) for single sur-
face reflection, and is known as the Gaussian form 
of the thin lens equation. Similarly, the magnifica-
tion for a thin lens is given by Eq. (7-3). (See Prob-
lem 8-5.) It should be noted that by our use of Eq. 
(7-5) we are restricted here to situations involving 
paraxial rays. 
For thin lenses, one can regard the refraction of 
individual light rays as taking place at a plane per-
pendicular to the axis and through the center of the 
lens rather than considering separate effects at the 
two surfaces. The resulting refraction is essentially 
an average of the effects at either surface, and for 
thin lenses introduces very little error. With this 
simplification, it is possible to analyze any thin lens 
graphically by means of three rays similar to the 
tSince air is essentially a vacuum; see Table 6-1. 
Figure 8-2 Refraction by a converging lens—image 
location by ray tracing. 
1. A ray from the object passing through (or to-
ward) the first focal point will travel parallel to 
the axis upon refraction. 
2. A ray parallel to the axis will be refracted so as 
to pass through (or appear to come from) the 
second focal point. 
3. A ray through the center of the lens will be 
essentially undeviated. 
The third ray is practically undeviated because the 
thin lens for a central ray approximates a parallel 
plate for which there is displacement without devia-
tion (Section 6-5). Since the displacement is propor-
tional to the thickness (Problem 6-5), it is clear that 
the displacement is negligible for a thin lens. For 
example, in Figure 8-2 an object is placed a distance 
ρ in front of a double convex lens of focal length /. 
The image is located as shown by constructing the 
rays indicated in the previous paragraph. Figiwe 8-3 
illustrates a similar graphical analysis for a double 
Figure 8-3 Refraction by a diverging lens—image 
location by ray tracing. 

Lens Combinations 
55 
concave (diverging) lens. It should be noted here 
that the first and second focal lengths are reversed 
compared to a converging lens. This is because of 
the signs of the radii of curvature in Eq. (8-4). 
8-2 
L E N S C O M B I N A T I O N S 
The effects of a combination of thin lenses are 
analyzed in the same manner as was refraction for a 
combination of surfaces. That is, the image formed 
by the first of a series of lenses is determined by 
Eqs. (8-4) and (8-5) or by a ray tracing technique. 
This image then becomes the object for the second 
lens (with proper account being taken of distances 
and signs relative to the second lens). This process 
is repeated for each successive lens in the combina-
tion. 
Example 1. Show that if two thin lenses of focal 
lengths /i and fi, respectively, are placed in contact, 
the effective focal length / of the combination is 
given by the expression 
S O L U T I O N 
Referring to Figure 8-4, Eq. (8-5) yields 
Ρ  
Q 
fx 
or 
i = l - l 
a 
fx 
P' 
But p ' "^-q (since the lenses are thin). Hence, 
P' 
q' 
fi 
becomes 
Ρ  
q 
fx 
fi 
But this is equivalent to a single lens of focal length 
given by 
In Figure 8-4, the rays for lens 1 are dashed as is the 
image for lens 1 alone. The ray parallel to the axis 
(ray 1) is refracted by lens 2 so as to pass through 
the second focal length, while the ray through the 
center of lens 1 (ray 2) is essentially undeviated by 
the second lens, thus locating the image. 
Example 2. A lens of focal length /i is placed at 
a point which is a distance d to the left of a second 
lens of focal length /2. An object of height y is 
located at a distance P i = 2/, to the left of the first 
lens. Describe the final image. Assume /i = /z = 
f>d, 
S O L U T I O N 
Applying Eq. (8-5) for the first lens. 
2 / % . 
r 
so that 
qx = 2f 
Now, applying Eq. (8-5) for the second lens, 
P2 = -{qx-d) 
= 
d-2f 
and 
or 
1 
d-2fq^ 
f 
Thus, the final image is real and inverted, features 
shown in Figure 8-5. The magnification is mtotai = 
mim2 = (- qxipx) x (- 
qilpi). 
^ 
- / - 2 / V - / ( 2 / - d ) \ / 
1 
\ 
f_ 
2/ A ( 3 / - d ) A-(2/-d)/- 
3 / - d -
Therefore, 
Figure 8-4 Refraction by two lenses in contact. 
yfinaj —  mtotajy —  —  

56 
Lenses and Optical Instruments 
Figure 8-5 Refraction by two lenses not in contact. 
8-3 
THE E Y E 
As an optical instrument, the eye possesses the 
features shown in Figure 8-6. Although the physiol-
ogy of the eye is extremely complicated, the func-
tions of the various parts of the eye can be broadly 
outlined. Light enters the eye through the transpar-
ent, curved shell called the cornea, C, and passes 
through the liquid substance (aqueous humor—A) 
immediately behind the cornea. It then enters the 
pupil, a circular aperture in the colored portion of 
the eye called the iris. The expansion or contraction 
of the pupil is controlled by the eye muscles in 
response to the brightness of the entering light. The 
beam of light then enters the crystalline lens, L, 
which is actually rather plastic and whose surface 
curvature can be changed by muscle action of the 
eye in order to achieve an appropriate focal length 
for viewing objects at various distances. The in-
terior portion of the eyeball contains a jelly-like 
material, the vitreous humor, V, while the inner 
lining of the back of the eyeball is a light sensitive 
lining called the retina, R. It is made up of millions 
of light receptors (called rods and cones because 
of their shapes) which are arranged normal to the 
retinal surface and which transmit signals through 
the optic nerve, O, when light is incident upon them. 
Thus, for proper functioning, light entering the eye 
should come to a focus on the retinal surface, which 
is spherical in shape. Although the major part of the 
focusing action occurs at the cornea, all of the trans-
parent portions of the eye are appropriately re-
garded as part of the lens system of the eye, as Table 
8-1 of approximate indices of refraction for these 
portions indicates. 
Table 8-1 
the Eye. 
Indices of Refraction for Portions of 
Comea 
1.351 
Aqueous humor 
1.336 
Crystalline lens 
1.437 
Vitreous humor 
1.336 
Figure 8-6 The human eye. 
Because of the adaptability of the eye, objects can 
be viewed distinctly over a range of distances, the 
extremes of which are known as the far point and 
the near point. For a relaxed normal eye, the far 
point is at infinity. Myopia, or nearsightedness, is 
the name given to the case for which the far point is 
at some finite distance, a situation that can be cor-
rected by means of a suitable diverging lens. 
The near point of the eye is determined by the 
flexibility of the crystalline lens and the consequent 
ease with which the eye muscles can increase the 
lens curvature for close distance viewing. This pro-
cess of accommodation, 
as it is called, is a function 
of age in that the crystalline lens loses flexibility 
with age. As a result, the near point is smallest for 
children and increases with age, going from about 
7 cm at the age of ten years to over 40 cm after 50 
years. This condition is called presbyopia, and is a 
normal change in the eye rather than a defect of 
vision. The near point for the normal adult eye for 
many years can be taken to be 25 cm. An eye for 
which the near point is beyond 25 cm during the 
adult years is said to be hypermetropic, 
or far 
sighted, and correction is obtained by use of a 
suitable converging lens. A third common type of 
defect is astigmatism, a condition indicated by dis-
torted images. It occurs when one or more of the 
elements of the eye (such as the comea or crystal-
line lens) are not perfectly spherical. Correction is 
obtained by means of a suitably placed cylindrical 
lens to correct the asynunetry of the eye elements. 

Telescopes 
57 
Example 3. A very nearsighted person cannot 
see clearly objects that are more than 30 cm away. 
What is the focal length of the lens required for this 
person to be able to see distant objects clearly? 
S O L U T I O N 
The required lens must form a virtual image at 
30 cm in front of the eye for an object at infinity. 
Thus, ρ = 00, q = - 30 cm, and Eq. (8-5) yields 
1 - J - 
= 
1 
00 
30 
/ 
or / = - 30 cm. 
8-4 
THE S I M P L E M I C R O S C O P E O R 
M A G N I F I E R 
8-5 
THE C O M P O U N D M I C R O S C O P E 
When one wishes to examine very small objects 
with great magnifying power, it is necessary to use 
a combination of two lenses each of which contri-
butes to the total magnifying power. Such a combi-
nation is called a compound microscope. It is 
shown schematically in Figure 8-7. As the figure in-
dicates, the object to be examined is placed at a 
point slightly greater than the focal length of the 
objective lens, which forms an enlarged or laterally 
magnified real image at a point just within the first 
focal point of the eyepiece or ocular lens.T The 
eyepiece lens then forms an angularly magnified 
image at a distance that can be anywhere between 
the near and far points of the eye. 
When one views an object with the unaided eye, 
the detail that can be seen is dependent upon the 
size of the retinal image, which in turn is dependent 
upon the angle subtended by the object at the eye. 
Thus, the nearer one places the object to the eye, 
the larger will be the angle subtended. However, 
the eye cannot focus sharply on an object closer 
than the near point, so that the maximum angle sub-
tended by a clearly focused object will occur when 
the object is at the near point, which we assume as 
above to be 25 cm. To increase the angle subtended, 
one places a converging lens in front of the eye to 
act as a magnifier or simple microscope. The lens 
thus used forms a virtual image of an object placed 
closer to the eye than the near point, and the eye 
then sees this enlarged virtual image which can be 
seen clearly anywhere between the near point and 
the far point. Most relaxed viewing would occur if 
the image were to appear at infinity (the normal far 
point); we assume this to be the case. 
The magnifying power Μ  of an optical instru-
ment is defined to be the ratio of the angle sub-
tended by the object when placed at the focal point 
of the lens (enlarged virtual image at infinity) θ' to 
the angle subtended by the object at the near point 
(unaided eye) Θ . Thus, 
(8-6) 
where y, the object height, and / are both in cen-
timeters. (Why?) The influence of aberrations for a 
simple double convex lens limits Μ  to about 2 or 3, 
although a corrected lens can have Μ  values as 
high as 20. 
Image 
Objective lens 
Figure 8-7 The compound microscope. 
The total magnification in this case is the product 
of the enlargement lateral magnification mo pro-
duced by the objective and the magnifying power 
Me of the eyepiece. Thus, 
Aftotal = 
ttloMe 
- T ^ X 
all distances in cm. 
Jo 
Je 
(8-7) 
8-6 
T E L E S C O P E S 
The telescope, like the compound microscope, 
consists of an objective and an ocular. However, 
since the telescope is commonly used for examining 
very distant objects, the image from the objective 
will be formed at the second focal point of the objec-
tive. If the final (virtual) image is to be at infinity, the 
first image must be at the first focal point of the 
ocular. As a result, the length of the telescope or, 
tPor brevity we shall use objective and eyepiece—or 
ocular—when discussing the objective lens and the 
eyepiece—or ocular—lens. 

58 
Lenses and Optical Instruments 
more accurately, the distance between objective and 
ocular, is the sum of the focal lengths, /ο  + Λ-
The magnifying power for the telescope is de-
fined as the ratio of the angle subtended by the final 
image (ÖE) to the angle subtended by the object at 
the unaided eye. Figure 8-8 illustrates the situation. 
Thus, the essentially parallel rays from the distant 
object subtend the same angle θο  with the objective 
as they would for the naked eye. The ray from the 
object through the first focal point of the objective 
proceeds parallel to the axis of the telescope, 
through the ocular and its second focal point to the 
eye, which is slightly to the right of this second 
focal point of the ocular. Therefore, it is clear that 
tan θο  = -h 
tan Θ Ε = 7". 
Je 
Because of the great distances involved, the angles 
and their tangents are essentially equal, so that 
' 
θο  
fs' 
(8-8) 
The minus sign in Eq. (8-8) indicates an inverted 
image, which is not a serious disadvantage for non-
terrestrial observations. For terrestrial telescopes, 
an erect image is to be preferred. This can be ac-
complished by the insertion of a third (erecting) 
lens between the objective and the ocular. The 
reader can show that if the erecting lens is located 
at a distance of twice its focal length from the 
image of the objective, and the ocular is located so 
that the image due to the erecting lens is at its first 
focal point, the total magnifying power is un-
changed. What would then be the length of the tele-
scope? 
If the increase in length is undesirable, the erec-
tion can be accomplished by means of a pair of 
45M5°-90** prisms, which totally reflect the light 
four times (from each pair of the inclined faces of 
the prisms) between the objective and the ocular. It 
is left as an exercise to show that the image is erect 
after the reflections described. This method of 
image erection is utilized in the prism binocular. 
8-7 
THE C A M E R A 
The photographic camera is a device for produc-
ing a real image of an object on a light sensitive film 
so that a permanent record of the object photo-
graphed may be obtained by appropriate chemical 
treatment of the exposed film. The camera is similar 
to the eye in many respects. Thus, the lens of the 
camera must properly focus light from the object 
on the film in the rear of the camera, just as the lens 
system of the eye must focus light from an object 
on the retina. Also, most cameras are equipped with 
a diaphragm allowing light to pass through an aper-
ture which is adjustable depending upon light 
brightness, just as the pupil in the iris does for the 
eye. If the camera is to be used for both distant and 
close objects, a movable lens system must be pro-
vided to achieve the accommodation that muscular 
control of the crystalline lens provides in the eye. 
Automatic cameras use a photocell detector to con-
trol the aperture size continuously. 
If the photographic record is to be satisfactory. 
Image at infinity 
eye 
Objective 
lens 
Eyepiece 
lens 
Figure 8-8 The telescope. 

Problems 
59 
the exposure of the film must fall within rather 
well-defined limits. Greater exposure will result in a 
record that is too dark, while underexposure will 
produce a faded, washed-out result. The light-
gathering ability of a camera is determined by the 
open area of the lens and the focal length of the 
lens. That is, the light that is brought to a focus on 
the film is equal to the light from the object times 
the solid angle subtended by the open area of the 
lens at a point on the film. From Figure 8-9, we see 
that the solid angle Ω  is approximately equal to 
π ά  VAf \ Thus, the light gathered by the lens system 
is proportional to (dlff. 
The /-value of the lens is 
written as //(N), where Ν  is the ratio of the lens 
focal length / to the lens opening diameter d. It is a 
measure of the speed of the lens system. The smal-
ler the /-value, the greater is the amount of light 
gathered by the lens system. As a resuh, the expo-
sure time can be reduced compared to a "slower" 
(higher /-value) lens system. The exposure time is 
thus proportional to the square of the lens opening 
diameter for a lens of given focal length. For con-
venience, many camera lens systems have dia-
phragm settings for which exposure times differ by 
factors of two. Thus, the exposure time for //6.3 
compared to //4.5 is given by 
(6.3)^, 39.7 
(Is? 
20.2' l . % « 2 . 
Figure 8-9 Light gathering of a camera lens system. 
Camera lenses commonly available have /-values 
ranging from //1.5 to //16. For the low /-value 
lenses the paraxial rays approximation is no longer 
valid, and the effects of chromatic aberration and 
various other lens aberrations must be removed by 
appropriate combinations of lens materials and lens 
configurations. 
P R O B L E M S 
1. An object is placed a distance 3R from the surface of a glass sphere, radius R and index 1.50. Determine 
the position of the final image. Then repeat the calculation using the (invalid here!) thin lens equation. 
This shows to some extent the errors that improper use of relationships can incur. 
2. An object is located (a) 10 cm, (b) 20 cm from a double convex lens of focal length 15 cm. Find by ray 
tracing and by calculation the position, orientation, and magnification of the image formed in the two 
cases. Tell whether the image is real in each case. 
3. A concave lens forms a virtual image which is 10 cm from the object. The magnification is f. What is the 
focal length of the lens? 
4. (a) Two thin watch glasses, radius of curvature R, are used to form a convex lens that is filled with 
water (n = 1.33). If an object of height RI3 is located a distance R to the left of the lens, describe the 
image formed. Neglect the effect of the glass, 
(b) Suppose now that the water in the lens is emptied and the air lens is immersed in a tank of water. 
Describe the image formed for the same object and location as in (a). 
5. Show that Eq. (7-3) correctly gives the magnification for a thin lens. 
6. A lens produces an image on a screen 25 cm from the object. The image is 4 times larger than the object. 
Find the focal length of the lens. 
7. Calculate the possible focal lengths for a lens of index 1.50 that is to be made using two surfaces with 
radii of curvature of 10 and 15 cm. 

60 
Lenses and Optical Instruments 
8. A convex and a concave lens, with focal lengths which both have of a magnitude of 10 cm, are arranged 
coaxially and separated by 3 cm. Find the distance between an object and its image when the object is on 
the axis and 15 cm from: 
(a) the convex lens, 
(b) the concave lens. 
Note: There are four possible answers. 
9. An object is located 200 cm from a converging lens of focal length 40 cm. A diverging lens of focal 
length - 60 cm is located 20 cm behind the converging lens. Find the position of the final image both by 
calculation and by ray tracing. Is the image real or virtual, erect or inverted? 
10. (a) Where is the near point of eyes for which eye glasses of focal length + 50 cm are required? These 
glasses shift the near point to 25 cm from the eyes, 
(b) What focal length eye glasses are required for eyes with a near point of 25 cm if objects at 25 cm are 
to be seen clearly? 
11. (a) The far point for a myopic eye is at 75 cm. What is the focal length of a lens that will allow distant 
objects to be seen clearly? 
(b) Where is the far point for an eye for which a diverging lens of focal length -100 cm is required for 
viewing distant objects? 
12. A jeweler, using an eyepiece to examine a watch, holds the watch about 5 cm from the lens. What is the 
magnifying power of the eyepiece? 
13. A thin lens of focal length 10 cm is used as a simple magnifier. What magnifying power is obtainable 
with the lens? What is the closest distance that an object may be brought to the eye when using the lens? 
14. A compound microscope has objective and eyepiece lenses of focal lengths 1 and 3 cm, respectively. An 
object placed 1.2 cm from the objective leads to a virtual image produced by the eyepiece that is 25 cm 
from the eye. What is the total magnifying power of the microscope, and what is the separation distance 
between the lenses? 
15. The image from a terrestrial telescope is to be made erect by insertion of an erecting lens as discussed in 
Section 8-6. Show that the total magnifying power is unchanged if the erecting lens is located at a 
distance of twice its focal length from the image from the objective and the ocular is located so that the 
image from the erecting lens is at its first focal point. What is the length of the telescope now? 
16. If a fixed amount of light requires 1/50 second exposure for a camera setting of //4, what will be the 
exposure time for the same light using a setting of //2.8? 
17. A camera lens forms the image of a distant 
point source of light on a screen at A. (See Fig-
ure 8-10.) When the screen is moved backwards 
a distance of 3 cm to B, the image on the screen 
becomes a circle of light with a diameter of 
7.5 cm. What is the /-value of the lens? 
Figure 8-10 

9 
The Nature 
of Light 
9-1 
Introduction 
61 
9-2 
A Particle Theory for Light 
62 
9-3 
A Wave Theory for Light 
63 
9-4 
The Speed of Light and its 
Determination 
63 
9-1 
INTRODUCTION 
In our discussion of optical phenomena, we have 
thus far assumed only rectilinear propagation, since 
this is in apparent agreement with the experimental 
facts presented up to this point. Using this assump-
tion, relationships have been developed making 
quantitative predictions possible for reflection and 
refraction phenomena. This success does not, how-
ever, provide insight as regards the nature of light. 
To progress further in our understanding, we must 
make additional assumptions. Such assumptions 
are known formally as hypotheses or postulates. A 
set of such hypotheses together with any logically 
derived consequences constitute a physical theory. 
Such a theory is then judged on the basis of its 
success in correlating a variety of known phen-
omena and in predicting the outcome of addi-
tional experiments not yet performed. 
It should be stressed that a physical theory is not 
a static system. Thus, if it has been successful in 
explaining a number of situations but fails to ex-
plain still another, 4he usual procedure is to seek 
modifications of the theory to include the additional 
situation rather than to discard the partially suc-
cessful theory in favor of an entirely new one. 
Thus, for example, the laws developed by Newton 
that represent the theory of classical mechanics 
(see Chapter 14 et seq.) are entirely adequate for 
discussing the motion of everyday objects such as 
automobiles, hockey pucks, or even satellites. They 
are not, however, suitable for motion involving mi-
nute objects such as mesons or accelerated elec-
trons that are moving with speeds approaching that 
of light (which is extremely high as we shall see in 
Section 9-4). For the latter category of phenomena, 
the theory of relativity (see Chapter 15) as de-
veloped principally by Einstein is required. This 
situation does not invalidate the classical or Newto-
nian theory for slow moving objects. In fact, for 
speeds that are small compared to that of light, the 
relativistic theory contains the classical theory as a 
limiting or special case. In this sense, relativity 
theory is a generalization of the classical theory 
rather than a total replacement. 
Another situation similar to this is encountered 
when we wish to discuss systems on an atomic or 
subatomic scale, where the "graininess" of the ma-
terial world becomes more apparent. It is perhaps 
not surprising that in this case the classical theory 
must be replaced by a quantum theory to properly 
account for the discrete nature of matter. Quantum 
mechanics provides a satisfactory theory of the ma-
terial world, but the theory embodies many con-
cepts and consequences that are at variance with 
our observations of the everyday (or macroscopic) 
61 

62 
The Nature of Light 
world. Again, however, this quantum mechanical 
formalism leads correctly to the classical (large 
scale) results when it is applied to classical situa-
tions. 
It should be emphasized that the role of a physi-
cal theory is to provide a basis for understanding 
the results of experiments rather than a search for 
the true or correct theory of nature. The successful 
prediction of the outcome of a given experiment is 
a measure, not of the correctness of a theory, but of 
its adequacy. To state that a theory is true or cor-
rect is to imply that it will be successful in predict-
ing outcomes for any and all experiments, and (ex-
cept in a negative sense) this is not subject to com-
plete verification. 
It is in fact frequently the case in physics that 
alternative theories have been proposed to corre-
late or explain the same group of phenomena. In 
such situations, further research and discussion are 
directed toward: (1) finding which of the alternative 
theories is most successful for the widest range of 
phenomena, and (2) finding the modifications of the 
most successful theory (or theories) that extend the 
applicability without unnecessarily increasing the 
number of basic postulates. 
The development of a physical theory is thus a 
process primarily of evolution, involving a continu-
ous interplay between theory and experiment. (We 
remark here that an introduction to the study of 
physics is incomplete if it does not include some 
opportunity to become acquainted with some of the 
equipment and techniques that the experimental 
physicist employs in testing and extending the pre-
dictions of the theoretical physicist.) 
Let us now return to the question of the nature of 
light. 
The 
phenomena 
presented 
thus 
far— 
rectilinear propagation, reflection, refraction, dis-
persion, and the related existence of the color 
spectrum—had all been discovered or studied by 
1670. Other phenomena (to be discussed in Chap-
ters 11 and 12), including diffraction, double refrac-
tion, and the colors of thin films, were known by the 
beginning of the eighteenth century. At roughly the 
same point in time, two theories existed which were 
claimed (by their proponents) to be capable of ex-
plaining the behavior of light. In the following sec-
tions, we shall present an outline of these theories 
and compare them as to their adequacy. In the final 
section, we will present briefly some methods used 
for the determination of the speed of light and the 
values obtained as a result. 
9-2 
A P A R T I C L E T H E O R Y FOR LIGHT 
It is Sir Isaac Newton that we associate with the 
corpuscular theory of light. This theory proposes 
that light consists of a stream of particles or corpus-
cles that move at constant speed in a uniform 
medium, suffer elastic collisions at a reflecting sur-
face, and are attracted by an optically more dense 
substance as they approach it. Because light can be 
reflected and refracted simultaneously, he also 
proposed the existence throughout space of "an 
ethereal medium, much of the same constitution as 
ah- but far rarer, subtiler, and more strongly elas-
tic." This "luminiferous ether" at a boundary be-
tween media is "less pliant and yielding than in 
other places," so that the light particles impinging 
upon the boundary set this ether into vibration in a 
manner analogous to the action of a stone cast into 
a pool of water. The rapid vibrations thus provide 
intervals of easy reflection and intervals of easy 
transmission, which occur at such a high frequency 
as to create the impression of continuous reflection 
and refraction. 
This theory provides an optical analogue to the 
behavior of material particles as described by the 
laws of classical mechanics. Thus, it explains rec-
tilinear propagation in a uniform medium as well as 
reflection and refraction. Snell's law can be derived 
by means of this theory; the relative index of re-
fraction can be shown to be equal to the ratio of the 
speed of the light particles in the second medium to 
that in the first medium. In this way, Newton con-
cluded that the speed of light was greater in an 
optically more dense medium. This conclusion was 
not satisfactorily examined experimentally for over 
one hundred and fifty years, at which time it was 
shown to be incorrect. However, it was the earlier 
failure of this theory to predict either the bending 
of a beam of light about an obstacle (diffraction) or 
the interference effects of two or more beams of 
light that made it obsolete. It was discarded in favor 
of a wave theory when it became clear that no 
satisfactory modification of the particle theory 
could be found that incorporated these phenomena 
discovered after its initial formulation. Thus, it was 
a wave theory of light which survived an unfortu-
nately persistent, often bitter, and intensely per-
sonal controversy extending through the eighteenth 
century, even into the nineteenth century. 

The Speed of Light and its Determination 
63 
9-3 
A W A V E THEORY FOR LIGHT 
The wave theory was developed initially by 
Christian Huygens, a Dutch contemporary of New-
ton. Like the particle theory, it assumes the exis-
tence of an ether in which the light waves propagate 
similar to water waves propagating on the surface 
of a pond. In this theory, the light waves propagate 
according to a rule known as Huygens' principle. 
(See Chapter 10.) Since reflection and refraction 
occur simultaneously for any wave motion, the 
laws of reflection and refraction are explained satis-
factorily. In addition, one can conclude that light 
travels more slowly in an optically more dense 
medium, which proves to be the case as implied 
above in Section 9-2. More important is the fact 
that this theory is completely successful in also pre-
dicting interference and diffraction effects. As we 
will see in Chapter 10, however, Huygens' principle 
would not seem to predict rectilinear propagation 
of light. This was in fact the principal reason New-
ton advocated the particle theory, since non-
rectilinear propagation of light is not readily 
observed in experiments involving reflection and 
refraction. 
In subsequent chapters, we shall discuss the 
properties of waves. As a result, it will be clear that 
the wave theory of light merits adoption as an ade-
quate theory regarding the nature of light. 
9-4 
THE S P E E D O F LIGHT A N D ITS 
D E T E R M I N A T I O N 
When one attempts to determine the speed of an 
object, two quantities must in some way be mea-
sured; the distance traveled by the object and the 
time elapsed while the movement takes place. 
Racing cars or aircraft seeking to establish new 
speed records must traverse a carefully measured 
course of specified length, and are timed with 
highly sophisticated precision timing instruments. 
In principle, the determination of the speed of 
light would be expected to be equally straight-
forward. Thus, Galileo and an assistant stood on 
separate hills in Rome, a known distance apart, 
equipped with shuttered lanterns. The experiment 
was quite simple: Galileo opened the shutter of his 
lantern allowing light to proceed to the assistant who 
opened his shutter as soon as he saw the signal from 
the first lantern. Galileo determined the length of 
time between opening his shutter and detecting the 
assistant's signal. This should have provided the 
necessary data were it not for the extremely high 
speed of light. Thus, the crudeness of the timing ap-
paratus, together with reaction time limitations, li-
mited the results to a conclusion that the speed of 
light was either instantaneous or at least immeasura-
bly great. It is this great speed that makes it neces-
sary to resort to light paths of astronomical mag-
nitude if one wishes to measure time intervals di-
rectly. Any other technique necessarily will be less 
direct if accuracy is desired. 
The earliest astronomical method of measure-
ment was explained in 1676 by Olaf Rö mer, a Dan-
ish astronomer, who was then at Paris making 
measurements of the times of revolution of the 
satellites of the planet Jupiter. The orbits of these 
satellites lie in essentially the same plane as the or-
bits of Jupiter and the Earth about the Sun. Conse-
quently, revolution times can be determined by 
measuring the time interval between successive 
emergences of the satellites from behind the 
planet's disk. 
From a lengthy series of observations of the in-
nermost satellite, Rö mer found that the time of 
revolution was more or less than the average result 
when the Earth was receding from or approaching 
Jupiter. He correctly concluded that the differences 
were due to the variable distance from the Earth to 
Jupiter, and the finite time required by the light re-
flected from the satellite to traverse this distance. 
The time required for Jupiter to make one revolu-
tion about the Sun is about 12 times greater than the 
time (1 year) required by the Earth. Thus, while the 
Earth moves through an appreciable portion of its 
orbit, Jupiter is essentially at rest, as illustrated in 
Figure 9-1, which shows the two planets at succes-
sively numbered corresponding positions. Any 
change in the time duration of eclipses will be due 
to a change in the Earth-Jupiter separation dis-
tance. In particular, the differences in eclipse 
periods occurring when the Earth is at positions 1 
and 3 will be due to a change in separation distance 
that is satisfactorily approximated by the diameter 
of the Earth's orbit about the Sun. The speed of 
light is then found by dividing the orbital diameter 
of the Earth by the difference in eclipse times men-
tioned above. In Rö mer's time, the Earth's orbital 
diameter was not accurately known, and the time 
difference measured by him was about 22 minutes 
(compared to the present value of nearly 17 min-
utes). Thus, the speed derivable from his work was 

64 
The Nature of Light 
2 
·— 
^ 2 
Satellite 
Μ  
•— 
W 
Sun 
— 
4 
Earth 
^^Jupiter 
Figure 9-1 Rö mer's method for the speed of light (schematic). 
of interest mainly because it established that it was 
indeed finite, though extremely large (about 2x 
lO^m/sec). Another astronomical determination 
made by James Bradley some 50 years later, in-
volving an independent phenomenon (stellar aber-
ration), resulted in a value of 3 x 10^ m/sec, thus 
confirming the order of magnitude indicated by 
Rö mer. 
Successful terrestrial measurements of the speed 
of light were first performed in the mid-nineteenth 
century by the French physicists Hippolyte Fizeau 
and Jean Foucault, contemporaries who collabo-
rated for a time using a technique devised by 
Fizeau and improved and modified in later indepen-
dent work by Foucault. In the Fizeau experiment, a 
properly focused beam of light directed toward a 
distant mirror passes through a toothed wheel. 
When the wheel is not rotating, the light reflected 
from the distant mirror passes through the same 
notch in the wheel and continues to the observer 
(see Figure 9-2). Now if the wheel is set in motion at 
a rotational speed that is slowly increased, the light 
will first be observed when the reflected light ar-
rives back at the toothed wheel, just as the second 
Light source 
Observer 
Toothed wheel 
Distant 
mirror 
Figure 9-2 Fizeau's toothed wheel experiment (schematic). 

Problems 
65 
notch has arrived at the position of the initial notch 
for the original passage of the light. At rotational 
speeds such that the third, fourth, etc. notch can 
rotate into position while the beam of light travels 
to the distant mirror and back, the light will again be 
visible to the observer. 
Thus, if one can accurately measure the rate of 
revolution of a wheel with equally spaced notches, 
a series of rotation speeds that are integer multiples 
of a lowest speed is obtained. The speed of light is 
then found from the relation 
ations beginning in 1931. The result of his efforts was 
a value for the speed of light which is known to be 
accurate to within 1 part in 10^ 
Other techniques that are based upon the fact 
that light waves are electromagnetic in nature (as 
are radar and radio waves—see Chapter 36) have 
been developed and refined over the past 50 years.t 
As a result of these efforts it was recently reportedt 
that the speed of light propagating in a vacuum (this 
speed is designated by the letter c) has the value 
c =299,792.5 ±0.3 km/sec. 
2 X wheel to mirror distance 
speed of light = number of notches which shifts to initial position 1 ^• 
\/revolutions of wheel 
V 
total number of notches on wheel 
seconds 
(9-1) 
In Fizeau's experiment, the wheel had 720 teeth or 
notches, the distant mirror was 8.63 km away, and 
the slowest rotation rate for letting light pass was 
found to be 25.2 rev/sec, leading to a result of 3.13 x 
10* m/sec. Foucault later replaced the rotating wheel 
by a rotating plane mirror, a modification which 
made possible the use of laboratory distances, and 
obtained a result of 3.01 x 10* m/sec in air, (See 
Problem 9-3.) He also determined by this method 
that the speed of light in water was less than this 
value, as predicted by the wave theory. The Ameri-
can physicist Albert Michelson refined the experi-
ments of Fizeau and Foucault in a series of investig-
In subsequent discussions, we will use the approxi-
mation (satisfactory for slide rule calculations) that 
c = 3 X 10* m/sec, 
a figure in error by less than 1%. 
tSee "The Speed of Light," Scientific American, 
August 1955, p. 62. 
^Report to the Commission on Nuclidic Masses and Re-
lated Atomic Constants of the I.U.P.A.P. by E. R. Cohen 
and J. W. M. DuMond, June 24, 1%3. 
P R O B L E M S 
1. Suppose that the lanterns used for signals by Galileo and his assistant were 6 km apart. If the error in this 
distance were negligible and the speed of light was to be accurate to within 10% of the accepted value, 
what is the maximum error permissible in the determination of the time between the opening of the two 
lantern shutters? 
2. (a) The Crab Nebula is a luminous remnant of a supernova (exploding star) event observed by Chinese 
astronomers in A.D. 1054. Assuming the exploding star was originally a sphere of negligible diameter 
and that it has been expanding uniformly at 10^ km/sec since the event was observed, estimate its 
diameter today. 
(b) The light year is the distance that light traveling in a vacuum could cover in one year. Find the 
distance in (a) in light years. 

66 
The Nature of Light 
3. In 1862, Foucault employed a rotating plane mir-
ror instead of a toothed wheel in determining the 
speed of light. With the speed of rotation at 
800 rev/sec, a ray of light that is reflected from 
the rotating mirror to travel 15 m to a distant 
plane mirror and be reflected back to the rotating 
mirror again encounters the altered position of 
the rotating mirror shown dashed in Figure 9-3. 
This results in an angular displacement of 3.45 
minutes of arc for the returning ray. Find the 
speed of light from this data. Hint: A reflected 
ray of light experiences an angular displacement 
2Θ  when the plane reflecting surface experiences 
an angular displacement Θ . 
15 meters 
Ray 
S' 
displacement 
Source S 
Figure 9-3 
4. The Sun is on the average 1.49 x 10" m from the Earth. How long does light take to travel this distance? 
5. In 1925, Michelson modified the rotating mirror method by using an 8-sided mirror that could be rotated 
at high speed. Light from a source was reflected from one face to a distant plane mirror, then reflected 
back to another face, and from there reflected to a telescope for observation. (See Figure 9-4.) 
A series of light pulses from the source hit face 1 
of the rotating mirror. If the rotation speed of the 
mirror is adjusted so that face 2 has moved to 
replace face 3 while the light was going down to 
and back from the distant mirror, then the light 
pulses will appear in the telescope. If the distant 
mirror was 37.46 km away and the rotational 
speed of the mirror was 500 rev/sec, what value 
was obtained for the speed of light? 
Figure 9-4 
Source-
1 2 3 
Viewing 
telescope 
Distant 
mirror 

10 
Waves and 
Wave Motion 
10-1 
The Nature of a Wave 
67 
10-4 
Reflection and Refraction of 
10-2 
A Mathematical Description of a 
Water Waves 
71 
Wave 
67 
10-5 
Huygens' Principle and Diffraction 
10-3 Superposition of Waves 
68 
of Waves 
73 
10-1 
THE N A T U R E OF A W A V E 
We present here a discussion of the properties or 
characteristics of waves or wave motion, using ex-
amples taken from our everyday experience. This 
will help us evaluate the assertion of the previous 
chapter that light can be satisfactorily understood 
as a wave phenomenon. In addition, the concepts 
developed here will be useful in studying other ex-
amples of wave phenomena in later chapters. 
Common examples of wave motion are easy to 
find: the ripples on the surface of a lake or pond 
produced by wind action or by casting stones into 
them, pulses traveling along a long string (or along 
helical springs such as the Slinky toy) fastened at 
one end, and the sound pulsations detected by the 
ear as the result of the ringing of a large bell. While 
these examples differ in detail, they all have certain 
features in common. In each case, there is a 
medium (the water, string or spring, or the air, re-
spectively) through which a disturbance (the water 
ripples, the pulses, or the sound pulsations) travels. 
While the disturbance is passing a given point in the 
medium, the particles of the medium in the neigh-
borhood of the point execute small displacements 
about their original undisturbed positions. After the 
disturbance has passed, however, the medium re-
turns to its undisturbed state. Thus, no net displace-
ment or alteration of the medium results from the 
passage or transmission of a propagating wave. Be-
cause of this fact, it is possible to direct our atten-
tion to a mathematical description of a general 
propagating wave without having to consider in de-
tail its specific physical nature or the medium in 
which it propagates. 
Note carefully that this is not equivalent to say-
ing all wave phenomena are similar in detail. In 
particular, we mention here two distinct types of 
wave motion (to which we shall return in Chapter 
13)—namely, longitudinal waves and transverse 
waves. A longitudinal wave is one in which the mo-
tion of the particles of the medium is parallel to the 
direction of propagation of the wave, such as a 
sound wave traveling along an air colunm. A trans-
verse wave is one in which the particles of the 
medium execute motion in directions perpendicular 
to the direction of propagation of the wave, for ex-
ample, the vibrating string along which a wave 
propagates. 
10-2 
A M A T H E M A T I C A L D E S C R I P T I O N OF 
A W A V E 
In order to avoid mathematical complications, we 
assume a periodic (regularly repeating) disturbance 
of the medium (for visualization, consider a surface 
67 

68 
Waves and Wave Motion 
wave on water). Specifically, we assume that a side 
view of the surface wave can be described 
mathematically by the relation 
y = Λ  sin θ. 
(10-1) 
Here θ is the phase angle of the wave and depends 
generally upon both the time of observation of the 
wave and the location in the medium. For a given 
phase angle, y represents the actual displacement 
of the water surface from the undisturbed condi-
tion, and A, the amplitude of the disturbance, rep-
resents the maximum displacement from 
the 
equilibrium or undisturbed condition. A real water 
wave would be a more complicated function of the 
variable θ than the sine function, but it is a remark-
able fact (discovered by the French physicist 
Joseph Fourier) that the true wave form can be 
obtained by the appropriate addition of a series of 
different sine and cosine terms of the type indicated 
in Eq. (10-1). We therefore concentrate on this sim-
ple wave, since it can be used as the basis for de-
scribing more complex cases. 
Figure 10-1 is a graph of the assumed wave form 
as a function of the angle θ in radians. The distance 
between successive identical displacements is call-
ed the wavelength 
and is designated by λ. Suppose 
that the wave is traveling to the right with a speed v, 
and that it takes a time interval τ , called the períod 
of the wave, for one complete wavelength of the 
wave to pass a given point. This means that the 
medium at that point will have experienced, during 
a time interval τ , all possible values of the displace-
ment that the wave can cause. Then the speed of 
propagation ν  is related to τ  and λ by the equation 
That is. 
ν =λΙτ . 
(10-2) 
,(^/sec) = A ( í = ^ ) / r ( 5 ^ ^ ) . 
ν  wave / / 
ν  wave / 
If we define the number of complete wavelengths 
passing a given point per unit time as the 
frequency 
/, it is clear that it is necessarily equal to IIτ . There-
fore, Eq. (10-2) may be rewritten as 
V =λ/. 
(10-3) 
The speed of propagation depends upon the physi-
cal properties of the medium. 
The displacement y of a given particle of the 
water surface will be a function of both the time of 
observation and the location in a direction along the 
surface x. That is, y = y(x,t). 
Let us now choose an 
origin for the χ  coordinate by requiring that y = 0 
for jc = 0 and t = 0. From Figure 10-1 it can be seen 
that when χ  = λ/2, y = O and when JC = λ, y = 0. 
This suggests that we write 0(JC,O) = 27Γ χ /λ; the 
desired sinusoidal wave form is thereby attained. 
Thus, a sinusoidal wave at ί = 0 will have the equa-
tion 
y(x,0) = A s i n ( ^ ) . 
(10-4) 
Now, at a later time i, the wave form will have 
traveled to the right a distance vt. The particle dis-
placement at an arbitrary point χ  at time t will 
necessarily be the same as that for a particle 
located at the same position JC' at ί = 0 if χ  is 
related to χ ' by the equation 
x' = X - vt. 
Therefore, 
so that 
y(jc',0) = A s i n ^ = y ( j c , i ) , 
y{x, t) = A sin η ^ ( χ  - 
vt) 
A 
= A s i n 2 7 r ( ^ - ^ ) 
= A s i n 2 . ( f - l ) 
(10-5) 
Figure 10-1 Sinusoidal wave form. 
is the equation for a transverse wave traveling to 
the right. The angle 0(x, ί) = 27τ /λ (χ  - vt) is, as 
noted after Eq. (10-1), called the phase angle of the 
wave. For a sinusoidal wave traveling to the left, a 
similar analysis yields the same expressions with ν  
replaced by (- v). 
10-3 
S U P E R P O S I T I O N 
O F W A V E S 
Let us now consider the consequences of sub-
jecting the same medium to two (or more) sinusoi-

Superposition of Waves 
69 
dal traveling waves at the same time. Experiments 
for many types of wave phenomena (but not all) 
indicate that for small wave amplitudes the net dis-
turbance of the medium at any point is obtained 
simply by adding algebraically the individual dis-
turbances at the point. This useful principle, which 
in any situation should be subject to experimental 
verification, is known as the principle of linear 
superposition. Assuming that linear superposition is 
valid, we can write the following expression for the 
net disturbance at a point χ  in the medium at a time t 
by two different sinusoidal traveling waves 
ynet = yi + y2 
= A, sin27r^^-/,i^ + A 2 sin27r (£-4 
(10-6) 
where individual disturbances have wavelengths λ, 
and frequencies 
For Ν  separate waves. 
ynet = ^ A n 
SLU Θ „ . 
where 
e.=2^(^f-fj). 
(10-7) 
Consider as an example two waves of identical 
frequency (/. = Λ  = / ) . Because the speed of prop-
agation depends on the medium r, = i?2 = u, so λ, = 
= A. For the first wave, Eq. (10-7) yields 
2 7 Γ 
y, = A, sin Ö, = A, sin-^(jc - vt). 
Let us also assume that the expression for y2 can be 
replaced by 
y2 = A2 sin ^^(x-vt)^φ  
(10-8) 
In Eq. (10-8) the quantity φ  is known as the phase 
angle difference between the two waves at the point 
JC at time i, or, in terms of the phase angles, φ  = 
02-Su 
Substitution of Eq. (10-8) in Eq. (10-6) to-
gether with A, = A2 = A and Vx = V2=v 
yields 
ynet = A, sin ^ ( j c 
- vt) 
+ A 2 
sin 
A 
γ(χ -ν ί) 
+ φ  
(10-9) 
Suppose now that φ  is an integer multiple of 2Τ Γ. 
Since sin (a ±2ττη ) = sin a for integer values of n, 
we obtain 
ynet =(A. + A 2 ) sin γ 
(JC - vt), 
(10-10) 
and the amplitudes add in magnitude. The waves in 
this case are said to be in phase and 
constructive 
interference results. On the other hand, if the phase 
difference is an odd integer multiple of π , then 
sin [Θ  ±(2n - \)TT] = - sin Ö; we obtain 
ynet = (A. - A 2 ) sin γ { χ - vt), 
(10-11) 
and the amplitudes combine subtractively in mag-
nitude. In this case, the waves are said to be out of 
phase and destructive interference results, produc-
ing a reduced amplitude. If A, = A 2 for this case, 
there is a complete cancellation and no net disturb-
ance results. For values of the phase differences 
other than the cases considered, Eq. (10-9) will 
yield a result intermediate between the constructive 
and the destructive limiting cases. 
At this point we might well ask what factors de-
termine the phase angle difference φ  for two given 
sinusoidal waves observed at (JC, t). To answer this 
question, we must consider the sources that pro-
duce the two waves. If the sources are said to be in 
phase, this means that they emit the wave distur-
bances simultaneously or in synchronization. On 
the other hand, if they are out of phase, we under-
stand this to mean that there is a delay of one half a 
cycle (or wavelength) between the emission of a 
wave by one source and the subsequent emission of 
a wave by the second. Suppose the two sources are 
located at the same point. Then the net disturbance 
produced by them at any other point will be the sum 
of the amplitudes of the two disturbances for 
sources in phase. It will be the difference between 
the amplitudes of the two disturbances if they are 
out of phase, since in this case the crest of one 
wave will occur at the same time and place as a 
trough of a wave from the other wave, which is 
either a half cycle ahead of or behind the first. More 
generally, one can specify this phase relationship 
between the two sources by specifying the lag angle 
δ  by which the wave disturbance from the one 
source lags behind the other. Then the expression 
in phase means δ  = 0, while out of phase means 
δ  = 7Γ . 
A second factor producing a phase angle differ-
ence between two waves is related to the locations 
of the two sources relative to the point at which we 
observe the net disturbance. To see this, consider 
first two sources that are in phase (δ  = 0), that are 
respectively at distances JC, and X2 from the point in 
space. For identical sources (equal wavelengths A 

70 
Waves and Wave Motion 
and frequencies / ) , one would expect an addition of 
amplitudes to occur if the difference in distances 
from the sources to the point in space is equal to an 
integer number of wavelengths. That is, if the 
difference in path length traversed by the waves 
can be written as 
AJC = JC2-JC, = η λ, η  = 0 , 1 , 2 , . . . , 
(10-12) 
we can expect an addition of amplitudes. On the 
other hand, when 
Δ χ =(2η  + 1)^, η  = 0 , 1 , 2 , . . , , 
(10-13) 
the amplitudes will subtract. 
Conversely, when δ  = π  (sources out of phase), 
the wave amplitudes will add when 
Ajc=(2n + 1)^, 
and subtract when 
Δ χ  = η λ, η  =0,1,2, 
(10-14) 
(10-15) 
(Why?) 
We can express the phase angle difference that is 
due to this difference in path length by the quantity 
Ιττά χ Ιλ. 
The total phase angle difference between 
the two sources discussed in the examples above is 
then given by the relation 
ψ = ^ + δ . 
(10-16) 
It is not difficult to see that the effect produced by a 
given path length difference can be cancelled out by 
an appropriate choice of lag angle δ , and con-
versely. In any case, then the interference of two or 
more sinusoidal waves at a point in space will de-
pend upon both the synchronization of the sources 
(lag angle δ ) and the difference in path length from 
the sources to the point in question Ιπ Αχ Ιλ. 
When 
these factors are known, the nature of the interfer-
ence can be predicted. Conversely, if the interfer-
ence is observed and the location of the sources is 
known, one can deduce the value of the lag angle δ . 
As a further example, consider two waves of 
equal amplitude but slightly different frequencies. 
If they are impressed simultaneously on the same 
medium, we can say that at some point (for con-
venience let us choose χ  =0) the total amplitude is 
ynet = A [sin lirfa + sin 27r/2Í]. 
(10-17) 
It is useful here to employ the identity 
sin α + sin /3 = 2 sin 
cos (^"2^) 
This yields 
y„et = 2A 
sin 2π  
cos 
(10-18) 
We can interpret this as a wave disturbance whose 
frequency is the mean of the two separate frequen-
cies and whose amplitude varies with time as 
2ACOS 2 7 r ( í ~ ^ ) í . 
Thus, there will be a maximum 
disturbance 
whenever the cosine term takes the values ±1. 
There will, therefore, be two such maxima per 
cycle of amplitude variation. Such maxima are 
known as beats, and the beat frequency is thus 
twice the frequency of amplitude variation. 
beat frequency = 2 L 2 J 
= / . - / . 
(10-19) 
For audible sound waves, the unaided ear can 
discern beat frequencies up to approximately 
10 cycles/sec. 
When two waves of equal amplitude and fre-
quency travel in opposite dh-ections in a medium, 
the resulting disturbance is called a standing wave. 
Mathematically stated. 
ynet = A sm 
+ sin 2π  
(10-20) 
Since sin (a ± β ) = sin α cos j3 ± cos a sin j3, the 
expression can be written 
y„et = [2A cos 2nft] sin 
(10-21) 
A 
We see that, for all values of the time, the net dis-
turbance is zero for χ  = η λ/2, where η  is any in-
teger. For all other values of JC, the amplitude varies 
sinusoidally with the time through the factor in 
square brackets. 
Consider as an example the case of a stretched 
string of length L with fixed ends. When the ends 
are fixed, yn« must vanish at χ  = 0 and at χ  = L, for 
any value of i. Equation (10-21) then shows that the 
only wavelengths allowed are those for which the 
equation 
λ„ = —, 
η  any integer, 
(10-22) 
η  

Reflection and Refraction of Water Waves 
71 
is satisfied. The corresponding frequencies are 
given by the equation 
. 
riv 
r 
(10-23) 
where /, = vl2L 
is called the fundamental fre-
quency or first harmonic, while the nth frequency 
is called the nth harmonic. Another type of nota-
tion refers to higher frequencies as overtones of the 
fundamental. In this notation, the second harmonic 
is the first overtone, etc. The speed of propagation ν  
arises through Eq. (10-3), ν  = /λ. 
10-4 
R E F L E C T I O N A N D R E F R A C T I O N 
OF WATER 
W A V E S 
We turn now to a physical discussion of what 
happens when waves encounter complete or par-
tial barriers or pass from one medium into another. 
For convenience of discussion, we choose water 
waves as observed by means of a ripple tank, a 
glass bottom tray of water illuminated from above. 
When a pattern of waves is generated in the tray, 
the crests of the waves tend to focus the incident 
light, while the troughs diverge it. As a result, a 
viewing screen placed below the tray will exhibit an 
alternating array of bright and dark bands, which 
are the images of the crests and troughs of the 
waves being studied. Provision must be made for 
the absorption of the waves at the edges of the tray 
if secondary influences are to be avoided. We as-
sume that this has been done for our discussion.t 
As an initial example, consider a plane wave 
train. Such a pattern of disturbance can be gener-
ated by periodically dipping a straight edge into the 
water, producing the result shown schematically in 
Figure 10-2. If a second plane wave train is de-
veloped simultaneously at right angles to the first, 
the resultant pattern will appear as indicated in Fig-
ure 10-3. This is what one would predict by means 
of the supeφ osition principle discussed in the pre-
vious section. 
Now consider the effect on a single plane wave 
train produced by placing a solid barrier diagonally 
in its path as in Figure 10-4(a). Experiments with 
ripple tanks show that the waves are reflected as 
tAn excellent article on the construction and use of 
such a tank appears in the Amateur Scientist section of 
the October 1%2 issue of Scientific American. 
Figure 10-2 Ripple tank plane wave (schematic). 
Figure 10-3 Ripple tank superposition of perpendicu-
lar plane waves (schematic). 
shown in Figure 10-4(b). The angles 0¡ and 0; are 
observed to be equal. As a result, the angles Ö, and 
or, which are measured with respect to the normal 
NN' to the barrier, are also equal. This is of course 
the law of reflection introduced in Chapter 6. 
To illustrate refraction with a ripple tank, it is 
necessary that there be two or more regions of the 
water in the tray in which the speed of propagation 
of the water waves assumes different values. This is 
accomplished by using the fact that for shallow 
water depths the speed of propagation varies in 
proportion to the square root of the depth. Thus, if 
one has a tank filled with water at two different 
depths, the propagation of water waves in this tank 
will be analogous to the propagation of light in a 
system involving two different media. In Figure 
10-5(a), we see the effect of a varying depth on a 
plane water wave propagating in a direction parallel 
to that of decreasing depth. Since the frequency of 
occurrence of wave crests is constant, and the 
speed of propagation decreases in shallower water. 

72 
Waves and Wave Motion 
Figure 10-4 Reflected waves in a ripple tank diagonal barrier (schematic). 
the wavelength must necessarily decrease in the di-
rection of propagation, as the figure indicates. 
In Figure 10-5(b), the plane wave direction of 
propagation is at an angle with respect to the re-
gions of different depth. It is clear that the direction 
of propagation is altered as the wave proceeds into 
the second "medium," so that refraction of the 
water waves occurs. By constructions similar to 
that in Figure 10-4(b), we can establish that Snell's 
law applies here as it does for light. Figure 10-6 rep-
resents successive wave crests in the vicinity of the 
interface with medium 1, the medium of greatest 
speed of propagation. Notice that the leading crest 
consists of two segments which, because they are 
in different media, propagate at different speeds. 
Since the plane wave crests are lines of constant 
phase angle, the time required for the second wave 
crest to travel to the location of the first is a 
constant independent of position along the wave 
crest. Thus, the time of propagation for a segment 
of the wave crest to proceed from Ρ  to O is identi-
cal to that for a segment proceeding from O' to P'. 
Let the speed of propagation in medium 1 be ui 
and that in medium 2 be V2. The time to traverse the 
; Shallow 
- | V2>V, 
Deep 
(a) 
Shallow 
Deep 
(b) 
Figure 10-5 (a) Effect of varying depth on wavelength, (b) Refraction of water waves in ripple tank due to varying 
depth. 

Huygens' Principle and Diffraction of Waves 
73 
/ 
Medium 1 
Medium 2 
\f2 
Figure 10-6 Wave crest refraction at a boundary between two media. 
distances OP and O'P' can then be written 
OP 
O P ' 
t =-
But from Figure 10-6, 
OP ^ 00' sin 
O'P' 
OO'sino/ 
(10-24) 
(10-25) 
Substituting Eq. (10-25) into Eq. (10-24), and re-
arranging, we obtain 
sm Oi 
Vi 
— 
7- = 
= M2 
Sin dr 
V2 
(10-26) 
We therefore obtain not only Snell's law but find 
in addition that the index of refraction is given by 
the ratio of the speeds of propagation in the two 
media. By analogy, we can infer from these wave 
theory arguments that if medium 1 were air, 
medium 2 water, and the wave disturbance were 
light waves, then the speed of propagation of light 
in water would be less than that in air since η  HJO air ^ 
4/3. This assertion was verified by Foucault in the 
mid-nineteenth century using a modification of the 
Fizeau toothed-wheel apparatus discussed in Chap-
ter 9. This is the reverse of the particle theory pre-
diction for light, and is therefore one of the strong 
points favoring the wave theory. 
10-5 
H U Y G E N S ' P R I N C I P L E A N D 
DIFFRACTION OF W A V E S 
In his study of the wave theory of light, Huygens 
developed a method for constructing by geometric 
means the behavior of a given wave disturbance for 
which the initial shape of a wave front is known and 
for which the variation of the speed of propagation 
with position is known. The principle upon which 
the method is based, known as Huygens' principle, 
can be expressed quite simply. The principle states 
that each point on the initial wave front can be 
regarded 
as a source of 
spherical 
wavelets 
(propagating radially in the forward directions) 
which travel with the speed of propagation of the 
wave disturbance in the medium. The envelope of 
these wavelet wave fronts at any subsequent time 
represents the wave front resulting from the initial 
wave front at the later time. In situations for which 
the speed of propagation varies with position, it is 
important that successive time intervals be chosen 
for which the propagation speed does not vary ap-
preciably in the space interval between wavelet en-
velopes. Figure 10-7 illustrates the application of 
Figure 10-7 Application of Huygens' principle for a 
propagating wave. 

74 
Waves and Wave Motion 
Huygens' principle for a wave front of arbitrary 
shape propagating in a uniform medium. 
If we apply Huygens' principle to the case of 
plane water waves striking a barrier with an aper-
ture, one finds (for wavelengths comparable to the 
width of the aperture) that the plane wave fronts 
exhibit a bending around into the "shadow" of the 
aperture, a situation described as diffraction in the 
case of light waves. This result, in contrast to the 
observed rectilinear propagation of light through 
slits with widths of 1 mm or more; was an apparent 
defect of the wave theory that led Newton to con-
sider it an untenable model for light. To resolve this 
difficulty, consider the series of ripple tank photo-
graphs in Figure 10-8(a-c). In this series, the aper-
ture width is held constant while the wavelength of 
the wave fronts is reduced from a value six-tenths 
that of the aperture width d [Fig. 10-8(a)] to a value 
three-tenths of d [Fig. 10-8(b)] to the last case for 
which the wavelength is one-tenth d [Fig. 10-8(c)]. 
It is clear that for wavelengths that are small com-
pared to the aperture width the diffraction effects 
become negligible. 
The resolution of our puzzle in the case of light, 
therefore, requires the assumption that the wave-
lengths of light are small compared to the dimen-
sions of 1 mm or more. For such a situation, one 
would not expect to observe optical diffraction 
effects unless the apertures were reduced sharply in 
width. In the next two chapters, we will consider in 
more detail the interference and diffraction of light 
waves. In Chapter 36, it is indicated that visible 
light represents a rather narrow band of wave-
lengths in the spectrum of electromagnetic waves 
extending from below gamma rays of very short 
(— 10"'° m) wavelength to beyond radio waves of 
long wavelength (~ 10^ m). In this spectrum, visible 
light possesses wavelengths ranging from 
~ 7 x 
10"' m (red) to ~ 4 x 10"' m (violet). Accepting for 
now the accuracy of this assertion, it is easy to 
understand how apertures of ordinary dimensions 
failed to exhibit diffraction effects. 
Figure 10-8 From PSSC Ptiysics. With permission 
from D.C. Heath & Company, Lexington, Mass., 1965 
and Education Development Center, Newton, Mass. 

Problems 
75 
P R O B L E M S 
1. A sawtooth wave form is in effect a linearized sine curve. (See Figure 10-9.) Using methods developed 
by Fourier, this wave form can be expressed as an infinite series of sine terms as follows: 
4 V 
TT n=0 
( - i r sin(2n + l ) g _ 4 
(2n + 1)^ 
sin 0 - p s i n 3 ö  + jjsinSÖ 
To appreciate how few terms of the series 
are needed for a reasonably good fit, plot 
(a) the first term, 
(b) the sum of the first two terms, and 
(c) the sum of the first three terms of the series 
for values of θ from 0 to 2 π  in steps of 7γ/4. 
Figure 10-9 
2. A sinusoidal water wave has an amplitude of 20 cm and a wavelength of 35 cm. If a photograph of the 
wave profile is made, what will be the distance between a given crest and a point farther along the wave 
where the phase angle has increased by 745°? 
3. An organ pipe emits a sound wave with a frequency of 20 cycles/sec. The velocity of sound is 344 m/sec. 
If the pipe is i of a wavelength long, what is its length in meters? 
4. A sound wave traveling in a given medium is described by the relation 
y = 10' sin 27γ(1.6Χ -4800, 
where y represents the variation in density of the medium (expressed in gm/cm') as the wave passes. 
What are the amplitude, frequency, wavelength, and speed of propagation of the wave? Time is in 
seconds, distance is in meters. 
5. A wave traveling to the left along a string has an amplitude of 2 cm, a wavelength of 5 m, and a speed of 
propagation of 20 m/sec. Write the equation of the wave, and determine the frequency of the wave. 
6. Find the amplitude of a wave made by superposing two waves of the same frequency and wavelength, 
traveling in the same direction. The first wave has an amplitude A. The second has an amplitude 2A, and 
lags the first by δ  = 7γ/3. 
7. Show that Eq. (10-9) can be written in the form 
ynet = A net Sltt 
where 
Anet = [ A , ' 4- 2A,A2 COS φ  + 
Α 2Ύ  
A , + A2Cos^ 
Then show that Eqs. (10-10) and (10-11) follow readily from this result. 

76 
Waves and Wave Motion 
8. Two sinusoidal waves of equal amplitude, frequency, and wavelength are traveling in the same direction 
in a medium but have a phase difference of π  12. Find an expression for the disturbance assuming linear 
superposition is valid. 
9. Two tuning forks with the same frequency are sounded simultaneously (in phase and equal amplitude). 
The frequency of the forks is 112 cycles/sec and the wavelength of the waves is 10 ft. 
(a) What will be the phase angle difference between the waves at a point 40 ft from one fork and 45 ft 
from the other? 
(b) What will be the sound amplitude at this point? 
(c) What will be the phase angle difference at a point equidistant from the two forks? 
(d) What will be the phase angle difference at a point 2.5 ft closer to one fork than the other? 
10. An oscillator generates waves of 4 m. A second oscillator is 10 m north of the first oscillator. It generates 
waves of the same wavelength and amplitude as the first, but it is 180° out of phase with the first. 
Measuring east from the first oscillator, find the locations where the resultant sound amplitude is zero. 
11. A vibrating tuning fork produces 11 beats/sec when placed near a standard fork of frequency 
512 cycles/sec. What is the frequency of the first fork? 
12. A stretched string 5 m in length has fixed ends. If the speed of propagation of a wave on this string is 
250 m/sec, find: 
(a) the fundamental frequency, 
(b) the frequencies for the first 3 overtones that exhibit nodes (zero disturbance) at the midpoint of the 
string, and 
(c) the beat frequency that would result if the 3rd harmonic frequency were compared to a tuning fork 
with a frequency of 768 cycles/sec. 
13. Suppose waves are excited on a string that has been formed into a circle of radius r. If the wavelength of 
the waves is λ, find the relation that must be satisfied if standing waves are to be formed on the string. 
14. The speed of propagation of waves on the surface of a certain tank of liquid is given by 
V = 1.5+Ir', 
where ν  is in m/sec and r is the perpendicular distance (in meters) along the surface from a reference 
line of the surface. Using a sheet of graph paper, use Huygens' principle to trace successive wave fronts 
for a plane wave front that is initially perpendicular to the reference line. 
15. Above a heated surface (such as a road bed or a desert heated by the sun), the speed of propagation of 
light is greater than in cooler air. Assume that this variation in propagation speed with distance from the 
surface is gradual rather than abrupt. Use this information to explain the mirage. This is a phenomenon 
in which light rays from an extended object travel to the observer directly from the upper portion of the 
object (through the cooler upper air) and also by curving path downward toward the surface and upward 
again to the observer. Thus, the observer sees both the actual object and an inverted image (why?) below 
it as though a reflecting surface lay between the two. In the desert, this apparent reflecting surface is 
interpreted by the thirsty traveler as a body of water. 

11 
Interference 
11-1 
Interference of Light Waves 
77 
11-2 Applications of interference 
79 
11-3 The Michelson Interferometer 
ö t 
11-1 
I N T E R F E R E N C E O F LIGHT W A V E S 
In Section 10-3, we discussed conditions leading 
to constructive or destructive interference when 
two or more sinusoidal waves simultaneously prop-
agate in a given medium. It was assumed in that 
discussion that the lag angle, δ , between any two 
sources is constant during the period of observa-
tion. Sources satisfying this condition are called 
coherent sources. If the lag angle, δ , and the am-
plitude of the net disturbance vary randomly, the 
sources are said to be incoherent. In the case of 
sound waves, two coherent sources could be ob-
tained by driving two loudspeakers simultaneously 
by a single audio oscillator (a device for generating 
audible sine waves). If, on the other hand, one of 
the loudspeakers were switched on and off in a 
completely random fashion, it is possible to show 
that the superposition of the two loudspeaker out-
puts will produce a net disturbance exhibiting a 
frequency identical to that of the driving audio os-
cillator. Now, however, the amplitude and the 
phase angle difference will be seen to vary in a 
completely random fashion; that is, the sources are 
incoherent. 
The relevance of this discussion to the case of 
light waves becomes clear when it is noted that light 
waves are produced when atoms that have been 
suitably excited (by heating in a flame, for example) 
make transitions to less excited states by the emis-
sion of light. The time during which a typical atom 
radiates light in such a transition is about 10"* sec-
onds. Although this would seem to be a very short 
time interval, it is in fact an interval long enough for 
a light wave to undergo about 10^ oscillations. (This 
follows from the fact that the frequency of light 
waves is about 10'' cycles/sec as determined from 
the value of the speed of light and the measured 
values of the wavelength for light waves.) Since in-
dividual atoms will be making transitions at random 
times if they have been excited thermally, it is clear 
that the light radiated by two such atoms will be 
incoherent in character and interference effects will 
not be observed. Similarly, two incandescent light 
bulbs will be incoherent light sources since each is a 
collection of atoms that are thermally excited by 
the passage of an electric current through a metallic 
ñlament (such as tungsten). 
Highly coherent sources of light can be produced 
by a different means of excitation, which was de-
veloped by Charles Townes and Arthur Schawlow 
in 1%0. This means of excitation is employed in the 
device called a laser, a name which derives from 
the expression /ight ampliñcation by stimulated 
emission of radiation. A complete discussion of 
laser characteristics involves the quantum theory 
of matter, which is beyond the scope of the present 
discussion. It is possible, however, to make several 
general statements regarding laser operation. First, 
the atoms constituting the active part of the laser 
77 

78 
Interference 
substance (such as chromium atoms in a ruby 
crystal) are raised to an excited state, not by ther-
mal means, but by a process known as optical 
pumping. In optical pumping, light of a higher fre-
quency than that which will be subsequently emit-
ted is absorbed by the active laser material. Now 
when light of the frequency to be amplified by laser 
action enters the laser device, the excited atoms are 
stimulated to emit light of the same frequency as 
they make transitions to a less excited state. The 
incident light and the emitted light are coherent, 
and the net disturbance represents an amplification 
of the incident light. If this coherent signal is made 
to travel repeatedly through the laser material (by 
positioning the laser material between perfectly 
parallel reflecting surfaces), the signal becomes 
enormously magnified and remains very nearly 
monochromatic due to the nature of the excited 
states of the active atoms. Finally, if one of the 
reflecting surfaces is actually only partially reflect-
ing, a substantial portion of the amplified signal will 
be transmitted in a well-defined direction, providing 
an intense source of coherent radiation. Two such 
laser beams can be used to provide striking de-
monstrations of interference effects.t 
To exhibit interference effects for light waves 
without employing laser sources, one might divide 
the spherical waves propagating from a point 
source into two or more beams. This can be done, 
for example, by placing an opaque screen contain-
ing two narrow, parallel slits in the path of the light 
waves. The beams transmitted by the slits are 
necessarily coherent at the plane of the screen 
since they are both portions of the same wave front 
from a single source (so that δ  = 0). Therefore, the 
effects due to superposition of the two beams at 
any given point beyond the slits would be expected 
to follow the predictions discussed in Section 10-3. 
This double slit experiment, first performed by 
Thomas Young in 1801, was in fact the earliest ex-
perimental evidence favoring the wave theory of 
light. 
Figure 11-1 is a schematic illustration of the dou-
ble slit arrangement. A point source S is located 
symmetrically with respect to the narrow slits S i 
and 52, which are separated by a distance d. The 
light emanating from the coherent "sources" S i and 
tThe reader seeking more details about lasers is refer-
red to articles by Schawlow appearing in the June 1961 
and September 1968 issues of Scientific American. 
Distant, 
source 
S 
^Δχ =λ/2 
O 
Figure 11-1 Interference at a double slit. 
Si is then allowed to strike a viewing screen (or 
photographic plate) placed parallel to the screen 
containing the two slits at a distance JR (much great-
er than d) from it. The point O is along an axis 
from S , which bisects the distance d between S i 
and S2. Thus, at point O, the path difference Ax =0 
for the two beams. Therefore, the two waves will 
interfere constructively at O to give a bright line 
image of either slit. Now suppose that the distance 
y from O to Ρ  represents the distance from this 
central bright line to the center of the nearest dark 
line that represents destructive interference. This 
requires that the distance S2P - S i P = λ/2. From 
the diagram, we deduce the relations 
S2P ~ SiP 
λ 
*^ 
s , n a = 
^ 
= 
(11-1) 
and 
t a n j 3 « | - . 
(11-2) 
Since ylR<l, 
the angles a, β  will be quite small, 
and we can assume α « sin α « tan β  « β . As a 
resuh, 
R 
2d' 
and, in general, 
y 
(2n + 1)A 
(11-3) 
2d 
n = 0 , l , 2 , . . . , 
(11-4) 
for dark lines. Similariy, the relation satisfied by the 
series of bright lines is found to be 
η  = 0 , 1 , 2 , . . . . 
(11-5) 
If a source of white light is separated by means of 
a prism into the colors of the spectrum, and the 
colors are individually used in this experiment, the 
conditions of Eqs. (11-4) and (11-5) lead to the ob-
servation that the wavelengths of visible light fall in 

Applications of Interference 
79 
a rather narrow range of values between about 4 x 
10"^ m at the violet end and about 7 χ  10"^ m at the 
red end. (The values cited are approximate since 
the limits of visibility depend upon the individual 
eye involved.) As asserted above, it follows from 
the speed of light and these wavelength values that 
the frequency limits for visible light lie between 
about IX 10'' and η  x 10" cycles/sec for violet and 
red light, respectively. Typical dimensions for the 
experiment are: slit widths of «0.1 mm, slit sep-
aration d « 0.2 mm and screen position 
R^lm. 
(We note that more convenient units for light 
wavelengths have been used over the years. Thus, 
the A n g s t r o m — i s equal to 10"'° m, so that a 
wavelength of 4 x 10"^ m becomes 4000 Ä. Cur-
rently, a more commonly used unit, the nan-
ometer—nm—which equals 10"'m, leads to the 
equivalence 4 x 10"' m = 4000 Ä = 400 nm.) 
There is another feature of light wave propaga-
tion that must be taken into account if a completely 
satisfactory interpretation of interference phe-
nomena is to be obtained in all cases. This is the 
fact that a light wave reflected from an optically 
more dense medium will undergo a phase shift of 
180° (corresponding to a shift of one-half a 
wavelength). On the other hand, a light wave re-
flected from an optically less dense medium is 
observed to undergo no phase shift. This is easily 
demonstrated by arranging a point source of light, 
a viewing screen, and a flat reflecting surface (with 
index of refraction greater than that of air) as 
shown in Figure 11-2. By a suitable positioning of 
the source relative to the reflecting surface, the 
path difference between a light wave traveling di-
rectly from the source to a point Ρ  on the screen 
and one which is first reflected before striking the 
screen can be made to approach zero. When this is 
done, destructive interference is observed, leading 
to the conclusion that a phase shift of one-half a 
wavelength occurred upon reflection. 
11-2 
A P P L I C A T I O N S OF I N T E R F E R E N C E 
The phase shift upon reflection from an optically 
more dense medium has practical application in re-
ducing the reflection of light from lens surfaces in 
optical instruments such as cameras and binocu-
lars. In a camera, for example, one would like to 
have all light incident on the lens system transmit-
ted to the photographic plate, providing maximum 
contrast, and at the same time minimizing "ghost 
effects" produced by stray reflections from the lens 
surfaces onto the plate. For a given wavelength, 
this can be accomplished if the surface of the lens is 
coated with a thin film that has an index of refrac-
tion less than the glass of the lens, yet greater than 
that of air. In addition, the thickness of the film 
should be equal to a quarter wavelength of the light 
involved as measured in the film, λ/. The reasons for 
these requirements are easily explained. First, there 
will be a phase shift of 180° at both surfaces of the 
film since at each surface the light proceeds toward 
a medium of greater refractive index. Therefore, a 
light ray reflected from the film-glass interface will 
differ in optical path length by 2 x λ//4 = λ//2 com-
pared to a ray reflected at the air-film interface, or a 
180° phase difference results. In other words, a 
quarter wavelength thickness of the film creates a 
favorable situation for transmission rather than re-
flection, as desired. However, since light entering a 
camera contains many wavelengths, a compromise 
must be made. Since the eye and many types of 
photographic plate are most sensitive to green light, 
the thickness condition is met for a wavelength in 
the green range of the spectrum. As a result, any 
reflection that does occur is a combination of blue 
and red so that "coated optics," as they are called, 
usually have a purplish appearance. Reflected light 
for coated optics is less than 1%, while the value for 
uncoated optics is from 4% to 5%. 
To determine the value of λ/ for a given Aair, we 
must recall that in our discussion of the nature of 
light (Chapter 9) it was stated that for a wave theory 
of light the relative index of refraction is given by 
the ratio of the speed of light in the less dense 
medium to the speed in the more dense medium. In 
our case, using this fact and Eq. (10-3), we obtain 
t^film 
A / / 
A/ ' 
(11-6) 
Figure 11-2 Lloyd's mirror experiment. 
since the frequency remains constant in proceeding 
from one medium to another. As a result, the thick-

80 
Interference 
ness of film required is given by the expression 
(11-7) 
^ _ λ/_ 1 Λ  air 
~ 4 ~ 4 η / air' 
Example 1. Magnesium fluoride (MgF:) has 
proved to be suitable for coating lenses to reduce 
reflection. If μ  = 1.38 for Aair = 5500 Ä , what thick-
ness is required? 
S O L U T I O N 
From Eq. (11-7), 
5500 A 
t = 4x1.38 
1000 Ä = 10'm. 
Another practical application of interference 
effects is related to the phenomenon known as 
Newton's rings. If a plano-convex lens with a 
radius of curvature R is placed on a plate of glass 
that is accurately flat (Figure 11-3), there will be a 
thin film of air between the plate and the lens whose 
thickness is a function of the radial distance r from 
the point of contact. If this system is illuminated 
from above by light of wavelength λ, a pattern of 
interference rings is observed, as in Figure 11-4. In 
this situation, the interference effects are due to the 
air film of thickness t at radial distance r. Light re-
flected from the bottom of the air film undergoes a 
phase shift of 180°, while there is no shift at the top 
of the air film. Thus, at the center of the air film 
(r = 0) where the thickness t is essentially zero the 
net phase shift is 180° and destructive interference 
occurs, as seen in Figure 11-4. On the other hand, if 
the condition 
2t = 
{m^\) λ.„; 
m = 0 , 1 , 2 , . . . , 
(11-8) 
\
r 
\
r 
1 
Figure 11-4 Newton's rings pattern viewed by reflec-
tion (courtesy of Martin Drexhage). 
is satisfied, then constructive interference will 
occur. From Figure 11-3, 
t = R-VR'-r' 
= R 1 - ^l-(-^y . 
(11-9) 
If rlR < 1, the binomial expansion yields 
2R' 
(11-10) 
so that the bright rings are seen to have radii given 
by 
r « y¡(m+^y^.R. 
(11-11) 
Figure 11-3 Newton's rings geometry. 
The observation of Newton's rings provides a very 
sensitive test as to whether or not the lens is sym-
metrically ground, since any departure from circu-
lar rings is an indication of high or low spots on the 
curved surface of the lens. In a similar way, a flat 
plate can be tested for flatness by placing it on a 
known flat plate and placing a thin slip of paper 
under one edge to provide a thin wedge shaped film 
of air. It is left as a problem to discuss the pattern 
of interference fringes in this case. 

The Michelson Interferometer 
81 
11-3 
THE M I C H E L S O N 
I N T E R F E R O M E T E R 
As a final example of interference phenomena, 
we now discuss the interferometer, invented by 
Michelson and used by him in some of the most 
sensitive measurements in the entire history of 
physics. These include the determination of the 
length of the international standard meter in terms 
of the red line of the spectrum of cadmiumt and the 
unsuccessful attempt to detect the luminiferous 
ether (see Chapter 15), which led to the theory of 
relativity. Figure 11-5 is a schematic diagram of the 
device. A beam of light from source S strikes a 
plate of glass A, whose upper surface has been 
lightly silvered. As a result, the original beam is 
split into two beams. One beam is reflected and pro-
ceeds to mirror Mi where it is reflected from the 
front surface back through the plate Λ to the ob-
server at O. (The glass plate Β is the same thick-
ness and is made of the same glass as A and 
inserted to insure that the two beams experience 
tToday the meter is defined to be 1650763.73 
wavelengths of the orange red light of the Krypton-86 
isotope as discussed in Scientific American, December 
I%0, p. 75. 
the same path length in glass. The reader should 
satisfy himself that this is accomplished.) The sec-
ond beam passes through plate A toward mirror M i 
and is reflected from the front surface back to A, 
where it undergoes another reflection at the upper 
surface, and emerges parallel to the first beam 
traveling toward O. 
If li = li (equal optical paths) one would observe 
a bright spot at O due to the fact that both beams 
have undergone the same phase shifts upon reflec-
tion. Now if mirror M i is moved backwards (as 
shown by the dashed lines in Figure 11-5) a distance 
λ/4, the bright spot at O will become a dark spot 
since the first beam will have experienced an opti-
cal path 2(λ/4) = λ/2 greater than the second beam, 
and destructive interference must occur. If the 
motion of M l is increased to λ/2, then a bright spot 
will reoccur. The apparatus is designed so that this 
distance can be measured directly. It is clear, there-
fore, that by counting a large number of alterna-
tions of bright and dark spots at O and measur-
ing the distance M i is moved, one can obtain 
wavelength measurements that are accurate to a 
fraction of a wavelength. We have considered only 
the central ray of light from S i , whereas the light 
beam incident upon A actually will be cone-shaped. 
Source S-
/ 
/ 
/ 
/ 
/ 
/ 
/ 
/ 
My 
V 
V 
My 
(Observer) 
Figure 11-5 The Michelson interferometer. 

82 
Interference 
As a result, rather than a single spot at O, one 
would expect a series of circular fringes analogous 
to those seen in the Newton's rings phenomenon. 
By suitable adjustment of the mirrors M i and M2 
and the source S, one can obtain a series of alter-
nating linear bright and dark fringes instead of 
rings. 
As an example of the type of measurement that is 
possible, suppose that a transparent material with 
index of refraction n(> 1) and unknown thickness t 
is inserted in the path of the beam that is reflected 
at mirror M2. Then the beam will traverse in the film 
an optical path length 
2t = NtXf 
η  
(11-12) 
where N, is some number. In the absence of the 
film, a thickness t of ah* (we assume n^r 
1) is 
equivalent to an optical path length 
2t = 
Ν αλ^ 
(11-13) 
where Na is also some number. Therefore, if the 
material is placed so that one can observe half the 
beam after traversing the material and half after 
traversing the same thickness of air, one would ex-
pect a shift of the fringe pattern that is equal to 
For thin films (Ν -Ν α)<\, 
so that fractional 
fringe shifts are observed. Knowing (N - Na), λα 
and n, the value of t is easily found. Figure 11-6 is a 
sketch of such a shift of fringes for a collodion film. 
In one case, λα = 5461 Ä, η  = 1.18, and the fringe 
shift (Nt - Na) was estimated to be about 5 of a 
wavelength. From Eq. (11-14), it follows that 
ί « 3 X 10"^ cm. 
Light rays 
through film 
Light rays 
through air 
Figure 11-6 Fringe shift in collodion film. 
( M - N J A a i r = 2i(n-l). 
(11-14) 
P R O B L E M S 
1. A monochromatic (single wavelength) point source of light is used to produce light that is incident upon 
two narrow slits separated by 0.5 mm. Adjacent bright fringes are 1.36 mm apart when observed on a 
screen 1 m from the slits. What is the wavelength of the incident light? 
2. What is the separation distance between the central bright fringe and the second bright fringe on either 
side when light of wavelength 5500 Ä passes through two slits 1.5 mm apart and is incident on a screen 
180 cm away? 
3. A thin layer of water (n « 5 ) on the surface of a layer of oil (n <|) shows interference colors. For 
normally incident light of wavelength 4000 Ä, what is the thickness of the water layer? 
4. Two pieces of flat glass are separated at one end by a piece of paper. When illuminated by a vertical 
beam of light of wavelength 7000 Ä, 60 dark fringes are observed across the upper plate from one edge 
to the other. What is the thickness of the paper? 
5. Explain why the interference pattern of bright and dark fringes observed in the transmitted light is just 
the opposite of that observed in reflected light in the Newton's rings phenomenon. 
6. A beam of white light is incident normal to a glass film (n = 1.50) that is 4x 10"^ m thick. What 
wavelengths in the visible spectrum will be most pronounced in the reflected beam? The film is held in 
air. 

Problems 
83 
7. A beam of light containing two wavelengths, 4500 A and 6000 A, passes through two narrow slits 
separated by 0.5 mm. A viewing screen is placed 3 m from the slits. 
(a) At what distance from the central bright fringe (of both wavelengths) will the bright fringe of one 
wavelength fall exactly on the dark fringe of the other? 
(b) What will be the orders of these overlapping fringes? The order of the interference is given by the 
values of η  in Eqs. (11-4) and (11-5). 
8. A system of Newton's rings is formed when a plano-convex lens resting on a flat glass plane is 
illuminated from above. The radius of the 25th bright ring is 1 cm when light of wavelength 6000 Ä is 
used. Find the thickness of the air film at the 25th ring and the radius of curvature of the lens surface. 
9. A system of Newton's rings are observed when a plano-convex lens resting on a flat glass plane is 
illuminated from above by light of 4800 Ä. If the radius of curvature of the lens is 10 m, and the diameter 
of the lens is 4 cm, how many dark rings are visible? 
10. A thin film of transparent material with an index of refraction η  = 1.50 is placed in one of the beams of a 
Michelson interferometer. It is observed that when light of wavelength 5000 Ä is used the insertion of 
the film produces a shift of 30 fringes. What is the film thickness? 
11. A Michelson interferometer can also be used to determine the index of refraction of a gas such as air. 
This is done by putting a cell with plane glass windows in one of the beams and adjusting the 
interferometer to get a fringe pattern. Then the gas is pumped out and the fringe shifts are counted. If 
the index of refraction for air is 1.000293, and the cell is 8 cm long, how many fringe shifts would be 
counted using 5000 Ä light? 
, 
12. A precision micrometer gauge is to be checked using a Michelson interferometer. Using red light from 
cadmium (λ = 6439 Ä), an observer counts 4640 fringes while one mirror is moved a distance of 
1.500 mm as measured by the gauge. What is the actual distance moved? What is the percentage error in 
the distance as given by the gauge? 

12 
Diffraction 
12-1 
Introduction 
85 
12-2 
Fresnel Diffraction 
86 
12-3 
Fraunhofer Diffraction from a 
Single Slit 
88 
12-4 
Fraunhofer Diffraction from a 
Double Slit 
89 
12-5 
The Diffraction Grating 
91 
12-1 
INTRODUCTION 
When the wave front of a propagating light wave 
passes by the edge of an obstacle it is observed 
that: 
1. within the geometrical shadow of the obstacle 
the light intensity is not zero, although it de-
creases rapidly with increasing distance into 
the shadow; and 
2. bands of unequal light intensity are observed 
in the region near the edge of the obstacle. 
This combination of effects is called a diffraction 
pattern and is due to the removal of a section of the 
incident wave front by the obstacle. The first expla-
nation of diffraction phenomena is due to Fresnel, 
who used Huygens' principle (with modifications) 
to show that there is no fundamental distinction be-
tween diffraction and interference. In the case of 
diffraction, the amplitude at any point on a prop-
agating wave front can be determined by regarding 
every point on a previous wave front as a source of 
secondary wavelets. Then, at the point on the new 
wave front the amplitudes of each arriving wavelet 
(with differences in phase properly considered) are 
added. It is however necessary (and reasonable) to 
assume that wavelets propagating to the point in 
directions that are increasingly oblique to the for-
ward direction of the wave front will make corres-
pondingly smaller contributions to the sum of amp-
litudes. 
Thus, diffraction emerges not as a process basi-
cally different from interference, but as an inter-
ference situation involving an indefinitely large 
number of sources (every point on a wave front) in-
stead of the cases involving small numbers of 
sources that were considered in Chapter 11. The 
main distinction that does arise in the analysis of 
diffraction is that because the number of sources is 
increased the mathematics involved can be rather 
complicated. We can, however, exhibit the essen-
tial features of diffraction phenomena without a full 
mathematical analysis. 
In Section 12-2, we discuss the diffraction pat-
terns observed along a plane beyond, but not 
greatly distant from, obstacles placed perpendicu-
lar to a plane wave front. Diffraction patterns 
created when the source or observing screen or 
both are near an obstacle are a type known as 
Fresnel diffraction. 
If the source of light and the pattern observed are 
both at great distances from the obstacle, there are 
significant differences in the features observed 
compared to Fresnel diffraction. This second type 
is called Fraunhofer diffraction, and is discussed in 
Section 12-3. 
85 

86 
Diffraction 
In Section 12-4, we consider the Fraunhofer dif-
fraction pattern due to a double slit. In the final sec-
tion, Fraunhofer diffraction by an array of a great 
number of slits very close together is given. This 
latter situation describes the diffraction grating, a 
device of great value for the analysis of a beam of 
light containing more than one wavelength. 
12-2 
F R E S N E L DIFFRACTION 
Figure 12-1 illustrates the connection between 
the amplitude at a point Ρ  on a new plane wave 
front BB' due to contributions coming from all 
points on an original plane wave front A A ' . When 
the point Ρ  is a distance R from the point O, the 
wave front A A ' is divided into concentric circular 
regions that are called Fresnel zones. These zones 
have radii r„ ΓΖ, η ,..., 
r„, where the subscript η  
specifies the respective zones. The radii are related 
to the distance R by the fact that the outer bound-
ary of each zone is one-half a wavelength (λ/2) 
greater than the distance from Ρ  to the inner 
boundary of the same zone. From the geometry of 
the figure, we can write for the nth radius 
r:=(R-.fJ-R^ 
= r^XR^{ff 
« η λΡ , 
(12-1) 
since \IR<1. 
The area of the nth zone is, there-
fore, given by 
An 
= τΓΓη ' - 7γγ'„-, « TTlnXR - (η  - 
\)\R] 
«τ τ λΡ , 
(12-2) 
so that the areas of all the zones are essentially the 
same. The significance of these relationships lies in 
the fact that by this construction we have obtained 
a series of adjacent zones containing corresponding 
points from which light of wavelength λ arriving at 
Ρ  will be 180° out of phase. That is, zones of even 
index will all be in phase, as will zones of odd 
index, but odd and even zones will differ in phase 
by 180° or λ/2. 
Now let the amplitude at point Ρ  due to the 
wavelets of zone 1 be represented by α», that due to 
the wavelets of zone 2—by ü í , etc. Because of the 
180° phase difference between adjacent zones, it 
follows that the total amplitude at point Ρ  due to 
the entire wave front A A ' is 
α total = fli - 02 + aj - 04 + 
· · 
which can be rearranged to the form 
(12-3) 
(12-4) 
Figure 12-1 Fresnel zones for a plane wave front. 
Because of the increasing obliqueness of wavelets 
arriving from successive zones, 
α ι > α 2 > α 3 > · 
· > α „ > 
· ·. 
Thus, the three terms in each parenthesis will con-
tribute almost nothing. As a result, the total am-
plitude at point Ρ  from the wave front at A A ' is no 
more than would be obtained if only one-half of the 
first zone were present! In Section 13-4, it is stated 
that the intensity of a wave is proportional to the 
square of the wave amplitude. Therefore, in the 
present situation, we must conclude that interfer-
ence effects from adjacent zones reduce the total 
intensity at point Ρ  to a value one-fourth as great as 
that which would be observed from the central 
zone alone. Paradoxical as this may seem, it has 
been demonstrated, not only with visible light, but 
also with radar waves whose longer wavelengths 
permit the use of more easily constructed Fresnel 
zones. 
When a screen with a circular opening (aperture) 
is placed perpendicular to a plane wave front, the 
pattern observed beyond the opening depends on 
the size of the aperture, the distance beyond the 
opening to the point of observation, and the loca-
tion of the observation point on a plane parallel to 
the blocking screen. As an example, suppose the 
observation point lies on a line normal to the screen 
through the center of the aperture, so that the Fres-
nel zones are concentric with the aperture of radius 
r. If the observation point Ρ  is at a distance R such 
that τΓΓ^ = π Rλ, only the first zone will be exposed. 

Fresnel Diffraction 
87 
giving an intensity that is (as we have already seen) 
four times that due to the entire unobstructed wave 
front. If the observation point is brought closer to 
the screen until τ γγ" = 2π λΚ , then two zones will 
pass through the aperture, giving darkness at Ρ  
(since the amplitude there is a, - 02). The same re-
sult could be obtained by keeping the observation 
point at the original distance R and increasing the 
radius of the aperture until the relation τγγ^ = 2n\R 
is again satisfied. By continuing this argument, it 
becomes clear that when any odd number of zones 
are passed by the aperture a centrally located 
observation point Ρ  will be bright, while for an 
even number of zones passed the point will be 
dark. 
Finally, consider the effect of shifting the obser-
vation point Ρ  along a line parallel to the blocking 
screen so that it is now no longer centrally located 
behind the aperture. The center of the Fresnel 
zones will also be shifted to lie along the same line 
normal to the aperture plane at P. As a result, the 
zones passing through the aperture are no longer 
symmetric with the aperture, and their contribu-
tions to the intensity at Ρ  depend strongly upon the 
location of P. Omitting further details, we state that 
the net result is a diffraction pattern made up of 
rings alternating in brightness and concentric with 
the center line normal to the aperture. Whether the 
center of the pattern is bright or dark depends upon 
the number of zones passed. Figure 12-2 shows the 
variation of intensity on a screen placed parallel to 
and behind a circular aperture of radius r with 
center at 0. 
Now suppose that the screen with a circular aper-
ture is replaced by a circular disk (or a spherical ob-
ject). In this case, the pattern observed will be simi-
lar except that the central region of the pattern will 
always be bright. The reason for this is that the first 
of the zones not obscured by the disk will make a 
positive contribution to the intensity that is reduced 
to a total value one-fourth as great by the interfer-
ence of the remaining zones, exactly as in the aper-
ture case. This bright central region was predicted 
by Poisson, who deduced its existence using Fres-
neFs modifications of Huygens' principle. It is his-
torically interesting that the possibility of such a 
situation seemed so absurd to Poisson that he re-
garded it as proof of the inapplicability of the wave 
theory of light. The subsequent demonstration of 
its reality by Arago (the bright spot is called the 
Arago spot by some, and by others—the Poisson 
spot!) and Fresnel essentially ended the con-
troversy that had shrouded the wave theory first 
proposed by Young. 
As a final example of Fresnel diffraction, con-
sider a plane wave front incident upon a straight 
edge that blocks part of the wave front. If the ob-
servation point Ρ  is directly behind the edge of the 
obstacle, as shown in Figure 12-3, it is clear from 
our previous cases that only the upper half of all the 
Fresnel zones will contribute to the amplitude at P. 
Since the total amplitude for the entire wave front 
was found to be «1/2, we see that at this point it 
becomes α 1/2. At points behind the obstacle (such as 
Pi), more than the lower half of the zones will be 
blocked, so that the total amplitude will steadily de-
crease to zero with a shift of the observation point 
more deeply into the shadow of the obstacle. Con-
versely, shifting out from the shadow toward points 
Figure 12-2 Fresnel diffraction pattern for a circular 
aperture. 
Figure 12-3 Fresnel diffraction-amplitude variation 
produced by a straight edge obstacle. 

88 
Diffraction 
such as P2 will expose additional portions of the 
lower half of the Fresnel zones, resulting in a total 
amplitude that is alternately greater or less than the 
unobstructed plane wave value of ajl. 
As the 
figure indicates, when the observation point be-
comes increasingly distant from the edge of the ob-
stacle, the difference in amplitude at successive 
locations becomes smaller and smaller; the total 
amplitude for points sufficiently distant assumes 
the value of ai/2, which is to be expected when the 
obstacle is so far distant that only outlying zones of 
negligible importance are affected by it. 
Figure 12-4 Fresnel diffraction at a single long rec-
tangular slit. 
It is not hard to understand that the Fresnel dif-
fraction pattern due to a single slit should have the 
features illustrated in Figure 12-4, since it is the re-
sult of bringing together the patterns due to two 
long parallel straight edges that were originally 
widely separated. Close to a slit of width d, which is 
large compared to the wavelength λ, we see that ex-
cept for positions very near the edges of the slit the 
amplitude is essentially constant. In addition, there 
is nearly total darkness within the geometric 
shadow, as would be expected from ray optics. For 
smaller slit widths, however, fewer zones remain 
unobstructed, and the diffraction features become 
more pronounced. When the viewing screen is 
moved to a greater distance behind the slit, an in-
crease in the number of zones that are obstructed 
results. The pattern that is then observed will show 
marked intensity variations even behind the center 
of the slit in analogy to the case of the circular aper-
ture. For example, distances behind the slit can be 
found for which the central position on the screen 
is totally dark or a maximum brightness. 
12-3 
F R A U N H O F E R DIFFRACTION F R O M A 
S I N G L E SLIT 
Let us continue our discussion of the diffraction 
pattern due to a long narrow slit. In particular, what 
features will the pattern exhibit when the secon-
dary wavelets (which originate from a plane wave 
incident normally on the slit) leave the slit parallel 
to each other? This is an example of Fraunhofer 
diffraction, a simpler mathematical case, even 
though there is no change in the physical situation. 
The experimental difficulty of determining light in-
tensities at great distances from the obstacle can be 
avoided by placing a converging lens behind the ob-
stacle and observing the light intensity at the focal 
plane of the lens. Similarly, the source need not 
be far removed from the obstacle if it is placed at 
the first focal point of a converging lens, so that the 
wave fronts striking the obstacle become plane 
wave fronts. Figure 12-5 shows a plane wave inci-
dent upon a single slit of width d. Diffracted waves 
leave the slit at an angle θ relative to the direction 
of the incident wave. The convergent lens indicated 
is used to bring the parallel wavelets leaving the slit 
to a focus at point Ρ  in the focal plane of the lens. 
The path difference for the wavelets traveling to Ρ  
from the upper edge of the slit coπ φ ared to those 
leaving the lower edge is equal to d sin Θ , from the 
figure. Now, if 
d sin θ = η λ, 
η  = 1,2,3,..., 
(12-5) 
then destructive interference will take place, and 
point Ρ  will be dark. To see this, consider the case 
η  = 1, so that the upper wavelets must travel a 
distance equal to one wavelength further than those 
at the bottom of the slit. When this is true, a 
Figure 12-5 Fraunhofer diffraction for a single slit. 

Fraunhofer Diffraction from a Double Slit 
89 
wavelet leaving the midpoint of the slit is one-half a 
wavelength out of phase with one leaving the bot-
tom of the slit when they arrive at P. As a result, 
they interfere destructively with each other. Pro-
ceeding from the lowest point of the slit, all 
wavelets leaving the slit can be paired (one from 
each half of the slit) so that their path difference re-
mains λ / 2 , so that complete destructive interfer-
ence exists at P. For η  = 2, the slit must be divided 
into four portions for each pair of which a path 
difference of λ / 2 exists. This creates a system for 
which two pairs of points each give destructive in-
terference. Since all points along the slit can be 
accounted for, total destructive interference oc-
curs. The proof of the validity of the assertion for 
subsequent integers is left to the reader. 
At point O, the angle θ is zero, and there is no 
path difference. As a result, all wavelets interfere 
constructively, giving a bright section in this central 
position in the pattern. Between each dark portion 
of the pattern there will be a bright section, whose 
intensity decreases with increasing distance from O 
on the screen. Although the proof is too advanced 
to be a part of this discussion, it is noted here that 
the variation in amplitude of the superimposed 
wavelets at a point Ρ  is given by 
A p = A o 
( - λ — ) 
ττά  sin θ 
(12-6) 
where A o is the amplitude observed at 0 in Figure 
12-6. In addition, the intensity at Ρ  is given by 
/ = /« 
". 
(π ά  sin θ\ 
ττά  sin θ 
(12-7) 
/ 
- 4 - 3 - 2 - 1 
( 
λ 
Figure 12-6 Intensity variation with diffraction angle 
for a single slit. 
Figure 12-6 is a sketch illustrating the variation of 
intensity predicted by Eq. (12-7). 
Exampie 1. Light of wavelength λ = 5 x 10"'m 
is incident as plane waves on a slit of width 
d = 5 mm, as shown in Figure 12-5. 
(a) Find the angular width of the central max-
imum in the pattern. 
(b) Calculate IlL 
for a diffraction angle of 
(1/7Γ )°. 
SOLUTION 
(a) From Eq. (12-5), 
. 
^ 
λ 
5 X10"' m 
^ ^ ^ , Λ - 4 
Thus, θ = 0.143°, and the angular width of the cen-
tral maximum is 
2Θ  = 0.286°. 
(b) From Eq. (12-7), 
ΓτΓ X f X 10"^ m 
l o ' 
sm L 5xlO"'m 
T r X l O ' 
. 
— 
xsin(iy_ 
25 - sm 
Since sin α 
α (in radians) for a less than 10°, we 
can write 
. /IV 
'nrad 
1 ^ 
Therefore, 
I 
—  
A
; 
' . /4007rV 
sm 
4007Γ 
180 
207Γ 
180° 
(rad)x — r a d 
7Γ 
20ir 
9 sin 400° 
L 
2 0 7 Γ 
3.1 X 10"' or about 3%. 
12-4 
F R A U N H O F E R D I F F R A C T I O N F R O M 
A D O U B L E SLIT 
It is logical to consider next the diffraction pro-
duced by two long parallel slits of equal width d, 
separated by a distance a, as illustrated in Figure 

90 
Diffraction 
12-7. Each slit creates a diffraction pattern that 
exhibits the features discussed in the last section. 
At a given observation point, in the present case, 
there will be a superposition of these separate dif-
fraction patterns. As a result, the intensity ob-
served will now be due to the interference of the 
two diffraction patterns. 
, 1 1 
d 
1 
1 
m 
• 
Figure 12-7 Plane waves diffracted by a double slit. 
For a given angle Ö, the amplitude due to either 
slit is given by Eq. (12-6). There will be a phase 
difference between the two, however, which cor-
responds to the path difference from point 3 to 
point 6 or, equivalently, from point 4 to point 5 in 
Figure 12-7. It therefore follows that the observa-
tion point will be dark if this path difference is equal 
to an odd integer number of half wavelengths: 
path difference,^ = (2n + 1) y 
η  = 0 , 1 , 2 , . . . . 
(12-8) 
From Figure 12-7, 
path differences-^ = a sin Θ , 
and our destructive interference condition becomes 
α sin 0 = (2n + 1) ^ 
(dark fringe condition). 
(12-9) 
On the other hand, when 
α sin 0 = 2 n y 
η  = 0 , 1 , 2 , . . . , 
(12-10) 
there will be constructive interference and bright 
fringes result. 
The resulting amplitude at the observation point 
is given by the expression (which is presented with-
out proof): 
Λ  = 2 
. 
/ T r d s i n ö V 
τ τ ά  sin θ 
cos 
^ττα sin 
(12-11) 
amplitude from a single slit 
and the intensity at the same point is 
/ ^ 7 r d s i n Ö \ \ 
/π α sin θ\ 
2 lira sin 
θ\ 
(12-12) 
where h is the intensity at d = 0 for a single slit as 
given by Eq. (12-7). We see that the effect of inter-
ference between the two single slit diffraction pat-
terns is one of modulation. In other words, the 
rapidly varying terms cos^(7ra sin θ/λ) produces 
variations in intensity due to interference, which 
are superimposed on the intensity variations due to 
the two sin¿e slit diffraction patterns. The resulting 
intensity distribution is shown in Figure 12-8. 
i 
^ 
Interference 
A^effects 
A 
Diffraction 
- 6 - 5 - 4 - 3 - 2 - 1 C ) 1 2 3 4 5 6 7 
a sin θ 
λ 
Figure 12-8 Intensity distribution for double slit dif-
fraction. 
Example 2. In a double slit diffraction pattern, 
the third principal maximum is missing because that 
interference maximum coincides with the first dif-
fraction minimum (zero). Find the ratio aid, 
S O L U T I O N 
Since the third interference maximum corre-
sponds to η  = 2 in Eq. (12-10), 
α sin θ = 2λ, 

The Diffraction Grating 
91 
the first diffraction zero corresponds to η  = 1 in Eq. 
(12-5), 
d sin θ = λ. 
Dividing the two relations, we obtain 
It is left as a problem to use this result and Eq. 
(12-12) to sketch the resulting intensity distribution. 
12-5 
THE DIFFRACTION G R A T I N G 
What sort of diffraction pattern should we expect 
from a diffraction grating, which consists of a very 
large number of extremely closely spaced long 
parallel slits of equal width? To answer the ques-
tion analytically, we could proceed along the lines 
suggested in Sections 12-3 and 12-4. However, to 
avoid the mathematical complexities, we present 
instead Figure 12-9, which illustrates the intensity 
distribution and the changes that take place as the 
number of slits is increased. (In the figure, the ratio 
of slit width to slit spacing is kept constant.) We can 
Single slit 
Double slit 
Three slits 
Four slits 
Five slits 
Six slits 
JUL 
make several observations about the patterns as the 
number of slits is increased: 
1. subsidiary maxima appear, with intensities 
that are much weaker than those of the princi-
pal maxima; 
2. the intensities of the subsidiary maxima be-
come progressively weaker; and 
3. the widths of the principal maxima become 
progressively narrower. 
If the number of slits is made very large (ruling 
engines have been developed that routinely pro-
duce several thousand lines or slit spacings per 
centimeter of ruled surface), it is observed that the 
pattern reduces to a series of extremely sharp lines, 
with no surviving subsidiary maxima. In addition, if 
the light incident on the slit system contains many 
wavelengths, there is a separation of the various 
wavelengths into a well-resolved series of spectral 
lines. This pattern is repeated at larger angles. In 
the pattern repetitions (called diffraction orders) 
overlapping of various colors from different orders 
occurs. The degree of overlapping increases in the 
higher orders. 
We can obtain a quantitative expression of these 
latter facts if we refer to Figure 12-10, which shows 
a plane wavefront of wavelength λ incident nor-
mally on a series of narrow parallel slits separated 
by a distance d. It is assumed that λ <^ d. As in the 
earlier discussion of Fraunhofer diffraction, we are 
interested in the phase relationship 
between 
wavelets arriving from the various slits at the focal 
plane of a converging lens (as in Figure 12-5). From 
Figure 12-1, we see that constructive interference 
between wavelets from all of the slits will occur at 
the screen whenever the angle θ is such that the 
path lengths traveled by the wavelets from adjacent 
slits differ by an integer number of wavelengths. 
Rgure 12-9 Effect of multiple slits on diffraction 
patterns—constant slit width to separation ratio. 
Figure 12-10 A diffraction grating. 

92 
Diffraction 
When this happens, the following relation holds: 
sm θ = n-r. 
a 
(12-13) 
The integer η  specifies the order of the spectrum 
that is produced by the grating when a non-
monochromatic beam is incident upon it. 
Thus, if the grating spacing is known, if experi-
mental provision is made for determining θ (or 
sin Θ ), and if the order η  is observed, one can iden-
tify the wavelengths that are characteristic of any 
given light source. Examples of such applications 
are the spectra produced by flame excitation of var-
ious inorganic salts or electric arc excitation of 
gases such as hydrogen or neon. In the field of 
chemistry, information of this kind can be used as a 
basis for either qualitative or quantitative analysis 
of unknown samples. In physics, the existence of 
intriguing numerical relationships between the vari-
ous wavelengths of the excitation spectrum for a 
given substance was known in the 1880s and led to 
the development of modern quantum theory. 
The equation indicates that for a given order a 
grating will produce a spectrum that has longer 
wavelengths (the red end of the spectrum) appear-
ing at larger angles than the shorter wavelengths. 
The equation can also be used to justify the over-
lapping of colors from successive orders that was 
asserted earlier. 
Example 3. Using Eq. (12-13), show that, regard-
less of the grating spacing, the third order violet line 
will overlap the second order red line for light inci-
dent normally on the grating. 
SOLUTION 
From Eq. (12-13), 
sin öviolet = 3Aviolet/ii, 
sin 0red = 2\rJd. 
Therefore, 
sin ö v 
3λν  
sin θ„  
2kr, 
6 
"τ  
This shows that övioiet <Öred, so that the second 
order red line overlaps the third order violet line, as 
asserted. 
Example 4. A diffraction grating has a spacing 
of 5000 lines/cm. Find the angular spread of the 
second order spectrum, assuming normal inci-
dence. 
SOLUTION 
In this case, 
d = 1/5000 = 2 X 10"' cm = 2 X 10"' m. 
Therefore, 
sin Övioiet«(2)(4 X 1 0 ' m)/(2 x 10'm) = 0.4 
sin Ö,ed«(2)(7 X 1 0 ' m)/(2 x lO*' m) = 0.7. 
As a result, 
Övio.et«24^ 
and 
θ r e a « 4 5 ^ 
SO that the angular spread of the second order 
spectrum is 
'2V. 
(λ. 
= 4 x l 0 - ' m , 
A,ed«7xl0-'m) 
P R O B L E M S 
1. Monochromatic plane waves of wavelength λ are incident normally on a screen with a circular aperture 
of diameter d. Along an axis through the center of the aperture and normal to the screen, the light 
intensity will pass alternately through maxima and minima. If the distance from the screen along this 
axis is R, show that the location of these maxima and minima are given by the relation 
4η λ' 
where η  is any integer. Show that maxima correspond to η  even, minima—to η  odd. 
2. (a) Referring to the results of Problem 1, what is the largest value of R for which the light intensity is a 
minimum? 
(b) What is the largest value of R for which the light intensity is a maximum? 

Problems 
93 
3. A Fresnel circular zone plate is a circular array of alternately transparent and opaque rings constructed 
so that each has approximately the same area. (See Section 12-2.) If the even numbered rings are 
opaque, the light transmitted by this zone plate will be much more intense than if all the light were 
transmitted. Explain why this is so. 
4. Plane waves of sodium light (λ = 5.893 x 10"' m) are incident normally on a circular aperture 4.60 mm in 
diameter, and observed at a point on the axis 3 m from the aperture. 
(a) Is the center of the diffraction pattern bright or dark? 
(b) Determine the minimum distance the point of observation should be moved along the axis to reverse 
the situation of (a). 
5. (a) Calculate the radius of a circular zone plate with a "focal length" of τ  m for light of wave length 
λ = 5.46 X 10"^ m. The plate has only 8 zones; zones, 1, 3, 5, and 7 are transparent, and the even 
numbered zones are opaque, 
(b) Repeat for infrared light, λ = 5.46 x 10"* m. 
6. Monochromatic plane waves of wavelength λ are incident normally on a rectangular slit of width d. A 
viewing screen is placed behind the slit as in Figure 12-6. Sketch the expected diffraction patterns when 
the screen is 
(a) near the slit, 
(b) somewhat farther from the slit, and 
(c) at a great distance from the slit. 
Note: it is assumed that λ <^ d. 
7. Equation (12-7) can be written in the form 
/ ^ sin' β  
Ιο  
β ' ' 
where 
_ 
Trd sin θ 
(a) Using this form, show that the minima of the Fraunhofer single slit diffraction pattern occur at 
values of θ such that β  = η ττ, where η  = 1,2,3, 
(Why is the case η  = 0 not included?) 
(b) Show that the maxima of the Fraunhofer single slit diffraction pattern occur approximately at values 
of θ such that β  =(2η - 
1)7γ/2, where η  = 1, 2, 3 , . . . , thus for odd integer multiples of π Ι2. Al-
though this is an approximation, it becomes increasingly accurate for large n, 
* 
(c) The exact values of β  for which Ilh has successive maxima can be shown (by calculus methods) to 
satisfy the relation 
tan β  = β ~. 
Using trigonometry tables, determine the errors involved in using the approximate values of (b) for 
the first 4 secondary maxima. 
8. Assume the focal length / of the converging lens placed just behind the slit in Figure 12-6 is large 
compared to the slit width d. Show that the locations of the maxima and minima on the viewing screen 
placed in the focal plane of the lens are given by 
y = / tan Θ . 
9. (a) Calculate the angles at which the first minimum would occur for the Fraunhofer diffraction pattern 
of a slit of width 2 x 10"* m on which plane waves of wavelengths λ, = 4 x 10"' m and 
= 7 x 10"' m 
are incident normally. 
(b) If the focal length / of the converging lens behind the slit is ¿ m, how far apart will the two minima be 
when viewed on screen in the focal plane of the lens? 
10. Repeat Problem 9 for a slit of width 2x10"^ m. 

94 
Diffraction 
11. When a single rectangular slit of width d is illuminated by two line sources of wavelength λ, there will 
be a diffraction pattern for each line source on the viewing screen. Depending upon the angular separa-
tion of the line sources, the two patterns will overlap to an extent that can make it difficult to say 
whether the total pattern is due to one or two sources. The two sources are said to be just resolvable 
when the first minimum of the one pattern coincides on the screen with the first maximum of the other. 
Draw a sketch illustrating this situation, and show that this limiting angle of resolution is given by 
θ -A 
(It is assumed here that 
\ld<l.) 
12. When two point sources of wavelength λ illuminate a circular aperture of diameter d, a more involved 
analysis leads to the resolution condition (known as Rayleigh's criterion) 
0c-1.22A/d, 
forA/d<^l. 
Two point sources of wavelength λ = 5 x 10"' m are 25 cm in front of a ch-cular aperture of diameter 
d = 4 X 10"' m. What is their minimum separation if they are to be resolvable? 
13. In the double slit diffraction example, it was found that a/d = 2. Use result and Eq. (12-12) to construct 
a plot of 
/ v e r s u s ^ , 
i o 
A 
14. Monochromatic light of wavelength A = 5.46 x 10"' m is incident normally on a double slit. The 
Fraunhofer diffraction pattern is viewed in the focal plane of a lens of focal length / = 2 m. The sixth 
principal maximum is missing, and the distance between the two minima next to the central maximum is 
observed to be 1 mm. Find d, the width of the slits, and a, the distance between them. 
15. When a/d = 1, the two slits are no longer separated (see Figure 12-8). Instead the system reduces to a 
single slit of width 2a. Show that Eq. (12-12) reduces to Eq. (12-7) with d replaced by 2a and 4 times the 
intensity Jo of a single slit. 
16. If a/d =4 for a double slit system, which of the interference maxima will be missing? 
17. A diffraction grating for which the slit spacing is unknown can still be used for quantitative work. For 
example, consider a plane wave containing a known wavelength A, and an unknown wavelength A, 
incident normally on a diffraction grating. It is observed that the maximum of order n. for the known 
wavelength occurs at the same angular position as the maximum of order 
of the unknown 
wavelength. 
(a) Show that 
\ 
(b) Show also that 
/sin^X 
{sin θ J ^' 
when the maxima are in the same order. 
(c) One of the visible wavelengths associated with the spectrum from atomic hydrogen is known to be 
A, = 6.560 X 10"' m. It is found that the third order maximum of this wavelength occurs at the same 
position as the fourth order maximum of an unknown wavelength Ax. Find Ax. 
(d) At what angle will the second order maximum of the unknown wavelength occur if the second order 
maximum of A. occurs at Θ , = 30°? 
18. A certain diffraction grating shows a maximum angular separation of visible light (4000 Ä to 70(X) Ä) in 
the first order. 
(a) Calculate the slit separation for the grating. 
(b) Calculate the angular spread of visible light for the grating. 

Problems 
95 
(c) If spacing of slits is increased, what changes will occur in the angular separation of visible light and 
in the number of orders that can be observed? 
(d) Repeat (c) for a decrease in slit spacing. 
19. In obtaining Eq. (12-13), normal incidence of monochromatic plane waves of wavelength λ was as-
sumed. If, instead, the light is incident at an angle i and the nth order maximum is observed at an angle 
Θ , show that Eq. (12-13) becomes η λ = d(sin i + sin Θ ). (A diagram analogous to Figure 12-11 will be 
very helpful.) Note that when i =0°, this result reduces as it should to Eq. (12-13). 
20. (a) Consider two wavelengths which differ only slightly, λ and λ + Δ λ, where Δ λ/λ <^ 1. Show that the 
angular separation ΔΟ  of these two lines in the nth order is given approximately by 
ΑΘ  
"^^ 
d cos Ö' 
Hint: Use the trigonometric identity 
sin (A + β ) = sin A cos Β + cos A sin B, 
and note that 
cosΔ 0 ^ 1, sinΔ (9 =^Δο , 
when ΑΘ  is small (less than 5°), and is expressed in radians. 
(b) If the nth order maximum for wavelength λ is observed at an angle Ö, show that 
Δ Ö « γ t a n a 
(c) The spectrum of sodium has two closely spaced lines, 5890 A and 5896 Ä. If the angular separation 
of these lines in the second order is 7.42 x 10"^ rad = 0.428°, determine the slit spacing of the grating. 

13 
Transverse Waves and 
Polarized Light 
13-1 
Introduction 
97 
13-2 
Transverse Waves 
97 
13-3 
Vectors and Vector Algebra 
99 
13-4 
Polarization and Polarized Light 
13-5 
Matter and Polarized Light 
t03 
tot 
13-1 
INTRODUCTION 
In the last two chapters, it was demonstrated that 
interference and diffraction phenomena involving 
light can be understood by assuming light has a 
wave nature. In Section 10-1, it was stated that two 
basic wave types exist: longitudinal waves—in 
which the vibrations take place in a direction paral-
lel to the propagation direction of the wave; and 
transverse waves—in which the vibrations take 
place in a plane perpendicular to the propagation di-
rection of the wave. Any given wave motion could 
therefore be either purely longitudinal, purely 
transverse, or some combination of the two types. 
The principle concern of this chapter is the appro-
priate type of description for light waves and the 
physical effects associated with it. 
As the chapter heading implies, light waves are in 
fact transverse. We therefore begin with a discus-
sion of transverse waves. For this purpose, we find 
it useful to introduce the concept of vector quan-
tities, and to provide a brief introduction to the 
algebra of vectors. (Further details are provided in 
the Appendix.) We then discuss polarization, and 
show how polarization effects can be used to de-
monstrate the transverse nature of light waves. In 
this way, we can avoid the mathematical complica-
tions involved in proving the principle of superposi-
tion for light waves. Finally, we consider a number 
of examples of the interaction of light and various 
materials that involve polarization. 
13-2 
T R A N S V E R S E W A V E S 
The vibrations associated with a transverse wave 
propagating along the ζ -axis must lie in planes 
parallel to the x-y plane. The simplest example of 
such a wave is provided by a transverse wave on a 
string, as described in Chapter 10. For wavelength 
A, propagation speed 
and vibrational amplitude 
Ay, and assuming vibrations occur in the y-
direction only, the vibration equation becomes 
y = A y sin γ(ν ί-ζ )-^φ  
(13-1) 
where φ  is a constant phase angle which depends 
upon initial conditions. (For example, if y 
=iAy 
when 2 = 0 , í = 0, then φ  = π  16.) 
If the same wave is related to vibrations in the 
jc-direction only, then Eq. (13-1) becomes 
X = Ax sin γ(ν ί-ζ ) 
+ φ  
(13-2) 
In either case, when one looks along the string 
(down the ζ -axis) toward the source producing the 
vibrations, one sees a linear disturbance along 
either the y- or χ -axis, respectively. This can be 
97 

98 
Transverse Waves and Polarized Light 
accomplished, for example, by setting one end of a 
long string in motion sinusoidally along the appro-
priate axis. 
Now suppose that sinusoidal vibrations along the 
JC- and y-axes are simultaneously imposed on the 
string. What is the configuration of the disturbance 
that would now be observed by looking along the 
string toward the source of the disturbance? As in 
Chapter 10, we can apply the principle of superpo-
sition provided we account for the fact that these 
vibrational displacements are perpendicular. It will 
not be possible for these disturbances to combine 
to produce a completely destructive or constructive 
interference. We present two techniques for deter-
mining the configuration of the disturbance—the 
first, a mathematical manipulation, the second, an 
equivalent graphical construction. 
To simplify the mathematics of our illustration, 
we introduce the substitution 
e.=Y(vt-z). 
In addition, we devote our attention to the point 
2 = 0 so that Eqs. (13-1) and (13-2) can be written 
y = Ay sin(ö o + <i)y) 
= Ay cos φ y sin Oo + Ay sin φ y cos θο  
(13-3) 
x= Α χ  sin(0o + ψχ ) 
= Α χ  cos φ χ  sin θο  + Α χ  sin φ χ  cos ö o, 
(13-4) 
where we have used the identity sin (a ± β ) = sin a 
cos β  ± cos α sin β . To determine the vibrational 
configuration independent of time, it is necessary to 
eliminate 
β « ( = ^ ( « ί ) = γ ( / λ ) ί = ω ί ) 
from the equations.t We do this by the following 
sequence of operations: 
1. Multiply Eq. (13-3) by A. cos φ χ , Eq. (13-4) by 
Ay cos ^y, and subtract the resulting equa-
tions. 
2. Multiply Eq. (13-3) by A. sin φ χ , Eq. (13-4) by 
Ay sin φ y, and subtract the resulting equations. 
3. Square and add the two equations obtained in 
steps 1 and 2. 
tNote that ω  = Inf. ω  is called the angular frequency. 
Since 2TT is the number of radians per cycle, and / is the 
number of cycles per second, ω  is the number of radians 
per second. 
4. Simplify the result by using the identities 
sin^ a + cos^ a = 1 and 
cos (α±β ) 
= cos a 
cos β  ± sin α sin β . 
When these steps have been performed (see Prob-
lem 13-2), we obtain the equation 
A.'y' + Ay'jc' - 2(A.Ay cos Αφ )χ γ = A/Ay' sin' Αφ , 
(13-5) 
where Α φ  = {φ χ  -φ y)is 
the phase angle difference 
between the two transverse waves. 
This equation may be recognized as the equation 
for a conic section. For special values of Α χ , Ay, and 
Αφ , it leads to several common situations. Thus, if 
either Α χ  or Ay is zero, the result is of course a 
linear disturbance as outlined above, with vibra-
tions along either coordinate axis. On the other 
hand, if neither Α χ  nor Ay equals zero and Α φ  = 0, 
Eq. (13-5) becomes 
or 
(A.y-Ayjc)' = 0 
y = Ä x ' ' 
(13-6) 
(13-7) 
This is a straight line disturbance also, but one 
which is inclined to the ac-axis by an angle ψ , where 
tan φ  = Ay I Αχ . If Α χ  = Ay, ψ  = π  14, and the distur-
bance is composed of equal amounts along the two 
coordinate axes. 
A circular disturbance results if A, = Ay and 
Αφ  = ± (2n + 1)7γ/2, 
where 
η  = 0 , 1 , 2 , 3 , . . . . 
Finally, if A, 5^ Ay and Α φ  Φ  O, the disturbance will 
be elliptical in form. These results are sununarized 
in Table 13.1. 
Now let us consider the second method, a graphi-
cal technique, for performing the same superposi-
tion of two transverse waves with perpendicular 
vibrations but identical wavelength and propaga-
tion speed. Figure 13-1 is a simultaneous plot of y 
versus ω t in the upper right part of the diagram, of 
X versus ω t in the lower left part, and in the upper 
left the resulting vibrational configuration is ob-
tained by plotting x, y values for identical values of 
ω ί. In this example. Α χ  # Ay and Α φ  = ir/2. The re-
sult is in agreement with the predictions of Table 
13-1. It might be noted that this configuration can be 
obtained on an oscilloscope if one applies the ap-
propriate sinusoidal voltages on the vertical and 
horizontal inputs of the oscilloscope. Such figures 
are known as Lissajous figures, and can be used to 
determine unknown frequencies if a signal of 
known frequency is available. (See Problem 13-3.) 

Vectors and Vector Algebra 
99 
Table 13-1 Summary of Vibrational Configurations. 
Conditions 
Configuration of 
Disturbance 
=0, 
Ay^O 
7Í 0, Λ> = 0 
Δ φ  =0, ±2η π  
Αφ  = ±(2η  - 1)7Γ 
Δ φ  = ± (2η  - 1)7Γ/2, Λ  = Λ , 
Δ φ  = ±(2π -1)7Γ /2,Λ ,7^Λ , 
η  = 1, 2, 3, 
(any integer) 
Λχ  TÍ Ay, Δ φ  any value 
not given above 
Linear, along y-axis 
Linear, along jc-axis 
Linear, at angle 
0 = t a „ - ( A ) 
from X-axis 
Linear, at angle 
^ = t a „ - ( : ^ ' ) 
from jc-axis 
Circular 
Elliptical, principal 
axes coincide with 
X- and y-axes 
Elliptical, principal 
axes inclined to the 
X- and y-axes 
Figure 13-1 Graphical addition of perpendicular 
sinusoidal vibrations (with equal wavelengths and 
propagation speeds) transverse to the same propaga-
tion direction. 
13-3 
V E C T O R S A N D V E C T O R 
A L G E B R A 
In the previous section, we saw that it is possible 
to "add" (superpose) disturbances and obtain re-
sults that are not consistent with simple algebra. 
That is, a positive disturbance and a negative dis-
turbance may yield a variety of results that are 
neither simple sums nor differences. The reason, of 
course, is that the quantities involved possessed not 
only a magnitude but also a direction, and both fac-
tors must be accounted for in the superposition pro-
cess. Quantities of this sort, which are of frequent 
occurrence in physics, are known as vectors. These 
are in contrast to scalar quantities (or simply sca-
lars) that have magnitude only. For example, the 
volume of a solid is a scalar, while the displacement 
of a particle of a string is a vector. By displacement 
of a particle, we mean the shift in location of the 
particle (due to some source of disturbance applied 
to the string, as above). 
A vector quantity is illustrated graphically by 
drawing an arrow from an origin to a point. The 
length of the arrow is the magnitude of the vector, 
and its direction specifies the direction of the vec-
tor. For example. Figure 13-2 shows the displace-
ment of a given particle of a vibrating string due to 
a transverse wave propagating along the string in 
the 2-direction at a given instant. For the situation 
shown, the particle is displaced from y = 0, χ  = 0 to 
the point y = A , , χ  = Α χ . From the Pythagorean 
theorem, the magnitude of the displacement is A = 
(Ax^ + Ay^)'^ and the direction is given by 
φ  = tan 
Any given vector consists of parts (called compo-
nents) representing the extent of the vector in three 
different (usually mutually perpendicular) direc-
__A 
A, 
1 
1 1 1 
J 
A, 
X 
Figure 13-2 Graphical representation of the two-
dimensional vector A. 

100 
Transverse Waves and Polarized Light 
tions. Thus, in Figure 13-2, A has no z-component 
while the jc- and y-components are Ax and Ay, re-
spectively. We can indicate that the vector A is 
obtained as a vector sum of the components if we 
introduce the concept of unit vectors. A unit vector 
is a vector of unit length. An arbitrary vector of the 
same physical kind and in the same direction as the 
unit vector is then simply a scalar number (in what-
ever system of physical units is required for the 
quantity) indicating the number of unit vectors that 
are equivalent in magnitude to the arbitrary vector. 
One convention is to denote unit vectors in the jc-, 
y-, and ζ -directions by i, j , and k, respectively. In 
this notation, the vector A in Figure 13-2 becomes 
or 
or, simply. 
A = A.i + A J + Ok 
A = Axi + Ayl 
(13-8) 
How do we add two vectors in general? The pro-
cedure is straightforward if the two vectors are 
written in the form of Eq. (13-8) because the addi-
tion of any two χ -components (and similarly for 
y- and ζ -components) follows regular algebraic 
rules. Physically, this must be true since we have 
seen that constructive or destructive interference 
can occur for superposition of two vibrational dis-
placements in the same direction. Thus, if C = 
A + B, we must have 
C i + Cy} -H C k = (A,i + Ay] + A.k) 
+ (Äi + ßyj + R k ) 
(13-9) 
Cx = Α χ -l· Bx, 
Cy = Ay + By, 
Cz = A, + R. 
(13-10) 
From the Pythagorean theorem, it again follows 
that the magnitude of C is 
|c| = ( G ' + C y ' + a y ' 
= [(Ax + B,)' + (Ay + By)' + (A. + B.ff" 
Example 1 . Given vectors A = 5i + 3j, Β  = 7i - 2j. 
Find C = A + B and D = A-B. 
S O L U T I O N 
From Eqs. (13-9), (13-10), and (13-11), 
C=12i + j , 
|C| = Vl45, 
D = -2i-H5j, 
|D| = V29. 
The results are also shown graphically in Figures 
13-3 and 13-4. Notice that the figures demonstrate 
that vector addition is commutative, that is. 
and 
A+B=B+A=C 
A + (- B) = (- B)-hA = D. 
A knowledge of the mathematics of vectors is 
essential for quantitative study in physics; the 
reader is therefore urged to study carefully the ma-
terial presented in the Appendix. For the remainder 
4 H 
3H 
2H 
-H 
-2H 
1 
1 
1 ^ ' ^ l 
1 
^1-..^^^^3 
4 5 6 7 
a 9/10 11 12 
Β  
Figure 13-3 Graphical representation of C=A+B. 

Polarization and Polarized Light 
101 
Figure 13-4 Graphical representation of D=A-B. 
of this chapter, the information provided above is 
sufficient, and we turn now to the subject of polar-
ization. 
13-4 
POLARIZATION A N D P O L A R I Z E D 
LIGHT 
The concept of polarization is used to identify 
the possible vibrational configurations that can be 
associated with transverse waves. From the discus-
sions of the previous sections, we can consider a 
transverse wave in terms of a vibrational displace-
ment vector that lies in a plane perpendicular to the 
direction of propagation and whose components 
along two perpendicular axes are sinusoidal func-
tions of time and position along the wave, with 
some phase angle difference between the two com-
ponents. The various configurations of Table 13-1 
are thus identified as different states of polariza-
tion. For example, the first entry in the table repre-
sents a wave linearly polarized along the y-axis, 
while the last entry represents an elliptically polar-
ized wave. From the analysis of Section 13-2, we 
see that transverse waves will have either elliptical, 
circular, or linear polarization. Furthermore, ellipti-
cal or circular polarization was shown to be the re-
sult of the vector addition of two linearly polarized 
waves. 
A transverse wave for which the vibrational dis-
placement vector points in a direction that varies 
completely randomly with time is called an unpolar-
ized wave since there is no preferred vibrational di-
rection. To produce a linearly polarized wave from 
an unpolarized wave, one needs a device known as 
a polarizer, which will allow only one polarization 
component to pass through it. For example, a trans-
verse wave on a string can be polarized by passing 
the string through the slot between the boards of a 
picket fence. Any vibrational displacement compo-
nent perpendicular to the slot will not pass through, 
while the parallel component will be passed. A sec-
ond slot aligned perpendicular to the first will stop 
any of the vibrational displacement component 
transmitted by the first slot. If the second slot is 
instead aligned parallel to the first, there will be no 
reduction of the vibrational displacement. Since the 
second slot can thus be used to analyze the polar-
ization of the vibrational displacements passed by 
the polarizer, the second slot is called an analyzer. 
For example, if a polarizer and an analyzer are used 
with their axes of transmission at right angles and a 
non-zero vibrational amplitude is observed, then 
the wave must have a longitudinal component of 
vibration. (Why?) 
Let us now consider the case of light waves. 
From the discussion above, if optical devices 
analogous to the picket fence slots (polarizer and 
analyzer) for waves on a string can be devised, then 
the vibrational character of light waves can be 
determined. There are in fact several ways of 

102 
Transverse Waves and Polarized Light 
accomplishing this. The simplest is to make use of 
materials which bear the general name polaroid 
(theh- inventor is Edwin Land). A polaroid material 
transmits nearly all the light with one linear polar-
ization but almost none having a linear polarization 
at right angles to the first. Thus, the axis of polariza-
tion of the polaroid is analogous to the slots in the 
picket fence of the string example. A polarizer and 
an analyzer each made of polaroid with crossed 
axes (polarization axes at 90° to each other) will 
produce complete extinction for any light beam in-
cident on the combination. This is consistent with 
our earlier assertion that light can be understood as 
a purely transverse wave phenomenon. Note, how-
ever, that using more than two pieces of polaroid at 
various angles gives results that have no analogue 
with the string-picket fence case. See Problems 
13-10 through 13-12. 
An unpolarized light beam passing through a 
polarizer will exhibit no variation in brightness as 
the polarization axis of the polarizer is rotated in a 
plane perpendicular to the propagation direction of 
the light. This is because an unpolarized beam has 
no preferred or prominent polarization features. 
The random nature of such a beam requires that the 
magnitude of the vibration component in one direc-
tion be no more (and no less) than that in any other 
direction (transverse to the propagation direction). 
On the other hand, a linearly polarized light wave is 
easily identified by this test since the brightness will 
decrease as the axis of the polarizer goes from an 
alignment parallel to the polarization axis of the 
wave to a perpendicular alignment. 
Consider the following situation. The polariza-
tion axis of an analyzer makes an angle θ with the 
polarization axis of the polarizer. From our previ-
ous discussion, it follows that the amplitude of an 
initially unpolarized light wave that is transmitted 
by the polarizer can be described by a vector A 
aligned parallel to the polarizer axis. Therefore, 
since the analyzer axis makes an angle θ with the 
polarizer axis, the amplitude of the light wave 
transmitted by the analyzer will be reduced to a 
magnitude A' = A cos 0 and will be linearly polar-
ized parallel to the analyzer axis, as shown in Fig-
ure 13-5. 
Quantitative measurements of the reduction of 
transmitted light in a polarizer-analyzer system are 
made by a variety of devices (photocells, for exam-
ple) that determine the intensity of light incident on 
the measuring device. The intensity of a light beam 
is given by the amount of energy that is incident on 
a given cross-sectional area of the detector per unit 
time. In MKS units, the intensity is measured in 
joules per meter squared per second. When such 
measurements are made, it is found that I (the in-
tensity transmitted by the analyzer) is related to Jo 
(the intensity transmitted by the polarizer) by the 
relation 
/ = /oCos'd, 
(13-12) 
where θ is the angle between the polarization axes 
of polarizer and analyzer. This relation is known as 
Malus' law, after its discoverer. We conclude that 
the intensity of a light wave is proportional to the 
square of the amplitude of the wave. That is. 
γ = cos' θ 
lo 
(13-13) 
since A' = A cos 0 from above. It should be noted 
that for an unpolarized wave h is only one-half the 
intensity of the wave incident on the polarizer. 
Incident 
unpolarized 
light 
A cos θ 
I 
A cos θ 
I 
Polarizer 
Analyzer 
Figure 13-5 Polarizer-analyzer effects on a normally incident unpolarized light wave. 

Matter and Polarized Light 
103 
since the directions parallel and perpendicular to 
the polarizer axis should be equally effective in car-
rying the energy associated with the unpolarized 
wave. 
13-5 
M A T T E R A N D P O L A R I Z E D 
LIGHT 
In addition to the use of polaroid materials, other 
methods for producing polarized light can be used. 
One of the simplest techniques is based upon the 
experimental fact that when light is incident upon a 
plane surface the reflected part of the beam will be 
linearly polarized to an extent that varies with the 
angle of incidence. That is, for initially unpolarized 
light one can regard the vibrations as being a combi-
nation of transverse vibrations that are perpendicu-
lar to each other and to the direction of propaga-
tion. It is conventional to identify the plane defined 
by incident ray and the normal to the reflecting sur-
face as the plane of incidence (as in Chapter 6). As 
Figure 13-6 indicates, the component of the polar-
ization parallel to the plane of incidence is largely 
refracted rather than reflected. When the angle of 
incidence has been adjusted so that a W angle re-
sults between the reflected ray and the refracted 
ray, complete linear polarization perpendicular to 
the plane of incidence results. When this is true, the 
angle of reflection Ö, = 90 - or, where Or is the angle 
of refraction. Using SnelFs law and letting the 
index of the reflecting medium be n, we obtain for a 
beam incident in air (n = 1): 
or 
sin Oi = η  sin or = η  sin (90 - Ö,) 
sin di = η  cos 0,. 
Therefore, when 
tan ft = n. 
(13-14) 
total linearly polarized light in the reflected beam 
results. This expression is known as 
Brewster's 
law, and the angle satisfying Eq. (13-14) is called 
the Brewster angle. 
Example 2. What angle of incidence is required 
to produce linearly polarized light by reflection 
using a flint glass plate (n = 1.65)? 
S O L U T I O N 
Using Eq. (13-14), 
so that 
tan ft = 1.65, 
ft «58.8°. 
That the reflected light for the Brewster angle is 
indeed linearly polarized can be demonstrated by 
causing it to impinge upon a second reflecting sur-
face of the same material so arranged that the angle 
of incidence is the same but with the plane of inci-
dence at right angles to that of the first surface. 
When this is done, the light beam is extinguished. 
Plane of 
incidence 
Reflected 
beam largely 
"polarized 
perpendicular 
to plane of 
incidence 
Refracted beam 
largely unpolarized 
Figure 13-6 Polarization of a light wave by reflection. 

104 
Transverse Waves and Polarized Light 
showing that the polarized light is parallel to the 
now-rotated plane of incidence. At the Brewster 
angle, total absorption occurs and no reflection is 
observed. (One could of course use a polaroid anal-
yzer to show the same thing, but it is not necessary 
to do so.) 
Another polarization phenomenon 
was dis-
covered by Bartholinus in the seventeenth century. 
Materials such as calcite, tourmaline, and quartz, 
for example, were found to be capable of separat-
ing a light beam into two components in the follow-
ing sense. If a flat plate of such a crystal were 
placed over a spot of light, two spots could be 
observed at the upper surface. If the plate were ro-
tated about an axis parallel to the normal, one spot 
remained ñxed in position, but the other rotated 
about the ñrst. It is conventional to refer to materi-
als exhibiting this property as doubly refracting or 
birefnngent. The ray giving rise to the non-rotated 
spot is called the ordinary ray, and the ray related 
to the rotating spot is called the extraordinary ray. 
Experiments with these materials have shown 
that there does exist one direction in such crystal-
line materials for which the ordinary and extraordi-
nary rays are not separated. This unique direction is 
called the optic axis. Huygens explained double re-
fraction by assuming that in the direction of the 
optic axis the wavelets associated with the ordinary 
or O ray spread out spherically, while those of the 
extraordinary or Ε ray spread out in ellipsoidal 
form, with the optic axis as an axis of revolution for 
the ellipsoids. For directions not parallel to the 
optic axis, the wavelets of the E- and O-rays travel 
in the same direction but at different speeds. Thus, 
there will be two different indices of refraction for a 
given wavelength. It is common to quote values of 
the indices of refraction for a direction at right ang-
les to the optic axis in which the Ε -ray wavelets 
have a maximum (or a minimum) speed. As Table 
13-2 indicates, the speed associated with the E-ray 
wavelets may be larger or smaller than that of the 
O-rays. 
Although we will not pursue the matter, it is 
worth noting that double refraction has its origin in 
Table 13-2 Indices of Refraction.t 
Πο  
Calcite 
1.6583 
1.4864 
Quartz 
1.5442 
1.5533 
tPor sodium light, λ = 5893 Ä. 
the anisotropy, or non-symmetric, arrangement of 
the atoms or molecules making up the crystalline 
material. Since a light wave propagates in a trans-
mitting material by excitation of the constituent 
electrons, any variation in environment encoun-
tered in various directions might be expected to 
influence the propagation character of incident 
light. 
The value of doubly refracting materials in the 
case of polarization rests on the fact that the ordi-
nary and extraordinary waves are found to be 
linearly polarized at right angles to one another. 
There are two methods for obtaining linearly polar-
ized light that make use of this fact. In the one 
method, a material such as calcite is cut into a 
particular shape, separated into two pieces, and 
cemented together again using a transparent mate-
rial (such as Canada balsam) with an index of re-
fraction that can cause total internal reflection of 
the O-ray while transmitting the E-ray, thus pro-
ducing a linearly polarized beam. Such a polarizer 
is known as a Nicol prism, and the interested reader 
will ñnd additional details in almost any advanced 
optics text. 
The second method relies on the phenomenon of 
dichroism—the ability of some doubly refracting 
materials to strongly absorb one component of the 
light passing through them, either the E-ray or the 
O-ray. The inventor of polaroid, E. H. Land, dis-
covered that needle-like crystals of herapathite 
(quinone iodosulfate), arranged to lie with their 
optic axes parallel and embedded in thin sheets of a 
cellulose material, are extremely dichroic. This dis-
covery and subsequent improvements have made 
possible the development of large surface-area 
polarizers which have had widespread applications 
in science and industry. 

Problems 
105 
P R O B L E M S 
1. Two transverse waves of equal amplitude, frequency, and wavelength propagate simultaneously along 
the same string. They are described by the relations 
jci = A sin 
and 
X2 = A sin 
γ ( ζ - ι; ί ) + 90° 
l ^ y ( z - t ; í ) + 30° 
(a) Using the principle of superposition, find the resulting wave disturbance on the string. 
(b) Then repeat the process for the case Ai = lAi = 2A. 
2. Starting with Eqs. (13-3) and (13-4), carry out the sequence of operations indicated to verify Eq. (13-5). 
3. Two sinusoidal transverse waves propagating in the direction of the ζ -axis have vibrations in the x- and 
y-directions, respectively. The frequency of vibration in the jc-direction is one-half the frequency of 
vibration in the y-direction (fx = jfy). Using the graphical technique illustrated in Figure 13-1, construct 
the Lissajous figures which result when Δ φ  = (a) 0, (b) 30°, (c) 45°, (d) 60°, and (e) 90°. These figures 
are what is observed with an oscilloscope for various Δ ψ . 
4. Repeat Problem 3 for the case 2fx = 3fy and Δ φ  = 90°. Deduce from the result that the frequency ratio 
fxify = rivlriH, where n„ is the number of times the figure touches a vertical edge of the oscilloscope 
screen and ΠΗ is the number of times the figure touches a horizontal edge of the screen. (Warning: As 
illustrated in Problem 3, there are values of Δ φ  for which this interpretation would give incorrect re-
sults. See, for example, the figure in Problem 3 for Δ φ  = 90°.) 
5. A vector A has components A„ Ay, and A^ in the χ -, y- and z-directions, respectively. If a, ft y are the 
angles that A makes with the χ -, y- and ζ -axes, respectively, 
(a) show that cos a = Α χ I A , cos β  = Ay/A, and cos y = AJA, where A = [A/ + Ay' + A^']''', and 
(b) show that cos' a + cos' β  + cos' γ = 1. This result shows that if two directional angles of the vector 
A are known, the third follows from this identity. 
6. Suppose A + Β  -I- C = 0 and also that A=B 
= C. Find the angle between A and B. 
7. Vector A is oriented at an angle θ with respect to vector B. For A = Axi +Ayj and Β  = Bxi, 
(a) show that |A + B| = [(A. + B.)' + Ay']'", and that 
(b) Ax = A cos Θ , Ay =A 
sin Θ . 
(c) Therefore, show also that 
|A + B| = [A' + β ' + 2AB cos 0]'". 
This relation is a slightly rearranged form of the law of cosines for trigonometry. In texts on vector 
analysis, it is shown that much of plane and solid geometry and trigonometry follows readily from 
vector operations. 
8. Suppose vector A lies in the x-y plane and makes an angle of 45° with the χ -axis. Give the magnitude 
and orientation of a vector Β  such that |A + B| = |A-B|. 
9. Given A = 5i-10j + 8k, 
and 
Β  = li + 20j + 12k. 
(a) Find the magnitude of A + B, A - B , and 2A + B. 
(b) Find the angles A makes with the χ -, y- and z-axes, respectively. (See Problem 5.) 
10. Two polaroids with crossed axes of polarization are placed in a beam of light. A third polaroid is placed 
between the other two with its axis of polarization at an angle θ with the axis of the first polaroid. If the 
incident light is unpolarized and has an intensity Jo, show that the intensity of the light transmitted by the 
third polaroid is given by 
J = |/osin'2a 

106 
Transverse Waves and Polarized Light 
'2(η ο -Η Ε ) 
2{NE-noY 
where the denominator is to be positive. This, of course, depends on whether ME > no (like quartz) or 
HE < no (like calcite). In a half-wave plate, there is one-half wave more of one ray in the plate than there 
is of the other ray. Thus, the phase difference Δ φ  between the ordinary and extraordinary rays is π  
when the light has traversed the plate between two polaroids. 
19. A half-wave plate of calcite (for sodium light) between two polaroids is placed in an unpolarized beam 
of sodium light. The plate is oriented so that the optic axis is at 45° to the polarization axis of the first 
polaroid. If the polarization axis of the second polaroid is parallel to the axis of the first no light will be 
transmitted by the system. 
(a) Explain why. Hint: See Problem 17. 
(b) What will be the situation when the two polaroids are crossed (but the optic axis of the half-wave 
plate remains at 45° to the axis of the first polarizer)? 
20. Repeat Problems 16 and 17 for a quartz quarter-wave plate. 
11. (a) For the situation described in the previous problem, determine the angle θ which provides the 
maximum transmitted light intensity, 
(b) Determine the maximum transmitted light intensity. 
12. A stack of 7 polaroids is assembled with the axis of polarization of each polaroid making an angle of 15° 
with the one preceding it. Thus, the first and last polaroids are crossed. If unpolarized light of intensity Jo 
is incident on the stack, what is the intensity of the transmitted light? 
13. In Problems 10 to 12, it was assumed that the polaroids are ideal—that is, that light is transmitted only if 
its polarization axis is parallel to the polarization axis of the polaroid. In practice, it is more nearly true 
that a polaroid transmits a fraction 
of the incident light if the polaroid axis and the polarization axis 
are parallel and a fraction 
is transmitted when the two axes are perpendicular. (For an ideal polaroid, 
p^=\ 
and 
= 0.) Unpolarized light of intensity h is incident on a pair of real polaroid sheets that have 
an angle θ between their polarization axes. Show that the transmitted intensity is given by 
/ = 5(p'+ s') cos' Ö + p ' s ' sin' Θ . 
Show also that this reduces to the ideal expression discussed previously. 
14. A beam of light in water (n = 1.33) is incident on ñint glass (n = 1.65). For what angle of incidence will 
the reflected light be linearly polarized? 
15. (a) Sunlight is reflected from the smooth surface of a pond (n = 1.33). At >yhat angle relative to the 
vertical should one view the surface through a polaroid if surface glare is to be eliminated most 
effectively? 
(b) Should the polarization axis of the polaroid be vertical or horizontal? 
16. Light from a sodium lamp (λ = 5893 Ä) is incident normally on a rectangular plate of calcite which is cut 
so that the optic axis is parallel to the faces of the plate. Determine the wavelengths λο , AE of the 
ordinary and extraordinary rays in the plate. 
17. When the plate of Problem 16 is cut to the proper thickness, the number of wavelengths of the ordinary 
ray in the plate will differ from the number of wavelengths of the extraordinary ray in the plate by 
one-quarter of a wavelength. Such a plate is called a quarter-wave plate. Determine the thickness of a 
quarter-wave plate of calcite for sodium light. 
18. Show that the thickness of a half-wave plate of a birefringent material for light of wavelength λ (in 
vacuum) is given by 
. 
A 
A 
t =7^ 
- T 
or 

14 
Particle 
Motion 
14-1 
Introduction 
107 
14-2 
Displacement 
108 
14-3 
Velocity and Acceleration 
108 
14-4 
Uniformly Accelerated Motion 
110 
14-5 
112 
Frames of Reference 
14-6 
Relative Velocity and the Galilean 
Transformation 
112 
14-1 
INTRODUCTION 
In this chapter, we develop a vocabulary for dis-
cussing motion in a quantitative manner. Motion is 
a concept with an imprecise meaning for us in terms 
of everyday experience. If we are to gain an under-
standing of the connection between motion and its 
physical causes, it is first necessary that we develop 
a clearer picture of motion in terms of geometric re-
lationships and time. Such a study of motion (exclu-
ding its causes) is known as kinematics, while the 
more inclusive study (which correlates motion and 
its causes) is called dynamics. 
The study of kinematics is made simpler initially 
by discussing the motion of a fictional or ideal ob-
ject, the point-particle. Such a particle is com-
pletely described kinematically if the location of 
the point in space is known as a function of time. 
By contrast, a real, or physical, particle occupies a 
volume in space, the dimensions of which as well as 
the location of which may vary with time in a 
complicated manner. The concept of a point-
particle is a simplification that is not only useful but 
is also a quite reasonable approximation for real 
situations in which the object in question has di-
mensions that are of negligible extent by compari-
son with other objects in its environment. For 
example, one may regard the motion of a Tiros 
satellite about the Earth as that of a point-particle 
because of the extreme disparity in dimensions. 
Furthermore, the point-particle concept can be 
useful for discussing a rigid body of non-negligible 
dimensions. In this case, the object is regarded as 
an aggregation of a large number of point-particles 
that move in a correlated way because of influences 
whose nature will be discussed at a later point in 
our study of physics. To see that this point of view 
is plausible, one need only recall that any physical 
object is composed of one or more extremely small 
entities known as molecules, each of which can be 
said to approximate the ideal point-particle. A more 
familiar example of a collection of point-particles 
moving together is provided by a football that has 
been kicked off and is in motion. To the receiver 
watching it move toward him, it is hrelevant to 
observe that the football is in reality many many 
molecules of "pigskin" bound together in such a 
way that they entrap and thus compel a great 
number of air molecules to accompany 
the 
"pigskin" in its motion downfield. To the receiver, 
it is only necessary that he control his motion so 
that his hands arrive at a given limited region of 
space at the same instant that the collection of "pig-
skin" and air molecules arrive at that same location. 
With these considerations as a motivation, we 
adopt the point-particle concept as a means of ac-
107 

108 
Particle Μ ο ί/ο η  
quiring a clear understanding of motion, deferring 
until later a consideration of the complications that 
can accompany the motion of a large aggregate of 
correlated (essentially) point-particles when the 
aggregate is of non-negligible dimensions. 
14-2 
D I S P L A C E M E N T 
In Chapter 13, we indicated that a point in space 
is uniquely located if its three spatial coordinates 
are given with respect to a particular origin of coor-
dinates. The vector specifying this point, called the 
position vector, in rectangular coordinate form is 
written as 
r = xi + yj + zk 
(14-1) 
where i, j , and k are the vectors of unit length in the 
JC-, y-, and z-directions respectively as in Chapter 
13, and jc, y, and ζ  represent the multiples of the 
unit lengths in the three directions, respectively. 
Let us consider a particle (henceforth in this 
chapter particle will mean point-particle) whose 
position is initially given by 
r, = x,i-l-y,j + zik, 
and which is then shifted to a new position given by 
r2 = ac2Í+y2J + Z2k. 
Such a particle is said to have experienced a dis-
placement, defined by the vector equation 
d = Γ2 - r, = (X2 - xi)i + (y2 - yi)j + (Z2 - z,)k. (14-2) 
From the laws of vector algebra (Chapter 13), the 
magnitude of d is given by 
|d| = |(d. d)'"| = |[(X2 - X . ) ' + (y2 - y.)^ + (z2 - zm 
(14-3) 
and the direction cosines of the displacement with 
respect to the coordinate axis are obtained by form-
ing the dot or scalar product of d with the respec-
tive coordinate axes. Thus, for example, the angle 
between d and the χ -axis is given by the relation 
or 
d-i = |d||l| COS (d,i) = ( x 2 - x i ) 
COS (d, l) = 
, 
(14-4) 
with similar expressions for the other two coordi-
nate axes. 
It should be stressed here that the displacement 
is not the same as the distance traveled by the par-
ticle in getting from ΓΙ to Γ2. The distance traveled is 
a scalar quantity that is vitally dependent upon the 
path taken in moving from FI to Γ2, while the dis-
placement is a vector quantity giving the magnitude 
and direction of the separation distance between ΓΙ 
and Γ2. To see the distinction, consider a particle 
that was originally located in New York City, that is 
moved to Chicago via turnpikes, and returned to its 
starting point by some alternative route. For this 
situation, the displacement is zero since the initial 
and final positions are identical. The distance 
traveled, however, is non-zero. Furthermore, the 
actual distance traveled cannot be given without 
identifying the complete route taken. 
Example 1. A particle moves along a circular 
path of radius R from point 1 to point 2 and on to 
point 3 as in Figure 14-1. Find the displacement and 
the distance traveled in moving from point 1 to 
(a) point 2, and (b) point 3. 
Figure 14-1 
System diagram for Example 1. 
SOLUTION 
Γι = 1?ί = Γ3, Γ2 = Α ]. Therefore, for case (a), 
d = ( r 2 - r , ) = Ä(j-i) so that d = Ä((l)' +(-1)')'" 
= V2Ä, and the distance traveled is given as s = 
ttR 12 since it represents one-fourth of the circum-
ference of the circle. 
For case (b); d = 
- ΓΙ = 0, and the distance 
traveled is s = 27γ1?, the circumference of the circle. 
14-3 
VELOCITY A N D A C C E L E R A T I O N 
Although changes in location are adequately de-
scribed by the displacement, one frequently is more 
interested in the manner in which the displacement 
varies with time or the extent to which the change 
in displacement with time is itself a function of 

Velocity and Acceleration 
109 
time. Accordingly, we define the velocity of a par-
ticle as the time rate of displacement. It is a vector 
quantity, since displacement is a vector quantity 
while time is a scalar quantity. 
If 
the particle 
displacement 
d = Γ2 - r, = Δ γ 
occurs in the time interval Δ ί = Í2 - ii, where At is 
finite, then the average velocity is defined to be 
Δ γ ^ U 2 - 
+ ( y 2 - 
y i ) i 
(22 - zi)k 
"Δ ί 
Í 2 - Í 1 
so that 
where 
V = Vxi-\-dy} 
+ tJzk, 
X2-X\ 
, 
V. = - — - , etc. 
ti-ti 
(14-5) 
(14-6) 
The magnitude of ν  is found by application of the 
scalar product to be 
\y\ = (vx'-^vy' 
+ v.y\ 
(14-7) 
Example 2. Suppose that the displacement in 
Example 1(a) takes place ia 10 seconds. Then the 
average velocity will be 
Γ 2 - Γ , ^ 
Í 2 - Í , 
10^^ 
'^' 
and the magnitude of the average velocity is 
Vir 
"
M
m
-
10 
If the time interval during which the displace-
ment occurs becomes very short, it is usually found 
that the magnitude of Ar/At approaches a constant 
value as a result of the physical fact that as the time 
interval decreases so also does the displacement. 
Thus, if we reduce the time interval until subse-
quent reductions do not change the value obtained 
for ArlAt, we have effectively determined the vel-
ocity at a particular instant of time within the inter-
val Í 2 - ii. This limiting process is, of course, more 
rigorously presented in the differential calculus in 
defining the process of differentiation. Thus, one 
writes that the instantaneous velocity ν  is defined 
by the relation 
Δ γ 
dr 
V = l i m 
= 
-T
-
Δ r^Δ í 
dt 
(14-8) 
Graphically, the distinction between ν  and ν  is 
clear. Figure 14-2 shows a schematic graph of r 
versus t for one-dimensional motion (along the x-
axis, for example). From our discussion above, ν  is 
represented by the ratio of the chord length be-
Figure 14-2 Schematic graph of one-dimensional 
displacement as a function of time. 
tween points 1 and 2 and the time interval Δ ί = 
Í2 - Í1. On the other hand, ν  is the value of the slope 
of the curve at a particular instant (or at least in a 
very small time interval in the neighborhood of that 
instant), for example, at point 3. At any interval 
about point 3 for which the chord has the same 
slope as the curve at point 3, the average velocity 
and the instantaneous velocity at point 3 will be the 
same. For example, if r is a linear function of t 
(Γ = Γο  + ο ί), then 
- ^ Ar ^ Γ2 - Γι ^ (γο  4- cti) - (γο  + er i) ^ 
Δ ί Í 2 - Í . 
Í 2 - Í 1 
c. 
But 
dr 
d(ro + cí) 
dt 
= c. 
Therefore, the average velocity and the instantane-
ous velocity are identical when the position of a 
particle varies linearly with time. On the other 
hand, if r is not a linear function of time but is 
known either as an analytic function of time or as a 
table of data relating r and t values, then ν  and ν  
can be found either by analytic or numerical (or 
graphical) techniques in a straightforward manner. 
It also should be noted that the speed of a particle 
in general is not equal to the magnitude of the vel-
ocity, since the speed is defined to be the distance 
traveled divided by the time, a scalar quantity 
which is path dependent, while the velocity is a 
path independent vector quantity. 
There are abundant examples in our daily lives of 
motion for which the velocity is non-constant. 
Thus, the scene available on any busy city street 

110 
Particle Motion 
will show several "particles" whose velocities are 
changing in various ways (to "beat the light" or 
avoid a dented fender, for example). To discuss 
such motions, we define the acceleration of a par-
ticle in a manner analogous to that used in defining 
the velocity. Thus, the acceleration is defined as a 
vector quantity that gives the change of velocity 
with respect to time. The average and instantane-
ous acceleration, respectively, are defined as 
and 
Δ ί 
Í 2 - Í , 
Δ ί 
\ Δ ί 
,. 
Δ ν  
d\ 
d'r 
a = hm - Γ - = -77 = -jTT. 
Δ ί-.ο Δ ί 
dt 
dt 
(14-10) 
Example 3. A particle obeys a displacement-
time relationship given by r = (5-l-2i-f 3í')i-l-
(4í-6í'+3í^)j, where r is in meters and í is in 
seconds. 
(a) Find the velocity and acceleration at í = 0 
seconds. 
(b) Find the velocity and acceleration at í = 2 
seconds. 
S O L U T I O N 
dr 
(a) v(í) = ~ - = [(2 + 6í)i + (4-12í+9í')j] m/sec, 
a(i) = ^ = [6i + (- 12 + 18i )j] m/sec'; 
therefore, 
v(0) = (2i 
4j) m/sec, |v(0)| = iVS m/sec, 
a(0) = (6i - 12i) m/sec', |a(0)| = ó Vs m/sec'. 
Similarly, for 
(b) v(2) = (14i + 16j) m/sec, |v(2)| = 2 ν Π3 m/sec, 
a(2) = (6i + 24j) m/sec', |a(2)| = ó Vl? m/sec'. 
The average velocity and acceleration between í = 
0 and Í = 2 seconds are found to be 
= (Si + 8j) m/sec, 
|v| = 8V2 m/sec, 
and 
á = ^ ^ í ^ ^ = i[14i+16j-(2l4-4j)] 
= (6i-f6j) m/sec', 
|ä | = 6V2 m/sec'. 
The reader can determine the appropriate direction 
of the various vector quantities by means of the 
scalar product as in Eq. (14-4). It should be noted 
that the motion in the x- and y- directions had dif-
ferent characteristics. The acceleration in the x-
direction was constant so that the velocity in the 
JC-direction was a linear function of time, while the 
acceleration in the y-direction varied linearly with 
time so that the y-velocity was a quadratic function 
of time. This is a result that is quite general in that 
the type of motion in one direction can usually be 
discussed independently of the motion in either of 
the two remaining directions in space, with the total 
motion being obtained by vector addition of the 
motion in the three space directions. Thus, for ex-
ample, 
^dv, 
dv,. dv, 
dt^^ dt^^ dt^ 
W^W^^dP^' 
14-4 
U N I F O R M L Y A C C E L E R A T E D 
M O T I O N 
Let us now consider the kinematical description 
appropriate for a particle for which the acceleration 
is constant. Then, since 
d^ 
^ 
. 
a = 
= constant. 
it follows that 
Therefore, 
_ 
V-Vo 
V = Vo + ai. 
(14-11) 
To determine the displacement in a time i, we first 
find the average velocity in the interval (i - 0) to be 
the average 
- 
v + vo 
Vo + ai+vo 
, ai 
v = -
^ 
= 
2 
= Vo + y . 
(14-12) 
(Equation (14-12) simply states that the average vel-
ocity for a particle undergoing constant accelera-
tion for a time í is given by the initial velocity plus 
the acceleration times one-half of the total time in-
terval ( Í - 0 ) . ) Next, the displacement during the 
time interval (i - 0) is found from the relation 
d = v(i - 0) = voi + ^ai'. 
(14-13) 
Since 
d = r - r o , 

Uniformly Accelerated Motion 
111 
we finally have 
Thus, 
r = Γο  + ν ο ί + | a í ' . 
(14-14) 
Alternatively, we can relate the initial and final 
velocities and the displacement by eliminating t in 
Eqs. (14-11) and (14-13). This is most readily seen in 
component form. Thus, for the χ -components, we 
have 
Vx = 
Vox 
+ 
ü xt 
and 
d x = X - X o 
= 
Voxt-^-Qxt 
, 
from which 
t = 
Vx - 
Vox 
ü x 
and 
. 
_ 
(Vx-vox) 
1 
(Vx-VoxY 
dx 
=VQX 
-^^cix\—-—I 
ü x 
2 
\ 
Qx 
J 
Vx-Vlx 
2ax 
' 
or 
Vx^vlx 
+2axdx. 
(14-15) 
Similar relations in the y- and z-directions combine 
vectorially to yield 
t ; ' = r o ' + 2 a d . 
(14-16) 
Returning to Eq. (14-13), we see that the displace-
ment is the sum of two terms. The first, voi, repre-
sents the displacement due to a constant velocity vo. 
The second term, (jat)t, 
represents the displace-
ment due to a linear change in the velocity from Vo 
to Vo + ai in the time interval (i - 0). A graph of ν  
versus t for one-dimensional motion with constant 
acceleration is shown in Figure 14-3. The two terms 
above correspond to the areas labeled 1 and 2 under 
the linear curve, Voi and k(vo + a i - v o ) , respec-
tively. Hence, d = v o í + 5 a í ^ as we have already 
found. 
This graphical interpretation is more generally 
applicable than just to the case of uniformly accel-
erated motion. To see that this is true, consider the 
situation illustrated in Figure 14-4(a). Here, a is not 
constant since ν  is a non-linear function of t. How-
ever, one can regard the curve as a sum of curve 
segments for time intervals small enough to regard 
V as a linear function of time in that interval as 
shown in Figure 14-4(b). Then the total displace-
ment is the sum of the displacements for each small 
interval during which a assumes a constant value. 
d « Σ  d. = Σ  [ν(ί. + Δ ί) + v(í,)]y. 
(14-17) 
It is at least reasonable that greater accuracy will 
result if Δ ί is made increasingly smaller. In the limit 
as Δ ί becomes vanishingly small, so that the 
number of terms Ν  in the sum becomes infinite, the 
expression 
d = lim2[v(í.+Δ í) + v(í.)]^^ 
f-^ i=0 
-¿ 
becomes identical to the definition of the definite in-
tegral, 
d = |^ v(í)dí. 
(14-18) 
Therefore, if the velocity is known as an analyti-
cal or graphical function of time, the displacement 
Vo + at 
Vo + at 
V 
Vo 
\ \ \ \ ! 
t 
t 
Figure 14-3 Velocity versus time for uniformly accel-
erated motion. 
t 
(a) 
Figure 14-4 Velocity versus time for non-uniformly 
accelerated motion. 

112 
Particle Motion 
can be obtained as the integral of the velocity over 
the appropriate time interval (or, equivalently, by 
determining the area under the velocity-versus-time 
graph in the units of the graph). The reader might 
wish to show by an analogous argument that for an 
arbitrary a-versus-i relationship, \(t) can be found 
by the relation 
v ( i ) = 
a(í)dí+Vo 
(14-19) 
or, equivalently, by determining the area under the 
a versus t curve in the units of the graph. 
14-5 
F R A M E S O F R E F E R E N C E 
In Sections 14-3 and 14-4, we defined the quan-
tities displacement, velocity, and acceleration with 
respect to a given system of coordinates. Thus, 
when we write r = jc i -I- y j -I- zk, it is understood that 
X, y, and ζ  are the spatial coordinates of a particle 
with respect to a given origin (for example, the 
center of the Earth). It is clear that the values for x, 
y, and ζ  of the same point in space will be different 
if the point is referred to a new origin of coordi-
nates (such as a particular point on the Earth's sur-
face). Should observations be made of the particle 
by observers at rest at two different locations, they 
can correlate their information only if they are able 
to translate (both literally and figuratively!) their 
measurements from one origin to the other or vice 
versa. Such a translation corresponds to a transfor-
mation from one frame of reference to another. To 
be specific, suppose an observer is at rest at the 
origin of a system of coordinates (a reference 
frame) labeled Si, and another observer is at rest at 
the origin of another reference frame labeled Si. 
Suppose, further, that the origins of the two refer-
ence frames are at rest and separated by a displace-
ment Γ21 (the subscripts are read as: the location of 
the origin of S2 with respect to the origin of Si). 
Thus, it follows that Γ2Ι = - ΓΙ2. N o w a particle at 
point Ρ  in space would be described in Si as having 
a position vector Γ Ρ Ι , and in S2 as having a position 
vector Γρ2. As illustrated in Figure 14-5, the transla-
tion equation relating these position vectors is cor-
rectly given as 
Γ ρ ι=Γ ρ 2 + Γ2ΐ. 
(14-20) 
Example 4. Two students at opposite ends of a 
laboratory table are at rest with respect to the table 
when they observe a flash of light across the room 
Si 
Si 
0 
Figure 14-5 Translation of a position vector from one 
reference frame to another. 
in a direction parallel to the length of the table. One 
student reports a distance of 20 m to the point 
where the flash occurred, while the other observed 
a distance of 17 m. What is the length of the labora-
tory table? 
SOLUTION 
From the information given, one can write Γ Ρ Ι = 
20im, Γ ρ2= 17im. Therefore, from Eq. (14-20), 
20im= 17im + r2i 
or 
r2i = 3im, 
so that the laboratory table is 3 m in length. Notice 
that we could have alternatively written rpi = 17i m, 
Γρ2 = 20i m, which would lead to Γ21 = - 3i m. The 
physical interpretation here is still that the labora-
tory table is 3 m in length, but in this case the ob-
server in Si is nearer to the flash of light than the 
observer in S2, and the observer in S2 is therefore 
3 m behind him. (One would not conclude that the 
laboratory table was 3 m less than zero length!) In 
similar situations, one should always consider the 
physical situation involved to avoid any apparent 
mathematical paradoxes. 
14-6 
RELATIVE VELOCITY A N D T H E 
G A L I L E A N 
T R A N S F O R M A T I O N 
Now we consider the problem of transforming 
observations made by an observer in Si if the ob-
server in S2 is in motion relative to him. To be 
specific, suppose that at some instant of time the 
origins of Si and S2 coincide, and that the two ob-
servers synchronize their watches. The reference 
frame S2 and the observer at the origin of S2 are 
moving with a constant velocity ν  relative to 
another observer at the origin of Si. A somewhat 
frivolous example of such a situation could be ap-
proximated by two students in a laboratory cor-

Relative Velocity and the Galilean Transformation 
113 
ridor. One of the students is at rest while the other 
is moving down the corridor on a skateboard at a 
constant velocity ν  relative to the first student. At 
the instant the two are side by side, they synchron-
ize their watches to read zero. We now assume that 
the motion does not alter the synchronization of the 
watches. (We shall see in the next chapter that this 
assumption breaks down if ν  is comparable in mag-
nitude to the velocity of light.) It then follows that 
the separation of the two origins is given by 
or 
r,2 = -vi. 
(14-21) 
To continue the example, suppose that at a time t 
the two students simultaneously observe a station-
ary rabbit down the corridor. The situation is de-
picted in Figure 14-6. It is clear that in this case 
Γ Ρ Ι = Γρ2-Ι-Γ2ΐ = Γρ2 + VÍ 
or, conversely 
Γρ2 = Γ ρ , - ν ί . 
(14-22) 
(14-23) 
In the diagram, we imply that the motion is one-
dimensional (along the corridor), but we could con-
sider a more complex situation (for example, a 
hummingbird hovering at a definite point in space 
along the corridor while the student on the 
skateboard moves at constant speed diagonally 
along the corridor). In the more complicated case, 
we would simply decompose the vector motion into 
its spatial component equations and consider them 
separately. Thus, Eq. (14-22) would become the 
three equations 
Xp\ = Xp2-^ 
Vxt, 
ypi = yp2-+-ü yí, 
Zpi = 
Zp2+V,t. 
s, 
m 
vf 
m 
Tr 
* J n 
s, 
o 
i 
o 
1 
Figure 14-6 Diagram for the system described by Eq. 
14-22. 
We see, therefore, that there is no essential in-
crease in conceptual difficulty if the relative motion 
is not one-dimensional. 
To see how the relative motion of the two obser-
vers will affect a kinematical description, let us now 
assume that the rabbit moves from a point observed 
in Si given by rpi(ii) at time ii to a point rpi(Í2) at 
time Í2. The corresponding locations as observed in 
S2 will be found by means of Eq. (14-23). Now we 
ask: What relationship is there between the average 
velocities of the rabbit as measured by each ob-
server? From Eq. (14-5), the average velocity ü i as 
measured in Si is given by 
Ü P . = Arpi^Arpi(Í2)-Arpi(ri) 
Δ ί 
Í 2 - Í 1 
(14-24) 
From Eq. (14-22), this is equivalent to 
- 
_ (rp2(Í2) + VÍ2) - (Γ ρ2(ίι) - vil) 
Ö T^To 
^Γ ρ 2(Ί2)-Γ ρ 2(ί ΐ ) 
„ ^ Δ Γ Ρ 2 
( Í 2 - í l ) 
Δ ί 
= 
Ü P 2 + V 
or, conversely, 
Ü P 2 = 
Ü P 1 - V . 
(14-25) 
(14-26) 
Once again in the limit as At becomes vanishingly 
small, the average velocities in these equations may 
be replaced by the corresponding instantaneous 
velocities. Thus, we find that the observers obtain 
results that differ by a velocity representing their 
relative motion. Do they also obtain differing re-
sults for average and instantaneous accelerations 
observed for the rabbit? The answer is no and fol-
lows from the use of Eq. (14-9): 
- 
_Upi(Í2)-Upi(íi)_AUpi 
_(UP2(Í2) + V ) - ( U P 2 ( Í . ) + V) 
Í 2 - Í I 
^ AUp2 
Δ ί '• = ap2. 
(14-27) 
Similarly, in the limit as Δ ί ^ 0 , we obtain the result 
that 
ap, = ap2. 
(14-28) 
These results may be summarized by the state-
ment that for reference frames that differ by a 
constant relative velocity the acceleration remains 
invariant to (is independent of) a transformation 
from one reference frame to the other, while the 

114 
Particle Motion 
position and velocity transform according to Eqs. 
(14-22) and (14-25) or, conversely, Eqs. (14-23) and 
(14-26). These transformation equations are known 
as the Galilean transformation equations. They are 
satisfactory only if the magnitudes of the velocities 
involved are small compared to the speed of light. 
In the next chapter, we derive the transformation 
equations known as the Lorentz equations required 
for high velocity situations. Since there is ample ex-
perimental evidence that the Galilean transforma-
tion equations are adequate for low velocities, it is 
reasonable to anticipate that the Lorentz transfor-
mation equations must possess a form that reduces 
to the Galilean equations whenever the velocities 
are small in magnitude compared to the speed of 
light. This expectation, in fact, will assist us in 
finding the form of the Lorentz equations. 
We conclude this chapter with a final example of 
relative motion, which also proves to be useful in 
the next chapter in describing the Michelson-
Morley experiment. 
Example 5. Consider a stream of width w, which 
has a uniform velocity (relative to the shore) of 
magnitude v. Two swimmers A and Β who swim at 
a velocity u relative to the water are to engage in a 
contest in which A swims directly across the 
stream and back, while Β swims downstream a 
distance w and back again (see Figure 14-7). Who 
wins the contest, and what is the difference in times 
required by the two in completing their round trips? 
SOLUTION 
Swimmer B's velocity downstream relative to 
the shore is given by 
VBS = VBW + V W S = U + V, 
and his upstream velocity is 
VBS = U - V. 
Therefore, the time required by Β is given by 
itotalB — 
w 
w 
2 ^ u 
Μ  + Γ 
U-V 
For swimmer A, it is not sufficient to swim 
straight across the stream. If he did so, the move-
ment of the water would carry him downstream, re-
sulting in his covering a distance greater than 2w. 
He must, therefore, head upstream at an angle θ 
(see Figure 14-8) such that the component of his 
Figure 14-7 System diagram for Example 5. 
velocity (relative to the water) parallel to the shore 
is equal in magnitude but opposite in direction to 
the stream velocity relative to the shore. If he does 
so, his speed perpendicular to the shore is reduced 
to 
UAW^ = u cos Θ . 
Since u sine = v, it follows that 
UAW. = «(1 - sin' 0)·" = w(l -
As a result, his time of crossing one way will be 
ti = w 
UAW/ 
w 
u 
/ 
On the return trip, it will again be necessary to head 
upstream by a similar amount for the same reason, 
so the return time would be identical to i,. There-
fore, the total time for A is 
2 ^ u 
Now, the expression 
Figure 14-8 Velocity vector diagram for swimmer A 

Problems 
115 
Therefore, we conclude that swimmer A will take 
less time than B. The difference in elapsed times 
will be 
Δ ί = Í B - 
ÍA 
= 
2 ^ 
u 
- m 
1 
-iíí 
[Ϊ 72- 1 
(14-29) 
Finally, consider the case for which (v/uf<t\. 
Re-
call that the binomial expansion is written 
(l±jc)'' = l±njc + η (η -\)χ ' 
2! 
. n(n-l)(n-2)jc^ . 
^ 
3! 
^" 
For X < 1 , 
(\±xY 
^\±nx. 
In our case, the term [1-(r/w)']"'^' becomes ap-
proximately 
so that 
wv 
(14-30) 
While this result is not realistic for swimmers in a 
river, we will make use of Eq. (14-30) in our discus-
sion of the Michelson-Morley experiment. 
P R O B L E M S 
1. At time Í = 0, a body is traveling due east with a linear speed of 32.5 ft/sec. At time t = 4 seconds, the 
body is traveling due north with a linear speed of 32.5 ft/sec. What is the magnitude of the average 
acceleration experienced by the body? 
2. At the instant a motor patrolman waiting at an intersection starts ahead with a green light, an automobile 
passes him with a constant speed of 50 ft/sec. The patrolman accelerates at 10 ft/sec'. How far from the 
intersection will the patrolman overtake the automobile? 
3. A stone is dropped from the top of a cliff 256 ft in height and, at the same instant, another stone is 
projected upward from the bottom level with a speed of 96 ft/sec. At what height will the two stones 
meet, or pass each other? 
4. A dense ball is projected vertically upward at 30 m/sec. It rises higher than an observation point that is 
15 m above its starting point. 
(a) Compute the magnitude of its velocity as it passes the observation point on the way down. 
(b) Compute the time from the start until the ball passes the observation point on the way down. 
5. An automobile is initially traveling with a velocity of 40 ft/sec north. Two minutes later the automobile 
has a velocity of 120 ft/sec south. Determine the average acceleration experienced by the automobile. 
6. A ball shot vertically upward from the edge of a vertical cliff travels at 50 ft/sec when it reached a point 
25 ft below its starting point. 
(a) What was its initial velocity? 
(b) How high did it rise above the cliff? 
7. A ball is projected upward from the ground with a velocity of 60 ft/sec. 
(a) How high will it rise? 
(b) Find its velocity and height 3 seconds after it left the ground. 
8. A ball released to roll down an inclined track has constant acceleration of 30 cm/sec'. When it passes 
one mark on the track, the ball has a velocity of 40 cm/sec, and when it passes a second mark, it is going 
130 cm/sec. Compute: 
(a) the average velocity of the ball between the two marks, 
(b) the distance between the two marks, and 
(c) the time required for the ball to go from the first mark to the second mark. 

116 
Particle Motion 
9. It takes 30 seconds for a car to travel half way around a circular track from the eastern-most point to the 
western-most point of the track. The radius of the track is 1500 ft. 
(a) Determine the car's displacement during this 30 second interval. 
(b) Determine the car's average speed during this interval. 
10. An automobile accelerates uniformly to 15 ft/sec in 10 seconds. Compute: 
(a) the average speed, 
(b) the average acceleration, and 
(c) the distance traveled in the 10 seconds. 
11. A car having an initial velocity of 8 m/sec south experiences, by the application of the brakes, a constant 
deceleration of 2 m/sec'. Determine: 
(a) the velocity after 3 seconds and 
(b) the distance traveled and the average velocity during this time. 
12. A raindrop falls at a rate of 50 ft/sec, and is at the same time blown to the east at 10 ft/sec by the wind. If 
a car driving due east at 80 ft/sec has the drop graze its side window, find the angle its trace will make 
with the vertical. 
13. A ball is thrown vertically upward from the ground, and a student gazing out of the window sees it 
moving upward past him at 16 ft/sec. The window is 32 ft above the ground. 
(a) How high does the ball go above the ground? 
(b) How long does it take to go from a height of 32 ft to its highest point? 
(c) What was the initial velocity given to the ball? 
(d) Find its velocity and acceleration one-half second after it left the ground. 

15 
Special 
Relativity 
15-1 
Introduction 
117 
15-2 
The Michelson-Morley Experiment 
15-3 
Einstein's Special Theory of 
Relativity 
119 
15-4 
The Lorentz Contraction 
120 
117 
15-5 
Time Dilation 
120 
15-6 
Relativistic Addition of Velocities 
121 
15-1 
INTRODUCTION 
In this chapter, we briefly discuss the modifica-
tions required when transformation equations are 
needed for situations where the magnitudes of the 
velocities involved are comparable to the speed of 
light. As was asserted in the previous chapter, the 
Galilean transformation equations are then no 
longer adequate. The need for modifications was 
recognized as a result of the Michelson-Morley ex-
periment, which is described in the next section. In 
order to "explain" the results of this experiment, 
Hendrik Lorentz and George Fitzgerald indepen-
dently proposed an ad hoc solution by suggesting 
that moving objects suffer a contraction in their 
length in the direction of motion. Such a contrac-
tion would predict results in agreement with the 
Michelson-Morley experiment, but such a bizarre 
postulate was scarcely satisfactory philosophically. 
A more fundamental approach was provided by 
the special theory of relativity developed in 1905 by 
Albert Einstein. As a result of this theory, the de-
sired transformation equations, known as the Lor-
entz transformation equations, are readily derived 
as in Section 15-3. The remaining sections of the 
chapter are devoted to an examination of the 
kinematical consequences of the Lorentz transfor-
mation and the extent to which they are at variance 
with our usual experience. 
15-2 
T H E M I C H E L S O N - M O R L E Y 
E X P E R I M E N T 
In the 1880s, the wave theory of light (and, in 
fact, of electromagnetic radiation in general) was 
firmly established due to the experimental work of 
Michael Faraday (and later of Heinrich Hertz) and 
the theoretical work of Clerk Maxwell. A funda-
mental part of the wave theory was the assumption 
that an extremely rigid, yet transparent and highly 
permeable medium existed known as the luminifer-
ous ether (for brevity, we refer to it henceforth as 
the ether), which served as the means of transfer of 
solar and stellar radiation. Since sound waves do 
not propagate without air or other molecules to 
transmit the disturbance, it was argued by analogy 
that electromagnetic waves also required a medium 
for their transmission. 
Therefore, the ether was assumed to occupy all 
of space, and the Sun was assumed to be at rest in it 
while the planets moved freely through it. As a 
result, the ether (if its existence could be demon-
strated) could be regarded as a unique or absolute 
frame of reference to which all others might be 
referred. Michelson first and then Michelson and 
Edward Morley sought to demonstrate by interfer-
ence techniques that the ether did exist and possess 
the characteristics ascribed to it. The failure of 
their experiments to indicate its existence led first 
117 

118 
Spec/a/ Relativity 
to great consternation, followed by the efforts of 
Lorentz, Fitzgerald, and Henri Poincaré to theoret-
ically explain the null results. 
Figure 15-1 is a schematic representation of the 
Michelson-Morley experiment. A Michelson inter-
ferometer (discussed in Chapter 11, Figure 11-5) is 
arranged so that one of the light paths will be at 
right angles to the ether through which the Earth is 
moving, while the other light path is parallel to the 
movement through the ether. (From the point of 
view of the interferometer, the ether is streaming 
through the interferometer with a speed equal in 
magnitude to the orbital velocity of the Earth about 
the Sun.) This physical situation was regarded as 
analogous to the contest of the two swimmers (Ex-
ample 5 of Chapter 14) in that a difference in time 
for the light to traverse the two different paths 
would exist because of the movement relative to 
the ether. 
//////////////////////// Ml 
Source 
Partially 
reflecting 
surface 
1 
Observer 
Ether drift 
speed ν  
Figure 15-1 The Michelson interferometer as used in 
the Michelson-Morley experiment. 
This difference in times was to be observed di-
rectly in terms of a shift of the fringe pattern when 
the interferometer was rotated through 90° to re-
verse the paths with respect to the ether. The inter-
ferometer fringe pattern arises because the two 
mirrors Mi and M i are not exactly perpendicular. 
As the interferometer is rotated, a measurable shift 
of the fringe pattern was expected to occur. To esti-
mate the magnitude of the shift, we make use of the 
special case of Example 5, Section 6, Chapter 14. 
When υ is the ether drift velocity 
3 x 10^ m/sec 
and Μ  = c = 3 X 10* m/sec, the difference in time for 
the two paths becomes 
wv 
(15-1) 
For the interferometer, a time difference Δ ί is 
equivalent to an optical path difference Δ / through 
the relation 
Δ / = cat = η λ, 
(15-2) 
where Δ / represents η  fringes of light of wave-
length λ. Thus, the expected number of fringes will 
be 
Since each orientation will exhibit this fringe pat-
tern, the shift should be 
(15-4) 
By means of multiple reflections, Michelson and 
Morley were able to have w « 10 m and λ « 5000 Ä. 
As a result, they anticipated ritotai« 0.4, a value sub-
stantially within the experimental limits of preci-
sion. 
As indicated earlier, however, their result was no 
observable shift, a result which has since been ver-
ified repeatedly by other experimenters at various 
locations and at various times of the year. Later 
experiments by Roy Kennedy and Edward Thorn-
dike, utilizing an interferometer having inter-
ferometer arms of differing lengths, obtained the 
same null result. All these results demonstrate that 
the ether, if it exists, does not possess any experi-
mentally observable properties. As a consequence, 
its value in any theory must be severely limited. 
Furthermore, the experiments, including those of 
Kennedy and Thomdike, would strongly suggest 
that the most likely way of explaining the null result 
would be to assume that the speed of light in free 
space was not affected by any motion of the source 
or the observer. If this were true, then there would 
be no time difference for the two paths, which is in 
agreement with experiment. In this connection, it is 
interesting to note that when Einstein published his 
special theory of relativity in 1905 he was appar-
ently unaware of the Michelson-Morley experi-
ment. Instead, his paper dealt directly with the 
description of events as viewed in reference frames 
moving at high constant velocity relative to one 
another. 

Einstein's Special Theory of Relativity 
119 
The results of his analysis completely resolved 
the puzzle of the Michelson-Morley experiment 
without directly attempting to do so. 
15-3 
E I N S T E I N ' S S P E C I A L T H E O R Y 
OF RELATIVITY 
The conclusions Einstein reached in considering 
reference frames in constant relative motion are 
summarized in the two postulates that constitute 
the special theory of relativity. We state them as 
follows: 
1. The speed of light in free space will have the 
same value for all observers, regardless of 
their relative motion. 
2. The laws of physics must retain the same 
equation form for all frames of reference mov-
ing at constant velocity relative to one 
another. 
In this section, we derive the transformation 
equations obtained by Einstein using these pos-
tulates. They are identical to those obtained by 
Lorentz a year earlier in an ad hoc fashion. We 
begin by considering two reference frames Si and 
S2. S2 is in motion at a constant velocity ν  = ü i 
with respect to Si. (The restriction of relative mo-
tion along the χ -axis is for convenience only. It 
does not represent a fundamental restriction.) As 
noted in the previous chapter, the transformation 
equations obtained here must reduce to the 
Galilean transformation equations in the limit of 
vlc<l. 
We therefore assume (because of the 
simplicity obtained) that the transformation equa-
tions connecting Si to S2 have the form 
Xi = 
kix2-^vt2), 
Zi = 22. 
We note that for vie <\,k 
must go to 1 to give the 
Galilean transformation. By Postulate 2, we must 
have an identical form of transformation equations 
connecting S2 to Si, except that now ν  is replaced 
by - 1 ; because we are considering Si from the 
point of view of S2 instead of the former case. 
Therefore, we must require 
X2 = k(xi- Vti), 
y2 = yi, 
Z2= 2i. 
Now, however, we note an unexpected result. That 
is, the times measured in the two frames are not 
identical as they were taken to be in the Galilean 
transformations. To see this, substitute Eq. (15-5) in 
Eq. (15-6), to obtain 
or 
so that 
X2 = 
k{k(x2-^vt2)}-vti 
(k^-\)x2-^kvt2 
= kvtu 
(15-7) 
Notice once again that if vie < I so that 
^ 1, then 
Eq. (15-7) reduces to ii = Í2—the Galilean transfor-
mation relation. 
Next, we consider the following situation. As-
sume that at Í1 = 0, the origins of Si and S2 coincide, 
so that Í2 = 0 also [from Eq. (15-7)]. At this instant, 
a light flash is generated at the common origin and 
subsequently propagates as a spherical wave. Each 
of the observers will describe the spreading wave 
front as a spherical wave by Postulate 2. Since the 
relative motion is along the χ -direction only, the y 
and 2 relationships are trivial (y, = y2, 21 = 22). 
In Si, 
xi = CÍ1, 
(15-8) 
and in S2, 
X2 = ct2, 
(15-9) 
by Postulate 1. 
From Eqs. (15-6) and (15-7), Eq. (15-9) becomes 
k(xi-vti) 
= ckt 
kv 
Xi 
or 
vk\v±c)_ 
k\v + 
c)-c" 
(15-10) 
(15-5) 
But Eq. (15-8) requires that 
Xi = CÍ1. 
We see, therefore, that 
vk\v-^c) 
from which 
and 
' 
1 
(15-6) 
c'-t>' 
I-(vie) 
=[-(f)T" 
(15-11) 

120 
Special Relativity 
where we have chosen the positive root to satisfy 
the requirement that k 
1 when v/c <1. 
Thus, the Lorentz transformation equations are 
seen to be 
X2 = Γ 
/ vV 
1/2 
Η  
y2 = 
y i , 
Z2 = Zi, 
Í2 = Γ 
/ 
i/2 
(15-12) 
The inverse transformation from S z to S i can be 
shown by similar arguments to be given by the 
equations 
X\ = 
172. 
yi = y2, 
Z, = Z2, 
ti-^—T 
-(f) 
(15-13) 
We see that our assumed transformation equa-
tion form, in addition to being simple, reduces prop-
erly to the Galilean transformation equation as 
required. What we had not anticipated was the pre-
diction that not only position measurements but 
also time measurements are affected by the motion 
of the observer. In the sections that follow, we con-
sider some of the consequences of these transfor-
mation equations on measurements required for a 
kinematic description of physical events as seen by 
observers in constant relative motion. 
15-4 
THE L O R E N T Z C O N T R A C T I O N 
Consider the following experiment. An observer 
at rest in S i locates the ends of a rod also at rest in 
S i . He finds that its length is given by Lo = 
X\R - XiL, where the subscripts R and L refer to the 
right and left ends, respectively. A second observer 
in a frame S2 moving in the χ -direction with speed υ 
relative to S i measures the length of the same rod 
to be L = X2R - XiL' How do Lo and L compare? 
From Eqs. (15-13) we obtain 
^ 
_jMR±Vt2l 
X\R 
2η ι/2> 
XlL 
(X2L + ^^2) 
Therefore, 
Lo — X\R — XlL — (X2R 
-XIL) 
1/2 
or 
-(1)1 
(15-14) 
We see that an object measured by an observer at 
rest in the same reference frame will have a max-
imum length. An observer in motion relative to the 
object will regard the object as moving relative to 
him, and will obtain a smaller or contracted dimen-
sion along the direction of constant relative motion. 
Notice that since ν  appears only as a squared quan-
tity in Eq. (15-14), it is irrelevant which reference 
frame is regarded as the stationary one. Thus, an 
observer in S i will regard an object at rest in S2 as 
contracted, while an observer in S2 will similarly 
regard an object at rest in S i as contracted. The 
primary question is simply: What is the state of 
motion of the object relative to the observer? If 
they are in the same reference frame, their relative 
velocity is zero and no motional effects are noted. 
It is the relative motion that provides results at 
variance with our Galilean expectations. 
Example 1. A moving meter stick appears to be 
shrunk in length to I m as seen by a stationary ob-
server. What is its speed relative to the observer? 
S O L U T I O N 
By Eq. (15-14), L/Lo = | = Vl-(i?/c)\ so that 
V = xl^c 
^0.97 c. 
15-5 
T I M E 
DILATION 
Let us slightly modify the experiment of Section 
15-4. In this instance, a stationary clock in Si re-
cords a time interval Δ ίι = ii/ - iio. What is the time 
interval as measured by a clock at rest in S2, which 

Relativistic Addition of Velocities 
121 
is moving at speed ν  relative to Si along the x-
direction? 
Again, from Eq. (15-13), we obtain 
tif-
VXi 
. 
VX2 
t20 —  — T 
Δ ί ι = Í1/ - ίιο  = 
/ «VI 
1/2 
[•-( f)] 
'-(f)T 
Δί2 
/ 
1/2 
Η  
(15-15) 
We conclude, therefore, that the moving clock 
runs slow; that is, the elapsed time registered by it 
will be less than that registered by the stationary 
clock. This seemingly bizarre prediction actually 
finds confirmation in data related to μ  mesons 
(radioactive particles created in the upper atmos-
phere by cosmic rays), μ  mesons created in the 
laboratory by high energy accelerators have been 
found to have a mean lifetime before decay of 
about 2 microseconds. Those created by cosmic 
rays are observed (by an earthbound observer) to 
have speeds of about 0.998 c. We might therefore 
incorrectly conclude that during their mean lifetime 
after creation they travel a distance given by y = 
0.998c x 2 x 10"'«600m, whereas they are actu-
ally created at altitudes an order of magnitude 
higher and are still detected at the Earth's surface. 
We can resolve this paradox in either of two ways. 
First, in the rest frame of the μ  mesons (Si), the 
distance 
traveled 
is 
given 
by 
yi = 0.998 c x 
2 X 10"'« 600 m, since the mean lifetime was mea-
sured in the laboratory for which the μ  meson vel-
ocity was negligible. On the other hand, an earth-
bound observer would translate the distance 
traveled by the moving μ  meson to be 
yi = 
600 
Vl-(0.998)' 
* 9000 m, 
in agreement with observations. 
Second, the earthbound observer could measure 
the lifetime of the μ  meson to be not 2 x 10"' sec, 
but the longer time (since the μ  meson's "clock" is 
in motion) of 
2 X 10"' 
2 X 10"' 
Δ ί,= 
/ 
1/2 
['-( f)] 
[1-(0.9987F^ 3 X 10"' sec. 
from which y, = ϋΔ ίι « 9000 m as before. 
In either case, we find that the proper use of the 
Lorentz transformation equations leads to results 
in accord with experiment. At this point, we pause 
to recognize a common objection expressed by stu-
dents of physics encountering time dilation or 
length contraction for the first time. That is the feel-
ing that these results are in violent contrast to 
"common sense." However, one must realize that 
the term "common sense" necessarily refers to our 
everyday experience, which is (for nearly all of us) 
restricted to kinematic situations for which r <^ c. 
As a result, these seemingly absurd effects are not 
observable. Therefore, while the predictions are at 
variance with "common sense" as identified here, 
they are in no way rendered impossible. It would be 
more satisfactory in fact to redefine "common 
sense" to mean "in agreement with experimental 
results." In this way, the apparently unsatisfactory 
predictions should be as acceptable as the predic-
tions of the Galilean transformation equations. 
15-6 
RELATIVISTIC A D D I T I O N O F 
V E L O C I T I E S 
In this section, we consider the effect of the Lor-
entz equations in the addition of velocities as in Eq. 
(14-22). From Eqs. (15-12), we can write 
X2R ~ X2L — 
jXlR 
-X\L)-V(tlf-tlo) 
-(1)1 
and 
/vVl 
1/2 
[·-© ] 
tif — tio — 
Since 
_ XlR — XlL 
XlR — XlL 
M2x = 
— 
and 
Ml. = 
—, 
tif — tio 
tif — tio 
we obtain from Eqs. (15-16) the result 
Mix - 1 ; 
M2x = · 
1 ~ ^ : 2 -
(15-16) 
(15-17) 

122 
Special Relativity 
Similarly, we find that 
1 
VUlx 
(15-18) 
and 
U2z 
=• '-(f)" 
1 — r r 
(15-19) 
As before, the converse equations are obtained by 
the interchange of quantities from S2 with those 
from Si, and by replacing ν  by - v. In addition, we 
see again that for t;/c<^l, Eqs. (15-17) through 
(15-19) reduce to Eq. (14-26) as they should. 
On the other hand, the results are strikingly 
different from the Galilean predictions if the vel-
ocities involved are not small. For example, sup-
pose S2 moves at speed ν  = cl2 along the χ -axis 
relative to Si, and that the event observed is the 
transmission of a light signal from a source at rest 
in Si, so that Ui = c. Then, Eq. (15-17) predicts that 
c 
' - 2 
«2. = 
- T = c, 
which is, of course, the result required by the first 
postulate of special relativity. The reader can in 
fact show that for any sequence of observers, each 
moving at a constant velocity relative to the preced-
ing one, successive application of Eqs. (15-17) 
through (15-19) will provide the correct result. 
Example 2. An observer at rest sees two high 
speed objects approaching him from opposite di-
rections, each with a speed of 0.6 c. What is the 
relative speed of the two rockets? 
SOLUTION 
From the point of view of the rocket at the left 
(Si), the observer has a speed of 0.6 c to the left. 
The speed of the rocket at the right is denoted by 
relative to the first rocket. Then, we have 
_ U2x^v 
^ 2(0.6)c ^ 0.6 
,^U2.v 
1+0.36 
0.68^ 
1 
1.133 c « 0.89 c. 
Thus, the relative speed of the two rockets is 
« 0.89 c and not 1.2 c as predicted by Galilean 
relativity. 
Finally, we remark that this brief discussion by 
no means exhausts the new predictions that are a 
consequence of the special theory of relativity. It 
was our main purpose here to justify the assertion 
of the previous chapter that at high velocities the 
effect of motion on measurements is to produce 
differences in the results obtained, depending upon 
the state of relative motion. There is abundant liter-
ature at all levels (from the elementary to the ad-
vanced) on this topic, which is quite properly a 
subject for additional study in physics. In the fol-
lowing chapters, we shall only occasionally need to 
take further direct notice of special relativity 
theory, since our principal interest is in classical 
physics fundamentals. For this reason, we do not 
consider at all the general theory of relativity, 
which takes into account the motional effects that 
arise when one reference frame is in accelerated 
motion relative to another. 
P R O B L E M S 
1. Verify the result Wtotai« 0.4 fringes for the Michelson-Morley experiment. Use the data given in Section 
15-2. 
2. Show that applying the Lorentz contraction to the interferometer arm parallel to the ether drift velocity 
leads to a null result in the Michelson-Morley experiment. 
3. Verify Eqs. (15-13). Note that these relations follow from Eqs. (15-12) if one substitutes - r for r and 
interchanges subscripts 1 and 2 throughout. What is the physical significance of this procedure? 
4. Determine the value of the ratio vie for L/Lo = 0.95, 0.90, 0.75, 0.50, 0.10, and 0.05. 
5. What would be the value of the ratio L/Lo for a speed t; of magnitude 600mi/hr? 

Problems 
123 
6. A high speed runner runs at a constant speed over a distance he judges to be 432 m. His wristwatch 
indicates the time required was 2.88 x 10"' sec. 
(a) What was his speed? (This will be the same as the speed of the ground relative to him.) 
(b) What distance do stationary judges see him travel? 
(c) How long does the run take him according to the stationary judges? 
7. Two events occurring at the same time in Si are separated by a distance of 1 m along the χ -axis. In a 
frame S2 moving at constant speed along the χ -axis, the events are separated by a distance of 2 m. What 
is the time interval between the two events in S2? 
8. A cube with sides of length L moves with S2 at speed ν  relative to Si. 
(a) What is the volume of the cube as measured in S i ? 
(b) What is the value of v/c if the volume measured by Si is half that measured in S2? 
9. A uranium nucleus traveling at 10* m/sec away from Si ejects an alpha particle at 1.5 x 10* m/sec relative 
to itself toward Si. What is the velocity of the alpha particle as viewed in Si? 
10. A stick of length Li is inclined at an angle θι relative to the x-axis in Si. Find its length L2 and its angle 
of inclination Θ 2 relative to the x-axis as viewed by an observer in S2 moving at a speed ν  relative to S, 
along the x-axis. 
11. Starting with the relations 
U2x 1 
UixV 
1 " I r l -
and 
Í2 = 
/ 
1/2 
use the relations 
a2x = dU2x 
dt2' 
dtr dt, 
and the fact that ν  is constant, to show that 
duix dxi 
a2x 
3/2 
aix 
3 
This transformation relation for the χ -component of the moving object viewed from two reference 
frames with a constant relative velocity will be valid only as long as the velocities MI, and U2x remain less 
than c. 
12. A cosmic ray forms a particle having a lifetime of 10"^ sec when measured at rest. How far will it travel 
before decaying if its speed is 0.95 c when it is created? 

16 
Principles of 
Dynamics I 
16-1 
Inertia, Motion, and Forces 
125 
16-2 
The Concept of Mass 
125 
16-3 
Weight 
127 
16-4 
Inertia! and Gravitational Mass 
127 
16-5 
The Law of Universal Gravitation 
128 
16-1 
INERTIA, M O T I O N , A N D F O R C E S 
In this chapter, we shall be concerned with 
dynamics, the relationship between the motion of 
physical objects, and the causes producing the mo-
tion. Everyone has an intuitive understanding of 
the relationship, for we are aware that muscular 
effort in the form of pushes or pulls can cause mo-
tion. Furthermore, we distinguish between objects 
in terms of their relative response to a given push or 
pull. For example, a man who kicks with equal 
vigor first a beach ball and then a stone of similar 
dimensions will have no trouble identifying which 
one experiences a greater change in its state of 
motion. To describe this response, we introduce the 
property of inertia and say that, of two objects sub-
ject to the sanie push or pull, the motion of the one 
with the greater inertia will be affected least. Alter-
natively, we can say that an object whose state of 
motion is observed to remain unchanged is not sub-
ject to any net push or pull. That is, because of the 
inertia property possessed by physical objects, the 
natural state of motion is one of complete rest or 
constant velocity and changes from such a state of 
motion can only occur by the application of some 
outside effort. Such outside efforts need not be in 
the form of muscular pushes or pulls. In physics, 
the term force is applied to any form of effort that 
can produce a change in the state of motion of an 
object. At the present time, physicists recognize 
four fundamental kinds of force: gravitational, 
electromagnetic, strong nuclear, and weak nuclear 
forces. Gravitational forces are relatively very 
weak and will be discussed in Section 16-5. Elec-
tromagnetic forces 
include both electric 
and 
magnetic forces on charged particles and are much 
stronger than gravitational forces (see Chapters 29 
and 32). Strong nuclear forces refer to the forces 
protons and neutrons exert on one another within a 
given nucleus, while weak nuclear forces are re-
lated to the phenomenon of beta decay for some 
nuclei. These nuclear forces are stronger than elec-
tromagnetic forces. However, they are important 
only for very short distances («10"'^ m) and hence 
for nuclear particles. Since it is necessary to apply 
the principles of quantum mechanics in this do-
main, we are not prepared to discuss these forces in 
the remainder of the text. 
16-2 
THE C O N C E P T O F M A S S 
Let us return to the kicked beach ball and stone 
for the purpose of assigning quantitative signifi-
cance to the inertia property. Since we assumed 
similar dimensions, it is clear that the volume of an 
object is not a suitable measure of inertia. It seems 
intuitively clear that the quantity of matter involved 
in the case of the stone is greater than that of the 
125 

126 
Principles of Dynamics I 
beach ball, which would explain a greater inertia. 
Instead of pursuing the idea of quantity of matter to 
the microscopic or atomic level, we introduce the 
concept of the mass of an object to explain the 
inertia property. In particular, if we regard mass 
as a fundamental quantity (as we have already done 
with length and time), then we can define a unit 
mass and subject it to some reproducible force in 
order to determine the acceleration (change in mo-
tion) produced by the force. Any other mass can be 
measured in terms of the unit mass by determining 
the acceleration produced by the same force when 
applied to the unknown object. We have implied 
earlier that for a given force the change in motion is 
less for greater inertia. This suggests that the force 
should be proportional to the product of mass and 
acceleration or, since the force F is the same in 
each case, 
F = m a = (l)a„ 
(16-1) 
where 
and 
ai = the acceleration of the unit mass 
a = the acceleration of mass m. 
Thus, the mass m will be given by 
For any two arbitrary masses subject to the same 
force, the relation becomes 
(16-2) 
(16-3) 
In the MKS system of units, the unit mass is the 
kilogram (kg) and is, by definition, the mass of the 
platinum-iridium cylinder known as the standard 
kilogram, which is preserved in France. In addition, 
from Eq. (16-1), we can derive the unit of force in 
terms of the three fundamental quantities of mass, 
length, and time. In the MKS system, the force unit 
is the newton (nt). A force of one newton will pro-
duce an acceleration of one meter per second 
squared when applied to a mass of one kilogram, or 
1 nt = 1 kg X 1 m/secl 
We have previously noted that the second is de-
fined in terms of the atomic vibrations of cesium, t 
tSee the article "Standards of Measurement," Scientific 
American, June 1%8, p. 50. 
Specifically, 1 second = 9,192,631,770 Cs vibrations. 
Since non-atomic standards are perishable, it 
would be desirable to replace the standard kilogram 
by an atomic standard (for example, the number of 
atoms of the ordinary isotope of hydrogen in a 
particular state of excitation having a mass equal to 
the standard kilogram). For the immediate future, 
however, experimental difficulties make it impos-
sible to obtain a precision comparable to that avail-
able with the standard kilogram (1 kg masses can be 
compared to 1 part in 10* at the US National Bureau 
of Standards). Therefore, standards laboratories in 
the United States, Great Britain, Germany, etc. 
have carefully prepared replicas of the standard 
kilogram for use as secondary standards of mass in 
their respective countries. 
Although, for the most part, we shall employ 
MKS units, it is proper to present a brief discussion 
of the British engineering system of units because 
of its widespread use in many English-speaking na-
tions. In this system, the second retains the defini-
tion of the MKS system. The unit of length, the foot 
(ft), is also defined in terms of Kr*' orange-red 
wavelengths; 
1 ft = 12 in = 12 X 41,929.399 wavelengths of Kr** 
orange-red light. 
In this system of units, force is chosen to be the 
third fundamental quantity, and the unit of mass is 
then defined in terms of force, length, and time. The 
force unit chosen is the pound (lb), and the mass 
unit, the slug, is defined as the mass which will ex-
perience an acceleration of 1 ft/sec' when a force of 
1 lb is applied to it, or 
lib 
By defining the pound in terms of the pull of the 
Earth on a certain standard body at a certain place 
on the Earth, and comparing the standard body 
with the standard kilogram, it can be demonstrated 
that 
1 lb = (0.45359237 kg)(9.8066 m/sec') 
«4.448 nt. 
and that 
1 slug = 0.45359237 x 32.1740 kg 
«14.59 kg. 

Inertial and Gravitational Mass 
127 
Example 1. Two masses mi and m2 are placed on 
a frictionless, horizontal surface. Each is subjected 
to the same constant accelerating force. Both are 
initially at rest. After the time t in each case it is 
found that the distance traveled by mi is three 
times the distance traveled by m2. What is the ratio 
SOLUTION 
From Eq. (16-1), 
F = mifli = m2a2 = constant. 
Since the masses are constant, it follows that the 
accelerations are also constant. As a result, by Eq. 
(14-13) we can write 
d, = 
| α l í ^ 
d2 = 
^a2t' 
or 
Since 
Therefore, 
16-3 
WEIGHT 
d2 
02* 
di = 
3d2, 
^ = 3. 
02 
m, 
^ £ 2 ^ 
m2 
ax 
When an object is placed near the Earth's sur-
face, the gravitational force (Section 16-5) acts on 
the object. If it is released, it will fall to the surface 
with an acceleration designated by g—the accelera-
tion due to gravity. This gravitational force is called 
the weight (W) of the object. Both force and accel-
eration are vector quantities, while mass is a scalar 
quantity. Therefore, we write 
surface, ranging from a value of about 9.78 m/sec^ 
at the Equator to about 9.83 m/sec^ at the North 
Pole. At a given location, however, g is essentially 
uniform over a limited region of the Earth's sur-
face. Local variations in the value of g indi-
cate non-uniform mass concentrations (mascons). 
Geological surveys for oil and minerals make use of 
this situation. 
The equal arm balance utilizes this fact by letting 
the force due to gravity act upon two objects—one 
of which is the unknown, while the other is made up 
of known masses. When the known mass total on 
one side of the balance has the same mass as the 
unknown on the other side, the arm of the balance 
(see Figure 16-1) becomes horizontal, and the un-
known mass is thus determined. This process, un-
fortunately, is called weighing in practice; the 
known masses used are called weights, which leads 
to some confusion. The reason the masses are 
stamped on the so-called "weights" is that mass is a 
measure independent of location, while the weight 
is a measure of the force of gravity on the object, 
and it varies with location. For example, a mass of 
1 kg will weigh about 9.83 nt at the North Pole but 
only 9.80 nt in New York City. The mass will re-
main 1 kg anywhere, barring nuclear reactions 
(fission, fusion) or similar circumstances which we 
do not consider here. 
11 Μ  11111 η  
7 \ 
Unknown 
mass 
Standard 
mass 
Figure 16-1 The balance (schematic). 
W = mg, 
(16-4) 
which indicates that the gravitational force W, and 
the acceleration g it would produce on a mass m if 
it were released (that is, if W were the only force 
acting on m), both lie in the same direction. The 
magnitude of g varies with location on the Earth's 
16-4 
INERTIAL A N D G R A V I T A T I O N A L 
M A S S 
In Section 16-2, we indicated a procedure for de-
termining the mass of an object in terms of the 
acceleration produced by the application of a given 
force. The mass value thus obtained is said to be an 

128 
Principles of Dynamics I 
inertial mass. On the other hand, if we can deter-
mine both the weight of the same object (this is 
discussed in the next section) and the acceleration 
it experiences when released to fall near the Earth's 
surface, then we can calculate a gravitational mass 
by means of Eq. (16-4). It is reasonable, but by no 
means obvious, that the two values should be the 
same, and many physicists (including Newton) de-
voted much effort toward providing experimental 
verification for this assumption. That they are iden-
tical to within 3 parts in 10" was demonstrated in 
1964 by R. H. Dicke and collaborators at Princeton 
University.! 
This assumed equality was used by Einstein in 
1911 as the basis for the principle of equiva-
lence—an important part of his theory of general 
relativity. Briefly stated, the principle of equiva-
lence says that it is not possible to distinguish 
between an inertial (non-accelerating) frame of re-
ference in which uniform gravitational forces act, 
and a non-inertial (accelerating) reference frame 
where there are no gravitational forces but the re-
ference frame experiences an acceleration-g rela-
tive to an inertial frame. 
As an example of this principle, consider two 
elevator cars, each of which is sealed with an iden-
tical observer inside who cannot see outside. The 
first car is at rest in a location where the accelera-
tion due to gravity is uniform and equal to g. The 
other is removed to empty space beyond the galaxy 
where there is no acceleration due to gravity. If this 
second car is given an acceleration - g by the appli-
cation of the necessary force, the observer in this 
car will experience the same force as the observer 
in the first car. Furthermore, within their cars there 
is no experiment that either can perform enabling 
them to discover which car is which. 
One of the important consequences of the prin-
ciple of equivalence that has been observed experi-
mentally is the predicted bending of a beam of light 
by a strong gravitational force (for example, a light 
beam from a distant star bent by the Sun as it 
approaches the Earth). Unfortunately, it is not pos-
sible to pursue this fascinating subject here any 
further.* 
tSee Dicke's article, "The Eö tvö s Experiment," Scien-
tific American, December 1%1, p. 84. 
$The interested reader is referred to Chapter 3 in R. 
Skinner, Relativity, Blaisdell, 1969. 
16-5 
THE L A W O F U N I V E R S A L 
G R A V I T A T I O N 
In Section 16-3, we referred to the weight of an 
object at the Earth's surface as a gravitational 
force. This is because the object and the Earth are 
observed to attract (gravitate toward) each other 
unless the object is subject to a restraining force 
whose magnitude equals the weight but whose di-
rection is opposite to that of the weight. 
Example 2. An object rests on a bathroom scale 
which reads 10 lb. This reading indicates that the 
scale is exerting an upward force of 10 lb on the 
object. Since the object is at rest, the weight of the 
object is 101b and has a direction downward, so 
that no net force is acting on the object. 
The distance-dependence of the attraction of one 
mass for another was first discovered by Newton 
around 1670, but not published until 1686 (in part 
because of previous irritating controversies his dis-
coveries had aroused, and also in part because of 
mathematical questions which were not fully re-
solved until he had invented the methods of cal-
culus). Actually Newton's discovery, the law of 
universal gravitation, was of extreme importance 
for it related not just two masses but all masses in 
the universe. In words, the law states that: 
Every particle of mass in the universe attracts 
every other particle with a force which is 
directly proportional to the product of the 
masses, inversely proportional to the square 
of the distance between them, and which acts 
along the line joining them. 
In equation form, the law of universal gravitation is 
expressed as 
r 12 
(16-5) 
where 
F g = magnitude of the force exerted by particle 1 
on particle 2 and vice versa; 
m i = mass of particle 1; 
m i = mass of particle 2; 
Γΐ2 = separation distance of the two particles; 
G = universal gravitational constant; 
r = unit vector along the line joining the particles. 
There are some philosophically unsatisfying fea-
tures about gravitational forces that the statement 
of the law of gravitation and Eq. (16-5) fail to 

The Law of Universal Gravitation 
129 
clarify. Perhaps the biggest puzzle is, why do mas-
ses attract one another especially if they are some 
distance apart? Experiment will demonstrate that 
the attraction cannot be electrical in nature because 
gravitational forces are far weaker than such 
forces. (For that matter, why do electric charges 
attract or repel each other as observed?) The reader 
must understand that, in stating the law of universal 
gravitation, Newton made no attempt to explain 
why masses attract one another. What he did was 
assert a relationship for determining such forces, 
and then use it to explain a number of observed 
phenomena (see Section 18-5). Since Eq. (16-5) can 
be used to successfully predict experimental out-
comes ranging from satellite motions to the discov-
ery of hitherto unknown planets, it seems reason-
able to accept the reality of gravitational forces, 
even though their exact nature remains unknown. 
Certainly it is more fruitful to adopt without 
a 
fundamental derivation a law enabling us to obtain 
a more quantitative understanding of our universe 
than to reject this understanding because we cannot 
prove the "truth" of the law. (This attitude, in fact, 
was another of the many great contributions that 
Newton made to the development of physics.) 
Another feature that is apparently a source of 
difficulty is the assertion that every particle of mass 
in the universe is attracting every other particle of 
mass at every instant. If this is so, how can one 
possibly ever hope to isolate two masses and per-
form an experimental test of Eq. (16-5)? Further-
more, we have already defined units of mass, 
length, and time [so that the unit of force is also 
determined by Eq. (16-1)]. How then are we to find 
the value of G, the constant which makes an equa-
tion out of the proportionality asserted in the state-
ment of the law of gravitation? 
The first question can be answered in terms of an 
example. Consider a point mass m resting on the 
surface of the Earth. One can easily show that the 
force between any two particles is very small by 
hanging them from slightly separated long thin 
threads. Because the force each exerts on the other 
is very small, the strings will both hang vertically; 
after all, the mass of the Earth is so much greater 
than any one small particle. Since the attractive 
forces decrease as the square of the separation dis-
tance, it begins to be clear that except for very care-
ful measurements the attractive force due to the 
rest of the universe on a particle on or near the 
Earth's surface will be primarily due to the mass of 
the Earth alone acting on the particle. (Similarly, an 
astronaut on the surface of the Moon will experi-
ence a gravitational force that is almost exclusively 
due to the mass of the Moon.) 
As a result, to determine the gravitational force 
exerted on the particle, one has only to sum up 
(vectorially) the forces of attraction between the 
particle and each element of the Earth. Newton was 
able to show that if the mass of the Earth is a 
function only of the radial distance from the center, 
then the gravitational force exerted on a particle on 
or above the surface can be calculated by assuming 
the mass of the sphere is concentrated at its center, 
giving a single separation distance in Eq. (16-5). In 
this connection, recall that we have already called 
this gravitational force on the particle its weight W 
in Eq. (16-4). Therefore, if we combine Eqs. (16-4) 
and (16-5), we obtain 
WW 7 
Gm Ε  Λ  
W = m g = m ^ T - r , 
(16-6) 
where mE and Re are the mass and radius, respec-
tively, of the Earth. Since g and Re are easily 
found, it follows that a knowledge of G is essen-
tially equivalent to a knowledge of the mass of the 
Earth. For this reason, when Sir Henry Cavendish 
determined a value for G in 1798, he was said to 
have "weighed" the Earth in the sense of Section 
16-3. 
Let us briefly consider the Cavendish experi-
ment.t Figure 16-2 shows a light rigid arm bearing 
//////// 
^^^^^^QuaxXz 
fiber 
Mirror (to measure 0) 
^-^"""^''"^ 
^ ^ - ^ - ^ Heavy mass 
^ 
' ^-^'Z—Light mass 
Figure 16-2 Schematic diagram of the Cavendish 
experiment. 
tAn annotated version of Cavendish's description of 
his experiment is given in M. H. Shamos, (ed.). Great Ex-
periments in Physics, Holt, Rinehart, and Winston, New 
York, 1959. 

130 
Principles of Dynamics I 
P R O B L E M S 
1. Starting from rest, a 2 kg mass moves 9 m in 6 seconds due to the application of a constant force. What 
is the magnitude of the force? 
2. A 1 kg mass is observed to experience an acceleration of 3 m/sec^ when subject to a given force. What 
will be the acceleration of a 5 kg mass subject to the same force? 
3. It was suggested in Section 16-2 that the kilogram be defined (for example) as the number of atoms of 
ordinary hydrogen required to equal the mass of the standard kilogram. List objections that can be 
raised to such a procedure. 
4. By Eq. (16-4), the weight of a 1 slug mass at a point where g has a magnitude of 32.2 ft/sec^ is 32.2 lb. 
Calculate your mass in slugs and in kg. Also find your weight in newtons. 
5. For homogeneous materials, the mass m is directly proportional to the volume V of the material. The 
relation can be written in terms of the density p: 
m =pV, 
where ρ is a constant characteristic of the material. The density of water is 1 x 10^ kg/m\ and the mass 
of a water molecule is about 30 x 10"^^ kg. If the water molecules are in contact with one another, what 
is their approximate size? 
6. A particle of mass m is placed along a line joining two particles (masses Μ  and 3M) that are separated 
by a distance of 2 m. At what point will the forces due to Μ  and 3M just cancel each other? 
7. At what point along the line from the Earth to the Moon is the gravitational pull of the Moon equal to 
that of the Earth? Give your answer as a fraction of REM, the distance from the center of the Earth to the 
center of the Moon. The mass of the Moon is about m the mass of the earth. This is the point at which a 
moonship is said to "leave" the Earth's influence and enter the Moon's. 
8. The planet Mars has a radius about 0.52 that of the Earth, and its mass is 0.11 that of the Earth. 
(a) Find the ratio gu^JgE^h 
(at the surface). 
(b) If an astronaut in full equipment can just leap 2 ft upward on the Earth, how high will he be able to 
leap on Mars? 
9. Two homogeneous spheres of equal radii and made of a material of density ρ are in contact. Show that 
the force of gravitational attraction between them is given by 
10. Given the value of G from Eq. (16-7) and the value of g (take 9.80 m/sec^), use Eq. (16-6) to determine 
the mass of the Earth. The radius of the Earth is about 6.4 x 10"^ m. Then calculate the average density of 
the Earth. 
equal spherical masses at its ends, supported at its 
separation distance. By measuring the angle of 
center by a fine quartz fiber attached to a fixed posi- 
twist and separately measuring the force required 
tion. The large dashed circles represent two larger 
to cause that angle of twist, one can calculate G 
spheres that are brought close (symmetrically) to 
(since the masses and r are also known). The cur-
the smaller masses on the arm. The gravitational 
rently accepted value in MKS units is 
force between the pairs of masses causes the quartz 
fiber to twist through an angle such that the twisting 
G = 6.670 x 10'" nt mVkgl 
(16-7) 
or torsional force due to the internal structure of 
the fiber just balances the gravitational force due to 
The reader should show that the units given are cor-
the mass pairs that are now at a measurable (fixed) 
rect. 

Problems 
131 
11. Find an expression for the variation of the acceleration due to gravity as a function of height h above 
the Earth's surface in terms of go, the acceleration due to gravity at the Earth's surface, and RE, the 
radius of the Earth. Also show that at large height (h >RE), 
12. For an astronaut on the surface of the Moon, compute the ratio of the pull of gravity due to the Moon to 
the pull of gravity on him due to the Earth. The radius of the Moon's orbit about the Earth is about 60 
times the Earth's radius. Assume the astronaut is on the side of the Moon facing the Earth. 
13. A space explorer becomes separated by 6 m from the surface of a spherical asteroid he has come to call 
home. His mass including equipment is 90 kg, while the mass of his asteroid is 45,000 kg. If the radius of 
the asteroid is 30 m, estimate the greatest amount of time required for him to drift back due to gravitational 
attraction. Assume he starts from rest. 
14. To appreciate the weakness of the gravitational force, determine the magnitude of the force between a 
1 kg mass and a 10 kg mass separated by a distance of 0.10 m. 
15. For the system shown in Figure 16-3: 
m, = 1 kg, mz = 2 kg, and 
= 3 kg. 
(a) Find the force on mi due to mi. 
(b) Find the force on mz due to ma. 
(c) Find the vertical and horizontal accelera-
tion components of mi due to mi and mi. 
(d) Find the direction and magnitude of the re-
sultant acceleration. 
Figure 16-3 
1 · 
mi = 1 kg 
3m 
2 · -
m2=2 kg 
4m 
• • 3 
m3=3kg 

17 
Principles of 
Dynamics II 
17-1 
Static and Dynamic Equilibrium 
17-2 
Non-Equilibrium Motion 
134 
133 
17-3 Newton's Laws of Motion 
136 
17-4 
Action and Reaction 
137 
17-1 
STATIC A N D D Y N A M I C EQUILIBRIUM 
In the previous chapter, the mass of a particle 
was related to the applied force and the resulting 
acceleration by the relation in Eq. (16-1): 
F = ma. 
If the applied force is zero (equivalently, if the vec-
tor sum of all forces acting on the particle is zero), 
the resulting acceleration is also zero and the par-
ticle is said to be in a state of equilibrium. When the 
velocity is also zero, the particle is at rest or in a 
state of static equilibrium. When the particle is in 
uniform motion (constant velocity), the situation is 
one of dynamic equilibrium. The feature common 
to both types is a balance of forces resulting in no 
acceleration. 
Example 1. A particle of mass m is supported in 
a state of rest by two forces F, and F2, which are 
oriented as shown in Figure 17-1. In addition to 
these forces, w = mg, the weight of the particle is 
vertically directed downward. Find Fi and F2 in 
terms of the weight w. 
SOLUTION 
Since the particle is at rest, the horizontal and 
vertical components of the forces must separately 
sum to zero. (There is no way a vertical force can 
balance a horizontal force.) For the horizontal 
force components, 
F, cos 37^-F2COS 53° = 0, 
0.80F, = 0.60F2, 
or 
For the vertical force components, 
F, sin 3T + F2 sin 53** - w = 0, 
0.60F, + 0.80F2=>v. 
>^F2Sin53*' 
^Fi sin 37° 
>^F2Sin53*' 
^Fi sin 37° 
: 
yr^f^ 
F2 cos 53" 
"Fi cos 37° 
w = mg 
Figure 17-1 Force diagram for Example 1. 
133 

134 
Principles of Dynamics II 
Combining these relations, we obtain 
0 . 6 0 ( ^ F 2 ) + 0 . 8 0 F 2 = 
w. 
and 
so that 
or 
and 
1 . 2 5 F 2 = 
w 
F 2 = 0.80w 
F , = ^ F 2 = 0.60w. 
4 
Example 2. Figure 17-2 represents a particle of 
mass m moving to the right with a constant velocity 
Vx. It is acted upon by a pulling force equal in 
magnitude to the weight ( F p = w), a drag force FD, 
an upward force FN due to the surface and the 
downward gravitational force w. Find FN and FD in 
terms of w. 
SOLUTION 
Again, the horizontal force components and the 
vertical force components must separately sum to 
zero. Therefore, referring to Figure 17-2, we have 
and 
F p cos β  - FD = 0 
FN + F p sin Ö  - w = 0, 
so that 
F D = F p cos θ = w cos Θ , 
FN= 
w-Fp 
sin Θ . 
Substituting the values given, we obtain: 
F D = H ' x i = | l b 
1 
i 
fp sin βΟ** 
F o 
w = 
Fp cos 60° 
mg 
FN=W-
«0.13w. 
As these examples illustrate, the analysis of a 
system in equilibrium can be described as a book-
keeping process in that there must be a complete 
balance of forces. In cases involving many forces, 
the process is made much more systematic if the 
force components in the χ -, y-, and ζ -directions are 
entered in separate columns of a tabular form. 
Summing the entries in each column must produce 
a result of zero. The advantages of the tabular form 
are that it minimizes the chance of omitting one or 
more components and it facilitates checking the 
arithmetic work to eliminate errors. 
17-2 
N O N - E Q U I L I B R I U M M O T I O N 
From the discussion of the previous section, the 
reader may already have correctly concluded that 
non-equilibrium motion occurs when the vector 
sum of all forces acting on the particle does not 
equal zero. In this situation, Eq. (16-1) is used to 
determine the acceleration, from which the velocity 
and displacement as a function of time can be 
found as outlined in Sections 14-3 and 14-4. Alter-
natively, one can use a knowledge of the accelera-
tion to determine further details of the system of 
forces acting on the particle. 
Example 3. A particle of mass m is released 
from rest to slide down a smooth incline that is 
elevated at an angle θ above the horizontal. It is 
subject to a downward gravitational force mg and a 
normal force FN directed perpendicularly upward 
from the surface of the incline. Determine 
(a) the magnitude of FN and 
(b) the acceleration in terms of mg and Θ . 
SOLUTION 
Since the motion occurs along the surface of the 
incline, the normal force FN must be equal in mag-
nitude but oppositely directed to the component of 
the weight acting normal to the incline surface. 
Thus (see Figure 17-3), 
or 
Figure 17-2 Force diagram for Example 2. 
FN - w cos Ö = 0 
F N = mg cos Θ . 

Non-Equilibrium 
Motion 
135 
Motion 
w cos θ 
Figure 17-3 Force diagram for Example 3. 
The component of the weight acting in a direction 
parallel to the incline surface is the only other 
force. Therefore, the acceleration must be directed 
along the incline in the same direction. It is given by 
or 
mg sin θ = ma 
fl = g sin Θ . 
This example provides an opportunity to illus-
trate one method of checking the correctness of our 
solution. Thus, consider the extreme situations (i) 
θ = θ'*, and (ii) θ = 90°. In the first case, the incline 
would become horizontal, FN would equal mg, and 
a would become zero (static equilibrium). In the 
second case, the incline becomes vertical, FN 
would become zero, and a would become equal to 
g. The reader can check that the equations for FN 
and a yield these results, and can therefore be ex-
pected to apply as well to intermediate values of Θ . 
Thus, by considering special cases for which we 
know the answer, we check the general solution. 
In obtaining our solution to this problem, the 
physical requirement that the motion had to take 
place along the incline was used to simplify the 
analysis. In effect, a new set of axes parallel and 
perpendicular to the incline were used in place of 
the usual x- and y-axes. Let us now analyze the 
problem in terms of x- and y-axes to demonstrate 
that the results are the same. As Figure 17-4 illus-
trates, we can write the component parts of Eq. 
(16-1) as follows: 
FN sin θ = mux = ma cos Θ , 
(i) 
mg - FN cos θ = may 
= ma sin Θ . 
(ii) 
From (i), we obtain 
FN tan θ = ma. 
(iii) 
FN 
FN cos θ 
1 \ 
FN Sin θ k > 
(a) 
(b) 
Figure 17-4 Force diagram for Example 3 in x-y 
coordinates. 
Combining (ii) and (iii), 
mg - FN COS Θ  = FN tan θ sin Θ , 
mg cos θ = FN(COS^ Θ  + sin^ Θ ) 
or 
FN = mg cos Θ , 
so that from (iii) 
a = —(mg cos Ö)tan Ö = g sin Ö, 
m 
as before. 
Example 4. A particle of mass m = 1 kg is ob-
served to move up a smooth incline with an acceler-
ation of 4.9 m/sec^ as a result of a pulling force Fp 
applied at an angle of 37° above the plane of the 
incline (which is oriented at an angle of 30° above 
the horizontal). 
(a) Find Fp. 
(b) Find FN. 
SOLUTION 
As in Example 3, it is easier to consider motion 
parallel and perpendicular to the plane of the in-
cline. Therefore, we resolve Fp and w into compo-
nents in those directions. Equation (16-1) then 
gives: 
parallel components 
F p cos 37°-
perpendicular components 
F p cos37°-w sin30° = ma = - a ; 
g 
FN + F p sin 37° - w cos 30° = 0. 

136 
Principles of Dynamics II 
Fp 
J 
\ 
w = 9.8nt 
Figure 17-5 Force diagram for Example 4. 
Therefore, 
F p 
= 
w(sin30° + | ) 
cos 37° 
and 
COS 30*^ - ^sin 30° 4- |^tan 37° . 
Using the numerical values given above, we find 
that 
F p = 12.2 nt 
and 
FN = 1.14 nt. 
The reader should verify that the parallel and per-
pendicular force components were correctly deter-
mined. 
17-3 
N E W T O N ' S L A W S OF l\/iOTION 
In analyzing the examples in the last two sec-
tions, we have made use of a limited number of 
basic principles. First, Eq. (16-1) proved to be ade-
quate as a cause and effect relationship for both 
equilibrium (a = 0) and non-equilibrium (a ψ  0) 
situations. A less obvious relationship was em-
ployed when we identified the normal force (FN) 
exerted by a constraining surface in the various ex-
amples. A normal force exists when a particle of 
mass m is in contact with the surface. As a result of 
the contact, the particle exerts a net force on the 
surface due to other forces (gravitational, pulling, 
etc.) acting upon it. If the particle is to remain on 
the surface (rather than sinking into it or rising 
above it), a force directed outward from the surface 
of the correct magnitude must exist. In Section 
16-5, for example, the object resting on the scales 
exerted a force of 101b downward on the scales 
(because of its weight). Since the object remained 
at rest on the scales, the normal force exerted on it 
by the scales is an upward force of 10 lb. 
In the language of physics, the normal force 
exerted by the surface on the object is said to be in 
reaction to the net force normal to the surface 
exerted on it by the object. If the surface were to 
vanish, so also would the normal force and the con-
strained motion associated with it. In the next sec-
tion, we will discuss at greater length the concept of 
reaction forces (and the action forces that oppose 
them). The remainder of this section is devoted to a 
brief historical discussion of the above relation-
ships. 
Galileo (1564-1642) found flaws in the mechanics 
of Aristotle. His experiments involving moving 
bodies led to the idea that a state of motion can 
exist without an applied force, while a change in the 
state of motion can only occur as a result of an 
applied force. It remained for Newton (1642-1727) 
to formulate a complete basis for analyzing prob-
lems of classical mechanics, the law of universal 
gravitation, as well as his contributions to optics 
and mathematics. 
In 1687, Newton published his Principia, 
or 
Mathematical 
Principles of Natural 
Philosophy. 
The first two parts of the book deal with the mo-
tions of bodies by means of propositions, theorems, 
and corollaries. In addition, these parts contain his 
development of the principle of universal gravita-
tion. The third part of the book is devoted to the 
solar system as an example of the applicability of 
the principles presented earlier. His description of 
mechanics is summarized in three relations which 
have acquired the status of "laws" of motion. In 
formulating these laws, it was necessary for him to 
clarify the meaning of mass and force, which we 
have discussed briefly in Chapter 16. 
Newton's three laws of motion can be stated (in 
modern terminology) as follows: 
I Every object continues in a state of rest or of 
uniform motion in a straight line unless acted 
upon by a net applied force. 
II The product of the acceleration and the mass 
of an object is proportional to the net force 
applied to the object. 
Ill When object A exerts a force on object B, 
object Β exerts on object A 2l reaction force 
which is equal in magnitude but opposite in 
direction. 

Action and Reaction 
137 
Some comments are in order. First, Eq. (16-1) ex-
presses in equation form the statements of law I 
(a = 0) and law II (a ψ 0). In addition, however, one 
must recognize that a frame of reference (see Sec-
tion 14-5) must be specified if accelerations are to 
be unambiguously identified. A frame of reference 
with respect to which a particle is not accelerated is 
called an inertial reference frame, and law I is also 
known as the law of inertia. Finally, it should be 
noted that Newton's original statement of law II 
could also be expressed symbolically in the form 
F 
= 
dt 
· 
(17-1) 
If the mass m is not a function of time, Eq. (17-1) is 
identical to Eq. (16-1). On the other hand, if m does 
depend upon time, the two equations will not give 
the same result. We will return to this question in 
Chapter 19, where it will be seen that Eq. (17-1) is 
the most generally satisfactory form of law II. 
17-4 
A C T I O N A N D R E A C T I O N 
Law III can be interpreted as a statement that a 
single isolated force cannot exist. Whenever one 
object experiences a force, it does so because of 
the existence of another object which must neces-
sarily be subject to a force in reaction to its action 
on the first. Furthermore, one should understand 
that the two objects need not be in contact, as we 
assumed for the object on the scales. For example, 
the law of universal gravitation [Eq. (16-5)] applies 
to any two particles at any separation. It states that 
each exerts on the other a force of the same mag-
nitude but opposite direction. (Thus a self-centered 
individual could describe his weight as a force 
equal in magnitude but opposite in direction to the 
force acting on the Earth due to his mass.) 
It is important to stress that the action-reaction 
pair of forces act on separate objects and not on the 
same one. As a result when one applies law II to a 
given object, only one of the action-reaction pairs 
will be included in the vector sum of forces applied 
to the object. If this were not true, one would 
never have a non-equilibrium situation. A common 
paradox used to emphasize this fact is the follow-
ing: when a pitched baseball is hit by a bat swung 
by a batter, it is observed to move away from the 
bat. However, law III says the bat exerts a force on 
the ball equal in magnitude and opposite in direc-
tion to the force it experiences due to the ball. 
Therefore, since the vector sum of these two forces 
vanishes, a change of motion cannot occur. The 
paradox is easily resolved by recognizing that 
whether or not the motion of the ball changes de-
pends only upon the forces acting upon it and not 
upon any reaction forces it exerts on other objects. 
As a final example emphasizing this point, con-
sider once more the 101b object at rest on the 
scales. This system involves four distinct forces— 
the pull of the Earth on the object, the pull of the 
object on the Earth, the force of the object on 
the scales, and the force of the scales on the object. 
All of these forces have the same magnitude but the 
direction of the first and third forces are toward the 
Earth, while the second and fourth are in the oppo-
site direction. In applying law I to the object (since 
a = 0), only the first and fourth forces are involved 
since the other two forces are exerted on objects 
other than the 10 lb object in question. 
In analyzing situations involving more than one 
particle, it is useful to schematically subdivide the 
problem into separate parts (called free-body dia-
grams) and indicate all forces, whether action-
reaction or not. In this way, it becomes possible to 
determine missing details of the total system. 
Example 5. Figure 17-6 shows a system of two 
particles whose masses are mi and m2, respectively. 
They are in contact on a smooth horizontal surface, 
and have an acceleration a together to the right due 
to an applied force F. Find the magnitude of the 
force F21 that the first mass exerts on the second. 
SOLUTION 
Figure 17-7 shows the free-body diagrams for the 
system, with all the forces indicated. Since the mo-
tion is along the horizontal surface Fn, = mig, Fn2 = 
niig. Law II for the horizontal motion of 1 and 2 
yields, the following expressions: 
Figure 17-6 System diagram for Example 5. 

138 
Principles of Dynamics II 
a—— 
1 
a — 
1 
Fa, 
2 
Fl2 — F21 — m i F 
(iv) 
Rgure 17-7 Free-body diagrams for Example 5. 
Law III requires that F12 = - F21 (equal magnitudes 
but opposite directions, as shown). Therefore, 
adding Eq. (i) and Eq. (ii) gives 
or 
F = (mi + m2)a 
F 
a = mi + m2 
Substituting Eq. (iii) in Eq. (ii) gives the result 
(ü i) 
The reader can show by an identical argument 
that if the two masses are interchanged (see Figure 
17-8), then 
F,2 = 
m i F 
mi 4- m2 
Is this last result consistent with Eq. (iv)? 
Figure 17-8 System diagram of Example 5 with 
masses reversed. 
PROBLEIMS 
1. Figure 17-9 shows a body of weight W supported at the midpoint of a wire of length /. The forces at 
point Β exerted along and by the wire segments AB and BC (such forces are called tension forces) are 
opposed by the weight W. 
(a) Show that the tension forces Ti and T2 in 
the wire segments are equal in magnitude. 
(b) Find the magnitude of tension force Τ  in 
terms of the weight W and the angle φ . 
(c) For what angle φ  is the magnitude of the 
tension force Τ  equal to the weight W? 
Figure 17-9 

Problems 
139 
2. In Figure 17-10, φ . = 60°, <í>2 = 30°, and W = 
2001b. Find the magnitudes of the tension 
forces Γ ι and Γ 2. In this case, the wire lengths 
AB and BC are unequal. 
Figure 17-10 
/ / / / / / / / / / / / / / / / / / / / y / 
Λ λ φ ,=60° / 
\<t>2=3oy^c 
0 
w 
3. A 3 kg block resting on a smooth horizontal surface is subjected to a force of 8 nt. 
(a) What is the acceleration of the block? 
(b) If the horizontal force on the block is sup-
plied by hanging a weight of 8nt over a 
pulley and releasing the system, what will 
be the acceleration of the block and the ten-
sion Τ  in the connecting cord? See Figure 
17-11. 
Figure 17-11 
4. What average force is required to stop a 2241b fullback in a distance of 3 ft if his initial speed is 
24 ft/sec? 
5. An automobile weighing 4000 lb and traveling 60 mi/hr is to be brought to rest in 300 ft. What average 
force must be exerted on the car? 
6. A rope fastened to a 60 lb block that is on a smooth plane inclined at 30° with the horizontal extends 
upward parallel to the plane and over a pulley at the top of the plane. A 70 lb ball is hung on the end of 
the rope vertically below the pulley. Find: 
(a) the acceleration of the block and 
(b) the tension in the rope. 
7. A 1920 gm block rests on a frictionless table. A cord attached to the block passes over a pulley at the 
edge of the table. A 40 gm block is attached to the hanging end of the cord. Calculate: 
(a) the acceleration of the blocks and 
(b) the tension in the cord. 

140 
Principles of Dynamics II 
8. The blocks and the ball have the weights shown in Figure 17-12. The plane is smooth. Find: 
(a) the acceleration of the ball and 
(b) the magnitudes of the tension forces T, and 
Γ2 in the connecting ropes. 
Figure 17-12 
361b 
T, T, 
241b 
w = 36 lb 
9. Atwood's machine consists of two particles of masses mi and m2 connected by a light string, which is 
passed over a light, smooth pulley as shown in Figure 17-13. The magnitude of the tension Τ  in the string 
is the same on either side of the pulley. By applying Newton's laws of motion, show that the acceleration 
of the masses is given by 
a = (mi - m2)gKmi-^ miY 
Provide a physical interpretation of this result 
for the cases 
(a) 
m , > m 2 ; 
(b) 
m i < m 2 ; 
(c) mi = m2. 
Τ  
τ  
m i ^ 
i 
Figure 17-13 
10. Show that the tension in the string of an Atwood's machine has a magnitude given by the expression 
I 
\mi + m2/ 
Provide a physical interpretation of this result for the special case mi = m2. 
11. In an Atwood's machine experiment, the acceleration of the system is measured to be 0.49 m/sec^ The 
lighter particle has a mass of 1.9 kg. 
(a) What is the magnitude of the tension T? 
(b) What is the mass of the heavier particle? 
12. A 0.5 kg block is allowed to slide down a rough inclined plane set at 30° to the horizontal. It starts from 
rest and accelerates uniformly, traveling 2 m in 4 seconds. Find the magnitude of the constant retarding 
force exerted on the block due to the rough surface of the incline. This force acts in a direction parallel 
to the incline surface. 
13. A man weighing 160 lb enters an elevator, stands on a bathroom scale, and is accelerated upward at 
2 ft/sec'. 
(a) What weight is indicated by the scale in this upward accelerated motion? 
(b) What would be the reading of the scale if the acceleration were 2 ft/sec^ downward? 
(c) Suppose the elevator cable snaps while the car is in motion. What would the scale reading then be? 

18 
Applications of the 
Laws of i\/lotion 
18-1 
Problem Solving—A Plan of Attack 
141 
18-5 
Planetary Motion and Kepler's 
18-2 
Friction 
142 
Laws 
149 
18-3 
Projectile Motion 
144 
18-6 
Central Force Problems 
151 
18-4 
Uniform Circular Motion 
146 
18-1 
P R O B L E M S O L V I N G — A P L A N OF 
ATTACK 
If the motion of a particle (or system of par-
ticles) is to be analyzed in terms of Newton's laws 
of motion, there are several aspects of the problem 
to be dealt with. First, one must be sure to include 
all the forces that can have a measurable effect on 
the particle. Furthermore, the nature of each force 
must be understood at least well enough to assign 
approximate values to the various parameters upon 
which the force depends. For example, the effect of 
a gravitational force (due to one particle) that is 
acting on another particle cannot be fully evaluated 
unless the masses of the two particles and their sep-
aration distance are known. It is one of the aestheti-
cally pleasing aspects of physics that most of the 
important force laws are simple in form however 
powerful their consequences. 
Second, an inertial reference frame must be cho-
sen that is suitable for the analysis of the particle 
motion. In a non-inertial frame of reference, the 
laws of motion will not give satisfactory results un-
less one or more "fictitious" forces are introduced 
in the analysis. These forces have the effect of 
"transforming" the problem from the non-inertial 
reference frame to an inertial frame. For example, 
one can generally regard a frame of reference at-
tached to a stationary laboratory table as an inertial 
reference frame for particle motion on the table 
top. However, for a laboratory on the surface of the 
Earth, the table is in fact both rotating about the 
axis of rotation of the Earth and traveling in a 
nearly circular orbit about the Sun (see Section 
18-5). Precise measurements would show that the re-
sultant acceleration of the table leads to small dis-
crepancies between the observed particle motion 
and the predictions obtained from the laws of mo-
tion assuming an inertial frame of reference. In 
Section 18-4, we will consider the effect of the 
Earth's rotation upon the weight of an object on the 
surface of the Earth. Except for this example, we 
will not consider the complications of motion rela-
tive to non-inertial frames of reference. 
The remaining facets of the analysis required in 
applying the laws of motion have to do with what 
can be called procedural details, in contrast to the 
conceptual or physical aspects discussed thus far. 
As noted in Section 17-1, when the analysis has 
been systematized to resemble a problem in book-
keeping, one can have confidence that few errors 
will arise. Those which do occur will not be difficult 
to locate if the desired systemization has been ac-
complished. The following suggestions should be 
helpful in systematizing the analysis of any 
problem. 
141 

142 
Applications of the Laws of Motion 
1. Determine all the non-negligible details of the 
problem. In reading a "textbook" problem, it 
is useful to sketch in a diagram the essential 
details of the problem as they are encountered 
in reading. Show a symbol for every quantity 
involved, and, if numerical values are given, 
indicate them as well. Be sure that all similar 
quantities have common units—do not at-
tempt to add i)ounds of force to newtons of 
force. 
2. Determine what physical laws (gravitation, for 
example) are necessary in addition to the laws 
of motion. Substitute in these expressions the 
symbols of the relevant quantities from the 
sketch of step 1. 
3. Separate the vector equations where neces-
sary into algebraic component equations. 
Solve these equations systematically for the 
desired quantities in symbol form. 
4. Check to see that the resulting solution equa-
tions are dimensionally correct before per-
forming any numerical work. 
5. After the calculations are complete, try to de-
termine whether or not the results are reason-
able. This is a value judgment that becomes 
easier to make with experience. 
18-2 
FRICTION 
It is common experience that the external force 
required to cause (or maintain) the motion of an 
object kept in contact with a solid surface depends 
upon surface characteristics such as degree of 
cleanliness and smoothness. Thus, a rough surface 
will offer a relatively great resistance to a change in 
the state of motion of an object resting upon it. The 
same surface if smoothed or lubricated will become 
much easier to negotiate. One attributes this de-
creased resistance to motion to a lessening of what 
are called frictional forces between two bodies in 
contact. It is also observed that a less massive body 
on a given surface will require a smaller force to 
overcome friction forces than will a more massive 
body on the same surface. Once a body has been 
set in motion on a surface, it is usually observed 
that the force required to maintain motion at con-
stant speed is markedly less than the force required 
to set it in motion initially. Within rather wide 
limits, this force is independent of the speed of the 
object. Finally, it is found that the force required to 
overcome friction forces is relatively insensitive to 
the magnitude of the area of contact between the 
object and the surface. 
To attempt a detailed analysis of friction forces is 
out of the question at this point, for it necessarily 
involves the interaction of atomic or at least mi-
croscopic portions of the objects involved. Such in-
teractions are extremely sensitive to many factors 
such as crystal structure, impurities or imperfec-
tions in the structure, and surface contaminations 
to name a few. While the effects of these factors 
could in principle be calculated with modern quan-
tum mechanical techniques, it is experimentally not 
possible to obtain sufficiently accurate values of 
the necessary physical parameters, although much 
progress has been made.t 
It is possible, however, to satisfactorily account 
for the effect of frictional forces by introducing em-
pirical relationships based on the observations cited 
above. In doing so, we stress the fact that these re-
lationships must not be regarded as fundamental 
physical laws like Newton's laws of motion. They 
are valid only for limited ranges of the physical 
parameters involved, and their use is justified only 
by the realization that without them it would be 
difficult to obtain even approximate solutions for 
many common particle motion problems. One need 
not apologize for the use of empirical relations, pro-
vided care is exercised to avoid their use outside 
their range of validity. 
We proceed by identifying the important parame-
ters involved in a particularly simple situation, il-
lustrated in Figure 18-1. A block of mass m is at 
Figure 18-1 Force diagram for a system with friction. 
tSee, for example, the article, **Friction,'' by R. Palmer 
in the February 1951 issue of Scientific American. 

Friction 
143 
rest upon a horizontal surface, subject to four dif-
ferent forces. The weight of the block (W = mg) 
acts downward, pressing the block into contact 
with the surface. Since the block is at rest on the 
surface, it follows that there must be a force 
exerted upward on the block by the surface which 
has the same magnitude. Since this force is perpen-
dicular (or normal) to the surfaces that are in con-
tact, it is common to identify this as the normal 
force (N). The force designated (P) represents a 
pulling force, and the frictional force due to the 
contact of the block with the surface is labeled (F/). 
Since the block is at rest, the magnitude of (P) must 
equal that of (F/). Experimentally, one finds that Ρ  
can be increased in magnitude from zero up to a 
maximum value. Exceeding that maximum value 
will produce motion of the block. 
This indicates that, for a static situation (no mo-
tion occurs), the friction force (F/) can assume val-
ues ranging from zero to a maximum value. This 
maximum value is related empirically to the mag-
nitude of the normal force (N) by the relation 
|Ρ /.^ = μ ^|Ν |, 
(18-1) 
where μ , is a dimensionless constant called the 
static coefficient of friction. This relation just ex-
presses the observed fact that as the force of 
contact (N) between the surface and the block is 
increased, the friction force will increase in direct 
proportion. This relationship holds true almost in-
dependently of the area of contact. It does not re-
main valid for a heavy object whose area of contact 
is made to approach zero—for example, by grinding 
the bottom to a (balanced) point surface. The 
reason it fails in such a case is that when the entire 
weight of the object is applied to such a small re-
gion of the surface, a local deformation of the sur-
face can occur. Motion in such a case would then 
be similar to pulling a plow through soil rather than 
along the surface of the soil. An example of this 
type of deformation is the unfortunate creation of 
dents in tile (and even hardwood or aluminum air-
plane floors) caused by women wearing spike-
heeled shoes. 
To summarize the results of this discussion in-
volving Figure 18-1, the block will remain at rest so 
long as the pulling force (P) has a magnitude given 
by the relation 
Equation (18-2) is valid because |P| = |F/| and |N| = 
m|g|. When |P| = μ ^ΐΝ ], one is to infer that motion 
impends, that is, we are to regard |P| as being in-
finitesimally close to, but less than μ ,ΐΝ ]. Once 
motion has begun, it is usually observed that the 
pulling force required to maintain motion of the 
block at constant speed along the surface can be 
accounted for empirically by the relation 
|Ρ | = μ κ|Ν |, 
(18-3) 
where μ κ is the kinetic coefficient of friction. In 
general, μ κ < μ *, which can be understood qualita-
tively by referring to Figure 18-2, which is a rep-
resentation of an enlarged portion of two surfaces 
in contact. In general, an apparently smooth sur-
face is microscopically very irregular so that two 
such surfaces are not in contact everywhere but 
only at irregularly located points such as A, B, and 
C. At such points, the interatomic forces of attrac-
tion are quite large and can be thought of as atomic 
size "spot welds." To break these "spot welds" so 
that relative motion can begin, requires relatively 
large forces. Once motion is begun, the protruding 
portions of the two surfaces move relative to each 
other, which tends to reduce the strength of the 
"spot welds." Thus, once motion begins, a smaller 
pulling force will be required to maintain relative 
motion of the surfaces at a constant speed. As indi-
cated earlier, it is observed that within rather wide 
limits the force required to maintain motion is inde-
pendent of the speed. Thus, Eq. (18-3) requires the 
additional relation 
ΙΡ Ι = ΙΡ /| = μ κΙΝ |. 
(18-4) 
We conclude this section with an example in-
tended to emphasize that, while Ν  is perpendicular 
to the surfaces of contact and F/ acts tangent to the 
surfaces of contact and in a direction opposite to 
any actual or impending motion, it is not always 
true that |N| = m|g| and |F/| = |P|. 
|P|«M,|N| = ^.m|; 
(18-2) 
Figure 18-2 Schematic enlargement of two surfaces 
in contact. 

144 
Applications of the Laws of Motion 
Example 1. A block of mass m and weight mg is 
held against a vertical plane surface due to a hori-
zontal force Ρ  as shown in Figure 18-3. The static 
and kinetic coefficients of friction are μ , and μ κ, 
respectively. 
(a) Determine the magnitude of |P| if downward 
motion of the block impends. 
(b) Determine the magnitude of |P| if downward 
motion at constant speed is taking place. 
Figure 18-3 Force diagram for Example 1. 
SOLUTION 
(a) As indicated in the figure, in this case Ν  -I- Ρ  = 
0 and F/ + mg = 0, or Ρ  = N, F/= mg. Since 
Ff = μ ,Ν , it follows that mg = μ ,Ρ  or Ρ  = 
mg|μ s. 
(b) In this case, the vertical and horizontal forces 
are still balanced (since the speed is constant, 
a = 0). Therefore, we obtain by the analysis used in 
(a) 
In Table 18-1, we list some values for μ , and μ κ 
for various surfaces. Since friction forces depend 
markedly on microscopic details of both surfaces 
(surface cleanliness, crystalline imperfection, etc.), 
these values should be considered as typical values 
only. 
Table 18-1 Coefficients of Friction (dry surfaces). 
Substances 
μ » 
μ κ 
Tenon on teflon 
0.04 
0.04 
Copper on steel 
0.50 
0.35 
Aluminum on steel 
0.60 
0.50 
Lead on steel 
0.95 
0.95 
Copper on cast iron 
1.05 
0.30 
18-3 
P R O J E C T I L E M O T I O N 
There are many familiar examples of the type of 
motion known as projectile motion. A projectile 
fired from some type of cannon, a football put in 
motion by a kicker, and a golf ball sent in flight by a 
golf club all can be described by the same physical 
characteristics with only minor variations in details. 
Figure 18-4 illustrates the principal features: at the 
origin, a particle of mass m has an initial velocity of 
magnitude Vo directed at an angle θο  relative to the 
horizontal, brought about by the action of some 
force (explosive propellant, muscular effort of the 
kicker, or the force transmitted by the driven golf 
club for the examples cited). We now limit our dis-
cussion to low speeds and ignore the rotation of the 
Earth and air resistance. In this situation, the only 
force that acts upon the projectile after launch is 
the downward force of (or due to) gravity. New-
ton's second law for this situation, in vector compo-
nent form is, from 2F = ma, 
-mg} 
= m(aA-^ay}), 
(18-5) 
Equating components on both sides of this equa-
tion gives 
(18-6) 
From Eq. (18-6), it follows that 
= constant. Now 
from Figure 18-4, vox = Vocos θο  is the initial x-
component of the velocity. In the absence of any 
forces in the χ -direction, we see that Vx = Vox = 
Vo cos So. In the y-direction, Eq. (18-7) requires that 
(refer to Section 14-4) 
Γ  dvx = - g 
¡ 
dt 
J voy 
Jo 
or 
Vy = Voy-gt, 
(18-8) 
Figure 18-4 Schematic diagram of projectile motion. 

Projectile Motion 
145 
Since 
Voy = Vo sin θο , 
we have 
Vy = uosinö o-gi. 
(18-9) 
Similarly, we can find the x- and y- components 
of the displacement by noting that 
dx 
= ^ = Uo COS oo 
or 
and that 
j dx = VQ COS OO j dt, 
X = t;ocos0oi; 
(18-10) 
d y 
i^y = ^ = uo sm 00 - gt. 
or 
y = ü osin0oí-^gí'. 
(18-11) 
What is the physical content of our mathematical 
manipulations? First, the absence of forces in the 
X-direction yields a constant velocity component in 
the X-direction. As a result, the χ -component of the 
displacement is a linear function of the time. In the 
y-direction, the acceleration is constant because 
the force (due to gravity) in the y-direction is con-
stant. As a result, the y-velocity-component is a 
linear function of time; the y-component of the dis-
placement, therefore, is a quadratic function of 
time. We see further that the application of New-
ton's second law for our projectile motion example 
leads to a separation of the x- and y-components of 
the motion. (The extension of the problem to in-
clude 3 components adds no further complications.) 
Let us turn now to the questions: 
1. What is the maximum height (y^ax) attained by 
the projectile and how long does it take to 
attain it? 
2. What is the range (R), the surface-to-surface 
horizontal distance traveled by the projectile, 
and the time of flight (itotai)? 
3. What is the shape of the trajectory (the path 
traversed by the projectile)? 
It is clear, first of all, that Vy must be equal to zero 
when the particle is at the maximum height. (If it 
were not, either the particle would still be moving 
upward or would be in motion to a lower position. 
For either possibility, the y-displacement could not 
be y„^x.) Therefore, Eq. (18-8) gives 
0= 
Vo sin oo - giup, 
where iup is the time from launch (i = 0) to the at-
tainment of ymax. Thus, 
iup 
Vo sin θο  
(18-12) 
Substitution of t = iup from Eq. (18-12) and y = y„ 
into Eq. (18-11) yields 
ymax = -
vo^ sin^ θο  
2g 
(18-13) 
From Eq. (18-13), we can deduce (as might seem 
obvious from experience) that a maximum height 
for given vo and g will occur for a projectile fired 
vertically (oo = 90°). 
To determine the range R , we first note that when 
t —  ί total» y = O (the projectile has returned to the 
launch elevation). Thus, Eq. (18-11) requires that 
0= Vo sin θot - | g i ^ 
or 
0 = ( r o s i n Ö o - y ) i . 
(18-14) 
Equation (18-14) has two solutions: (a) t = 0, the in-
itial situation; and (b) t = itotai = 2vo sin θolg, the de-
sired time of flight. Notice that itotai is exactly twice 
the time required to attain ymax. This tells us that the 
flight is symmetric; that is, it takes as long to go up 
as it does to come down when one considers only 
the vertical force due to gravity. Substituting the 
value for itotai in Eq. (18-10) for the x-motion, we get 
R = Ivo sin θο  cos 00 
or 
R = VQ sin 2 0 0 
g 
(18-15) 
(18-16) 
For fixed values of Vo and g, the maximum value of 
R will be obtained when sin20o attains its max-
imum value, which is unity. Since this occurs for 
2 0 0 = 90°, we see that for maximum range the 
launch angle must be 45°. (The reader who is 
anxious to test the concept of "max-min" determi-
nations by calculus techniques can arrive at this 
same result by solving the equation dR |dθo = 0.) 
The shape of the trajectory is not difficult to de-
termine. First of all, we already know the path is 

146 
Applications of the Laws of Motion 
symmetric about the point (l?/2, y max) since í total 
= 
2iup. To determine the analytical equation, we sim-
ply eliminate the time parameter from Eqs. (18-10) 
and (18-11). Thus, Eq. (18-10) gives 
t =-Vo cos θο ' 
Substituting this result in Eq. (18-11) yields 
rosinö o^ 
1 
gx^ 
Vo COS θο  
2 Vo COS^ θο  
(18-17) 
gx 
= x tan ö o-;r-r—TT". 
2vo cos 00 
(18-18) 
Since y varies quadratically with jc, Eq. (18-18) pre-
dicts that the path of the projectile will be 
parabolic. It is left for the reader to show that Eq. 
(18-18) predicts a parabola that is synmietric about 
the point x=Rl2. 
(That is, show that when jc = 
RI2 + A x the same value of y is obtained for posi-
tive or negative Δ χ ) 
Finally, it should be noted that the analysis of 
projectile motion carried out here is also valid for 
such seemingly different situations as skiers leaving 
the horizontal end of a ski jump or golf balls driven 
from a tee to a green at a higher elevation. 
Example 2. A skier leaves the horizontal end of 
a ski jump with a speed of 20 m/sec. If the skier 
lands 40 m out from the end of the ski jump, how 
far below the end of the ski jump is the landing 
point? 
SOLUTION 
Referring to Figure 18-5, it is clear that since θο  = 
0°, Vox = Vo, Voy = 0. Therefore, from Eq. (18-17), we 
Vo = 20 m/sec 
Figure 18-5 System diagram for Example 2. 
find the time of flight to be 
^ ^ 4 0 
Vo 
20 • 
Using this result in Eq. (18-18) gives the result 
(The minus sign indicates a vertical displacement 
below the origin.) Notice that this result also fol-
lows from Eq. (18-11) for β ο  = 0°, as it should. 
18-4 
U N I F O R M C I R C U L A R M O T I O N 
A particle in uniform circular motion follows a 
circular path with a velocity that is constant in 
magnitude. Its acceleration can have no component 
tangent to the path since this would cause the mag-
nitude of the velocity to change. The changing di-
rection of the velocity requires an acceleration 
component perpendicular to the velocity vector 
(and the circular path)—that is, directed radially in-
ward. Such an acceleration is called centripetal 
(center-seeking). The force associated with such an 
acceleration (via the second law) is known as a 
centripetal force, and it can arise in many ways. For 
planetary motion discussed in the next section, the 
force is gravitational in nature, while a whirling 
stone on a string experiences a centripetal force 
due to the tension in the string. 
Before we consider examples of such systems, 
let us first determine the kinematical relationship 
between the centripetal acceleration and the veloc-
ity of a particle in uniform circular motion. Figure 
18-6(a) shows the particle at two different points 
along its path corresponding to positions separated 
by a distance R A Θ , which was traversed in a time 
Δ ί. The displacement A s between the two points 
can be viewed as the base of an isosceles triangle 
with equal sides of length R . The two velocity vec-
tors V and v ' are also of equal length v, and each is 
perpendicular to a radius vector of length JR. Figure 
18-6(b) illustrates the vector relation v-\-Av 
= 
v' 
that the velocities must satisfy. Here again we have 
an isosceles triangle whose equal sides are each 
perpendicular to one of the equal sides of the 
triangle of Figure 18-6(a). As you may recall, the 
two triangles are said to be similar when such con-
ditions exist—that is, with identical apex angles, 
which in this case are of magnitude Α Θ . Further-

Uniform Circular Motion 
147 
1 
'— 
λ 
/^θyΛ  
Χ /Δν  
ν ' 
(a) 
(b) 
Figure 18-6 Displacement (a) and velocity (b) vector 
diagrams for uniform circular motion. 
more, Ar will be perpendicular to As. In terms of 
the two triangles, therefore, we can write 
or 
R 
V 
Av 
=-^As. 
(18-19) 
(18-20) 
Since the change in velocity Av occurs in a time Δ ί, 
the average acceleration ä  is 
_ 
_Av_v_As 
^~ At~ 
RAt' 
(18-21) 
In the limit as Δ ί ->0, As will become tangent to the 
path so that 
As 
ds 
and [inj ä  = a. When Δ ί -»0, As becomes tangent to 
the path so that Δ ν  is directed radially inward as 
required. We see, therefore, that the expression re-
lating the centripetal acceleration and the velocity 
in uniform circular motion is 
V 
^=R' 
(18-22) 
In situations involving non-uniform circular mo-
tion, it is rather conunon to find the centripetal ac-
celeration denoted by a«, while the component 
tangent to the path (which is responsible for the 
change in magnitude of v) is given by 
ατ—the 
tangential acceleration. 
From the discussion above, we conclude that for 
a particle in uniform circular motion Newton's sec-
ond law takes the form 
lF=mñR. 
(18-23) 
The physical significance of Eq. (18-23) is that re-
gardless of the nature or number of forces applied 
to the particle of mass m, if it is in uniform circular 
motion, the vector sum of these forces must be 
directed radially inward. 
Example 3. A boy fastens a small stone of mass 
m to a light string of length /, raises it over his head, 
and sets the stone in motion so that it travels a 
horizontal circular path of radius r with a velocity 
having a constant magnitude. Determine the ten-
sion Τ  in the string and the magnitude of ν  in terms 
of m, g, /, and r. (See Figure 18-7.) 
SOLUTION 
As indicated in Figure 18-7, we first resolve the 
tension into vertical and horizontal components. 
Thus, 
and 
Tx = Τ  sin θ 
Ty = T cos Θ . 
(18-24) 
(18-25) 
Applying the second law to the vertical forces, we 
have equilibrium (no vertical motion) 
Ty-mg= 
0. 
For the horizontal motion, we obtain 
Tx = mar = 
mt; 
(18-26) 
(18-27) 
Figure 18-7 Force diagram for Example 3. 

148 
Applications of the Laws of Motion 
Combining these last four relations yields 
mv 
Since we can also write 
mg 
υ 
gr' 
tan θ = 
it follows that 
(18-28) 
By combining Eqs. (18-24) and (18-27), we also find 
T = 
mgl 
mg 
(18-29) 
Notice that our solution [Eq. (18-29)] indicates that 
T-^oo as r ^ / . This means that it is not physically 
possible to swing the stone so that the string is 
horizontal. If it were possible, Eq. (18-26) would 
have to become 
-mg 
=0, 
because there would now be no vertical component 
of the tension to counteract the force due to gravity 
on the stone. Alternatively, Eq. (18-28) would re-
quire that i; ^ 00 as Γ 
/, which is also not possible. 
As another example of uniform circular motion, 
we now consider an object supported by a spring 
scale at some point on the surface of the Earth. The 
problem is to determine W, the reading of the 
spring scale that will be the effective weight of the 
object. At first glance, it would appear reasonable 
to say that due to equilibrium the reading of the 
scales will be equal to the product of the mass of 
the object, and 
GME 
the acceleration due to gravity at the Earth's sur-
face when rotation is neglected. However, because 
the Earth rotates about its axis, a point on the sur-
face will be in uniform circular motion about the 
axis of rotation. As a result, the object experiences 
an acceleration toward the axis of rotation. In other 
words, a point on the surface of the Earth cannot be 
used as the origin of an inertial reference frame. In-
stead, we must describe the motion from an inertial 
frame (with an origin at 0) that is fixed in space and 
relative to which the Earth rotates about its axis. 
Figure 18-8 shows a particle of mass m at rest at 
point Ρ  on the surface of the Earth at an angle θ 
(the latitude angle) above the Equator. At this loca-
tion, the magnitude of uniform circular velocity of 
the particle due to the Earth's rotation will be the 
circumference of its path 2'7γγ divided by the time Τ  
corresponding to the rotation of the Earth (24 hours 
or 8.64 X 10^ sec). Since r = RE c o s 0, it follows that 
V = -
2π RE 
cos 
θ 
(18-30) 
This means that there will be a resultant accelera-
tion directed inward along r toward the axis of 
rotation given by 
ar=j = Att^RE c o s Θ  
ψ , 
. 
(18-31) 
For the Earth, RE « 6.4 x 10^ m, Τ  = 8.64 x 10^ sec, 
so that ü r « 3.34 x 10"^ cos θ m/sec^ 
Now let us consider the forces acting on the par-
ticle. Figure 18-9 illustrates the situation with W 
representing the force supplied by the spring. W 
can be replaced by component forces 
WR^ = 
W cos a and WT = W sin a, respectively. Simi-
larly, the centripetal acceleration directed along r 
can also be replaced by components along OP and 
tangent to the surface at P. Thus, the component 
along OP has a magnitude 
= ü r cos θ « 3.34 χ  10"' cos' 0, (18-32) 
\ 
/ 
V/ 
7s 
^ 
Equator 
Figure 18-8 Schematic diagram of a particle at rest 
on a rotating earth. 

Planetary Motion and Kepler's Laws 
149 
W cos a <" 
Figure 18-9 Force diagram for the system of Figure 
18-8. 
and the magnitude of the component tangent to the 
surface is 
ατ  = ar sin θ « 3.34 x 10"^ cos 0 sin 0 
« 1 . 6 7 x 1 0 ' s i n 20. 
(18-33) 
Now we can apply Newton's second law in compo-
nent form to the radial and tangential parts just 
evaluated. Thus, 
W cos a - mgE = - maR^ = - mar cos 0 
(18-34) 
and 
W sma= 
mar = mar sin 0. 
(18-35) 
By eliminating W from Eqs. (18-35) and (18-36), 
we can deduce that a must be rather small. 
Eliminating W yields 
tan a = 
ar sin 0 
gE - ar cos 0 
or 
Since 
and 
1—^cos0 
gE 
(18-36) 
Or 
gE 
3.34x10" 
9.832 
0 ^ 0 ^ 9 0 ° , 
= 3.3x10" 
it follows that tan α « 3.3 x 10"^ sin 0. We see, 
therefore, that even when sin 0 is unity (P is at the 
North Pole), a will be substantiaUy less than Γ so 
that cos α « 1 . Returning now to Eq. (18-34), we 
have 
W' - mgE « - mar cos 0. 
Using Eq. (18-31) and rearranging gives 
W « m(gE - 3.34 X 10"'cos'0) 
or 
W 
= mgeff « m (9.832 - 3.34 x 10"' cos' 0). 
(18-37) 
This analysis shows that the effective value of g is 
reduced (except at the North Pole, where cos 0 
and, hence, r vanish). Therefore, the use of New-
ton's laws assuming that the Earth is an inertial 
reference frame cannot give a correct result unless 
a fictitious "centrifugal" force F c of magnitude 
given by mar cos 0 and directed outward (from O to 
Ρ  in Figure 18-9) is assumed to act. When this is 
done, Newton's first law (assuming an inertial 
reference frame with origin at P) yields 
or 
W' + 
Fc-mgE=0 
W = mgE - mar cos 0, 
which is just Eq. (18-37), as it should be. As noted 
in Section 18-1, we will not pursue further the com-
plications due to motion relative to non-inertial 
frames of motion. As a final remark, it should be 
noted that this approximate analysis provides an 
expression [Eq. (18-37)] for the variation of gcñ 
with latitude, which is in good agreement with 
experimentally determined values, ranging from 
9.780 m/sec' at the Equator to 9.832 m/sec' at the 
North Pole. 
18-5 
P L A N E T A R Y M O T I O N A N D K E P L E R ' S 
L A W S 
To begin this discussion, we consider a simplified 
version of planetary motion. Figure 18-10 shows a 
star of mass M, about which a single planet of mass 
m orbits in uniform circular motion. The radius of 
the orbit is P, and Τ  is the period or time required 
to complete one orbit around the star. We assume 
for further simplicity that the planet does not rotate 
about its axis. For the Earth-Sun system, this 
would involve a time error of only « 240 seconds in 
one year, so it is not a serious departure from 
reality. 
If it is further assumed that the star does not 
move, then we can place the origin of an inertial re-
ference frame at the center of the star and use this 

150 
Applications of the Laws of Motion 
f 
Ms 
m 
Table 18-2 Solar System Data. 
Figure 18-10 Schematic diagram of a planet orbiting 
a star in uniform circular motion. 
frame to apply Newton's laws to analyze the mo-
tion of the planet. The only force acting on m will 
be the gravitational force due to M„ which will act 
along the line joining the centers of the two masses 
toward Ms. Since the orbital velocity is tangent to 
the path and constant in magnitude, the resultant 
acceleration is also radially inward. Therefore, 
combining Newton's second law and the law of 
gravitation gives 
GMsin 
mv^ 
R 
(18-38) 
Since 
V = -
IjrR 
Τ  
' 
we can rearrange Eq. (18-38) to yield 
(18-39) 
Now, if we extend our model by assuming not one 
but several planets orbit the star without perturbing 
the motions of each other, then Eq. (18-39) shows 
that for any two planets regardless of their masses. 
:R7 1?· 
(18-40) 
Data obtained for the solar system (see Table 18-2) 
show all the planets have nearly circular orbits 
(only those of Mercury and Pluto are appreciably 
non-circular). However, the large masses of the 
planets lead to some small but significant depar-
tures from simple circular orbits for the outermost 
planets. Historically, in fact, the barely visible dis-
Mean orbit 
Planet 
Mass (kg) 
Period (sec) 
radius (m) 
Mercury 
3.28x10^^ 
7.60x10* 
5.79x10'° 
Venus 
4.83 X 10^ 
1.94x10" 
1.08x10" 
Earth 
5.98x10^^ 
3.16x10" 
1.49x10" 
Mars 
6.40x10" 
5.94x10" 
2.28x10" 
Jupiter 
1.90x10'" 
3.74x10' 
7.78x10" 
Saturn 
5.68 X 10'* 
9.30x10' 
1.43x10" 
Uranus 
8.67 X 10'' 
2.66x10' 
2.87 X 10" 
Neptune 
1.05x10'* 
5.20x10' 
4.50x10" 
Pluto 
5.37 X lO'^t 
7.82 X 10' 
5.91 X 10" 
tUncertain 
tant planets Uranus, Neptune, and Pluto were each 
discovered only after the minor variations in the 
orbit of a neighboring planet were accounted for by 
assuming the existence of another planet further re-
moved from the sun. 
The reader may well be puzzled by the fact that 
planetary masses are given in Table 18-2, when it is 
clear from Eq. (18-39) that the planetary mass is not 
involved in the period-orbital radius relation. How-
ever, astronomers have measured periods and orbi-
tal radii for the satellites (moons) of the planets and 
they are observed to also move according to Eq. 
(18-39) with M 5 now understood to be the mass of 
the planet. For example, the mean radius of the 
Moon's orbit about the Earth is 3.84 x 10* m, and its 
period is 2.36 x 10"^ sec, so that 
^ 
4 7 r U 8 4 x l O Y 
_ . ^ ο . . , π 2 4 , 
^ ^ = 6 . 6 7 Χ
1 0 - ( 2 . 3 6 Χ
1 0 ^ ) ' - ^ · ^ ^ ^ ^ ^ 
(A similar calculation shows that the mass of the 
Sun is 1.98X lO'^kg.) 
Historically, Newton used the empirical relations 
governing planetary motion that Johannes Kepler 
had obtained (by careful analysis of astronomical 
data gathered by Tycho Brahe) to deduce the law of 
universal gravitation. We therefore state Kepler's 
"laws" and try to confirm that such a connection 
exists. There are three relations and they can be 
stated as: 
I The planets describe elliptical orbits, with the 
Sun at one focus. 
II The position vector of any planet relative to 
the Sun sweeps out equal areas of the orbital 
ellipse in equal times. 

Central Force Problems 
151 
III The squares of the orbital periods are propor-
tional to the cubes of the average distances 
of the planets from the Sun. 
The reader may recall that a circle is a special 
case of an ellipse (where the two foci coincide). 
Then laws I and II are obeyed by our simple model, 
and we have already seen in Eq. (18-40) that laW III 
is valid as well. It is not as easy to demonstrate the 
fact that the law of universal gravitation satisfac-
torily predicts Kepler's laws for a general elliptical 
orbit. However, since the difficulties are connected 
with the geometry of ellipses rather than the physi-
cal principles involved we shall not pursue the 
question here. There are numerous sources to 
which the interested reader may refer.t 
Example 4. An Earth satellite moves in a circu-
lar orbit at a height of 6 x 10^ m above the Earth's 
surface. Determine: 
(a) the period of revolution and 
(b) the magnitude of its velocity. 
Note that the mean radius of the Earth is 6.40 x 
10'm. 
SOLUTION 
(a) Since we were given the period and radius of 
the orbit of the Moon about the Earth in calculating 
the mass of the Earth, we can use them in applying 
Eq. (18-40): 
χ  10" 
Τ . = 2 . 3 6 Χ 1 0 ' ( 3 ^ ) ' 
= 5.80 χ  10' sec 
~ 97 min. 
IttRs 
27Γ(6.40 + 0.60)Χ 10' 
— = 
5.80x10^ 
Ίττ Χ  10' 
2.9 
17,000 mi/hr. 
= 7.58 χ  10' m/sec 
18-6 
C E N T R A L F O R C E P R O B L E M S 
We have seen in the last section that the law of 
universal gravitation and Newton's laws of motion 
are sufficient for analyzing the motion of planetary 
systems. The analysis can be made far more general 
by noting that the form of the gravitational force is 
rather special in that the force acts along a line join-
ing the two interacting bodies. Thus, the direction 
of the gravitational force on the planet was taken to 
be toward the center of the star, so that even 
though the planet moved in a circular orbit, the 
force always pointed toward the same central point. 
Although we have not proved it, when the orbital 
motion is elliptical rather than circular, the interac-
tion force must still be centrally directed. A force 
that is always directed toward a fixed point is called 
a central force. It can be shown that motion analo-
gous to that described by Kepler's laws will result 
regardless of the exact nature of the central force. 
There are many situations in nature for which the 
forces are either central or very nearly central. As a 
result, understanding the characteristics of plane-
tary motion can lead to the analysis of any new situ-
ation if the interaction force is basically central in 
form. 
Although we will not discuss them here, two im-
portant cases of this sort from the field of atomic 
physics are: (1) the motion of an electron about a 
proton in a hydrogen atom is due to a central force 
that is attractive and electrostatic in nature, and (2) 
the scattering of alpha particles (helium nuclei) by 
gold atoms in a thin foil is due to an essentially cen-
tral force that is repulsive and electrostatic in 
nature. 
One should not feel that there is little merit in 
studying the motion of planets (or satellites) simply 
because such studies have been continuing for over 
300 years. After all, Rutherford used the alpha par-
ticle scattering technique to deduce the nuclear 
structure of atoms in 1911, and the present space 
exploration activities of the USA and the USSR are 
based on the satellite motions that are in principle 
no different from those of the planets around the 
Sun. 
tSee, for example, M. Alonso and E. Finn, Fundamen-
tal University Physics (Vol I), Addison-Wesley, 1967, 
Chapter 13. 

152 
Applications of tt)e Laws of Motion 
P R O B L E M S 
1. A 4.8 lb trolley moves along a horizontal track for which the coefficient of friction is I It is pulled by a 
string that passes over a frictionless pulley at the end of the track attached to a 1.6 lb piece of lead that 
hangs freely at the end of the string. Find: 
(a) the acceleration of the trolley and 
(b) the tension in the string. 
2. A 39 lb block A is placed on an inclined surface and connected by a cord to a 25 lb hanging block Β as 
shown in Figure 18-11. The coefficient of friction between the block and the surface is 0.10. Compute: 
(a) the acceleration of blocks A and Β and 
(b) the tension in the cord. 
Figure 18-11 
3. Find the acceleration that will be imparted to a 10 kg block lying on an inclined plane by a force of 100 nt 
parallel to the plane, if the plane is inclined at 30° to the horizontal and the coefficient of friction is 0.10 
between the block and the plane. 
4. A cord extends horizontally from an 8 kg block on a horizontal plane over a pulley and then vertically 
downward with a 5 kg ball at its lower end. The kinetic coefficient of friction between the block and the 
plane is 0.250. Compute: 
(a) the acceleration of the block and 
(b) the tension in the cord. 
5. In Figure 18-12, mi is 1 kg and m i is 2 kg. The coefficient of static friction between mi and m i is 0.40. The 
coefficient of sliding friction between m i and the table is 0.15. The body ma has that mass which, when 
m3 is released, gives the system the maximum acceleration possible without mi slipping relative to m i . 
(a) What is the maximum acceleration? 
(b) What is the tension in the cord when the 
system has the maximum acceleration? 
(c) What is the mass of ma? 
Figure 18-12 

Problems 
153 
6. An 80 lb block on a long plane inclined 30° to the horizontal, starts from rest and slides down the plane. 
The coefficient of sliding friction between the block and the plane is 0.25. Compute: 
(a) the resultant force on the block, 
(b) the acceleration of the block, and 
(c) the time for the block to slide 20 ft along the inclined plane starting from rest. 
7. A block weighing 20 lb is on a plane inclined at 40° with the horizontal. The coefficient of friction 
between the block and the plane is 0.50. 
(a) What is the minimum push on the block, applied parallel to the plane, that will keep the block from 
starting to slide down the plane? 
(b) What push applied parallel to the plane will start the block sliding up the plane? 
(c) What push applied parallel to the plane will give the block a constant acceleration of 8 ft/sec' up the 
plane? 
8. A 20 lb block on a long plane inclined at 30° to the horizontal is attached to a hanging ball of JC lb by 
means of a cord and pulley as shown in Figure 18-13. The coefficient of sliding friction between the 
block and the plane is 0.25. 
(a) When the block is sliding up the plane at 
constant speed, compute the weight of JC. 
(b) Later, the cord is cut, allowing the block to 
slide down the plane. Find its acceleration. 
Figure 18-13 
9. (a) A baseball player throws a baseball with a velocity of magnitude 96 ft/sec in a direction 30° above 
the horizontal. If the baseball leaves the thrower's hand at a height of 6 ft above the field, find the 
maximum height attained and the time it takes to reach the maximum height, 
(b) Find the horizontal distance traveled by the ball before it strikes the ground. 
10. A basketball player releases a ball 6 ft above the floor at 60° above the horizontal. It passes through the 
basket 2.1 seconds later, after reaching a maximum height of 29 ft above the floor. 
(a) What was the speed of the ball as it left the player's hands? 
(b) How high above the floor was the basket? 
11. (a) The angle of elevation of an anti-aircraft gun is 70°, and the muzzle velocity of the shell is 
2700 ft/sec. If the shell is fused for 50 seconds, what is the greatest height the shell can reach before 
explosion? 
(b) What are the x- and y- coordinates of the shell 25 seconds after it was fired? 
12. A golfer hits a ball and imparts to it an initial velocity 10 m/sec at an angle oo with the horizontal. At the 
apex of its path, it just clears the top of a tall tree 50 m high. Find: 
(a) the time for half the total flight, assuming a level golf course and 
(b) how far away on the golf course the ball lands? 
13. A bullet is fired from ground level. The horizontal component of the muzzle velocity is 1600 ft/sec and 
the vertical component is %0 ft/sec. 
(a) How long does it take for the bullet to reach the highest point in its trajectory? 
(b) To what maximum height will the bullet rise? 
(c) What is the horizontal range of the bullet? 

154 
Applications of the Laws of Motion 
14. A golf ball is driven with a velocity of 200 ft/sec at an angle of 37° above the horizontal. It strikes a green 
at a horizontal distance of 800 ft from the tee. 
(a) What was the elevation of the green above the tee? 
(b) What was the velocity of the ball when it struck the ground? 
15. A car starts from rest on a circular race track of radius 1000 ft, and increases its speed at the rate of 
4 ft/sec'. 
(a) How long a time will be required to attain the speed at which the tangential and radial components of 
the car's acceleration are equal? 
(b) How far does the car move along the track before the radial and tangential components of its 
acceleration are equal? 
16. A satellite travels in a circular orbit of radius 7.20 x 10^ m with a constant speed of 7 m/sec. 
(a) What is its acceleration? 
(b) What is the average rate of change of speed? 
17. A small box is placed 4 ft from the center of a horizontal rotatable platform. If the coefficient of static 
friction between the box and the platform is 0.50, what is the maximum number of revolutions per 
second the platform can make without having the box start to slide? 
18. On a horizontal frictionless surface, a 500 gm body revolves in a circle whose radius is 90 cm. Find the 
magnitude of the centripetal force on the body if it makes one revolution per second. 

19 
Momentum 
19-1 
Impulse and Momentum 
155 
19-2 
Conservation of Momentum 
156 
19-3 
Elastic and Inelastic Collisions 
158 
19-4 
Motion of Systems with Variable 
Mass 
159 
19-5 
Further Considerations 
161 
19-1 
I M P U L S E A N D M O M E N T U M 
In the previous chapter, a number of physical 
situations were analyzed satisfactorily by direct 
application of Newton's laws of motion. In each 
case, it was possible to identify the forces involved 
without regarding them as explicit functions of 
time. For projectile motion, the force due to gravity 
was constant in magnitude and directed downward 
throughout the motion. In the case of uniform cir-
cular motion, the centripetal force was constant in 
magnitude and always directed toward the center of 
the circular path. Thus, it was possible to analyze 
the motion without regard to the fact that the cen-
tripetal force acts in a direction that varies with 
time. 
The situation is less agreeable when we attempt 
to study cases involving forces that are explicit 
functions of time, especially those acting for very 
short time intervals and having magnitudes that 
vary markedly during the interval. Using the sec-
ond law, as we did in Chapter 18, will then be 
difficult. This is so because now the acceleration 
must also become an explicit function of time, 
which means that to determine the equation of mo-
tion (r as a function of time) will require two inte-
grations with respect to time. This can be seen 
when we consider a mass m subject to a resultant 
force F(0- Newton's second law of motion requires 
that 
2F.(í) = F(í) = ma, 
^ d v ^ F ( 0 
* 
di 
m · 
(19-1) 
We can now determine r(i) by two integrations 
with respect to time as indicated in Eqs. (14-18) and 
(14-19). 
It is not hard to see that if F(i) is a complicated 
function of time, it will be difficult to determine the 
equation of motion. Furthermore, there are many 
instances for which it is not possible to state the 
time dependence of the resultant force in functional 
form, which additionally complicates the problem. 
As an example, consider the force imparted to a 
golf ball when it is struck by a golf club. Until the 
club contacts the ball, the magnitude of the force is 
zero. It then rises rapidly to a very large peak value 
corresponding to the most highly contracted or 
compressed state of the golf ball. As the ball ex-
pands and begins to move away from the club, the 
magnitude of the force falls rapidly to a value of 
zero as contact is lost. Thus, a graph of the mag-
nitude of the force as a function of time might re-
semble Figure 19-1. 
For forces of this type, we shall timi to an alter-
native approach for studying the motion. To begin, 
we return to the second law and integrate it with 
respect to time. 
155 

156 
Momentum 
Figure 19-1 Schematic graph of an impulse force as 
a function of time. 
F(t)dt = 
m adt. 
The integral JO F(t)dt, is defined to be the impulse of 
the resultant force F(i). As we have seen, this in-
tegral may be difficult to evaluate. The integral of 
the RHS offers no difficulty, however, for 
m sidt = m 
dv 
= m(v-vo). 
(19-3) 
The quantity m ν  is defined to be the momentum of 
a particle. In many texts, it is given the single sym-
bol p. Thus, we could write the relation 
p = mv. 
(19-4) 
In terms of momentum, Eq. (19-3) states that the 
impulse of the resultant force acting on a particle 
for a time t is equal to the resulting change in 
momentum of the particle during that time. It 
should be noted that both impulse and momentum 
are vectors. 
Example 1. A golf ball of mass 0.05 kg is initially 
at rest on a tee. It is then struck by the head of a 
golf club which is in contact with the ball for 5 x 
10"^ sec. If the ball leaves the tee with a speed of 
80 m/sec find: 
(a) the magnitude of the momentum for the ball 
just after the contact, and 
(b) the impulse delivered to the ball by the club. 
(c) Suppose the ball had acquired the same initial 
speed by means of a constant force Κ  acting for 
the same time interval as the variable force due to 
the golf club. What would be the magnitude of the 
constant force? 
SOLUTION 
(a) From the definition of momentum, we obtain 
ρ =mv= 
(0.05 kg)(80 m/sec) = 4 kg-m/sec. 
(b) By Eq. (19-3), we have 
Fdt 
=mv-0 
= 4 kg m/sec. 
(c) If the final momentum is to be the same, then 
the change in momentum will also be the same. 
(The ball still starts from rest.) Therefore, 
Fcdt = Fcj^ di = ( F c ) ( i ) = 4 k g m / s e c . 
(19-2) 
Hence, 
_ 4 kg m/sec _ ^ 
^ ^ " 5 x 1 0 - ^ ™
" ^ ^ " ^ · 
sec 
19-2 
C O N S E R V A T I O N OF M O M E N T U M 
The relation between impulse and momentum 
[Eq. (19-3)] does not appear to have brought us any 
closer to our goal since it is in reality only a thinly 
disguised form of Newton's second law. To see 
how it can be put to good use, consider the collision 
of two particles A and Β that have masses mA and 
mfl, respectively. Initially, A and Β have velocities 
y A and Vß, respectively. At time t after the collision, 
their velocities have become 
and VB as shown 
in Figure 19-2. If the only force acting on either par-
ticle is the resultant interaction force, then the 
force on A due to B, FAB, is (by Newton's third law 
of motion) equal in magnitude but opposite in direc-
B 
AS 
Β  
y/A 
VB 
FAS 
fa 
Before 
During 
v. 
collision 
collision 
After 
collision 
Figure 19-2 One-dimensional two-body collision. 

Conservation of Momentum 
157 
tion to the force on Β  due to A, FBA. That is, 
FAB = - FBA, By Eq. (19-3), the impulse of FAB is 
F A B dt = mA\A- 
mA\A, 
(19-5) 
while the impulse of FBA is 
FBA dt = me V B - mfl VB. 
(19-6) 
Since 
1^ FABdt=-j^ 
FBAdt, 
Eqs. (19-5) and (19-6) yield 
mA V A - mA VA = mB \ B - me V B 
(19-7) 
or 
mA V A + mBVfl = W A V A + mßVB. 
(19-8) 
Equation (19-8) indicates that for a system of two 
particles subject only to an interaction force (no ex-
ternal forces) the total momentum of the system be-
fore the interaction is equal to the total momentum 
after the interaction. Since there has been no 
change in total momentum, Eq. (19-8) is a statement 
of what is called the law of conservation of momen-
tum. Therefore, Newton's laws of motion require 
that, for a system of two particles subject to no 
external forces, the total momentum of the system 
must be constant (or conserved). 
Example 2. A particle of mass mi = 1 kg and a 
velocity i;o = 5m/sec to the right collides with a 
second particle of mass mi = 2 kg, which is initially 
at rest. In this collision, mass m, is deflected from 
its original direction by an angle θι = 53°, and its 
speed after the collision is 
= 3 m/sec. Find: 
(a) the angle θι for the direction mass mi moves 
with respect to the original direction of mi, and 
(b) the magnitude of the velocity of the second 
mass (vif) after the collision. 
SOLUTION 
Before the collision, the total momentum of the 
system is m,i;io = 5kg m/sec, and it is directed 
along the x-axis. From Eq. (19-8), therefore, the 
total momentum afterward must also be along the 
X-axis. However, U I / has both x- and y-components, 
so it follows that vy must also have x- and 
y-components. Figure 19-3 illustrates the situation 
before and after the collision. We have shown mi 
moving downward when mi moves upward. This 
must be true if there is to be no y -component for the 
total momentum. If the final momenta for mi and mi 
Figure 19-3 System diagram for Example 3. 
are broken into x- and y-components, Eq. (19-8) and 
Figure 19-3 require that 
mii^io = miUi/cos 01 + m2Ü2/cos 02 
(i) 
and 
0 = mit;i/ sin 0i - ^2^2/ sin 02. 
Combining Eqs. (i) and (ii) yields 
tan 02 = 
Uj/sin 01 
Vio-Vif 
COS 01* 
(ii) 
(iii) 
Substituting numerical values in Eq. (iii) shows that 
02 = 37°. Using this result in Eq. (ii) gives 
t;2/ = 2 m/sec. 
At this point, we digress briefly to mention that 
one of the goals of the physical scientist is to dis-
cover the various conservation laws governing the 
material world. Other conservation laws will be en-
countered as we continue our study. For example, 
it is found that, regardless of the nature of interac-
tion involving a system of electric charges, the net 
(positive, zero, or negative) charge of the system is 
constant. We say, therefore, that the electric charge 
of a system of particles is conserved. (It is this prin-
ciple that guides the chemist in balancing a chemi-
cal reaction involving a number of positive and 
negative ions.) By determining all the physical 
quantities that must be conserved in a given in-
teraction, it may be possible to determine fully the 
later behavior of the system without direct applica-
tion of Newton's laws of motion. What is more, the 
application of conservation laws not only may suc-
ceed where the direct approach fails, but will fre-
quently be quite easy to accomplish. 
At this point, the reader may have deduced 
another apparent conservation law involving col-
liding particles. We refer to the fact that when bil-
liard balls collide their total mass remains constant 

158 
Momentum 
during the collision, so that one is led to propose a 
law of conservation of mass for a system of par-
ticles subject to no external forces. In this instance, 
however, the conservation "law" is applicable only 
for particles whose speeds are small compared to 
that of light. It was deduced by Einstein in 1905 
(and verified with terrible consequences in 1945 at 
Hiroshima) that the mass and energy of a system of 
particles are not separable quantities. Thus, it is 
possible for the mass of a system to decrease (or 
increase) provided that there is a simultaneous re-
lease (or absoφ tion) of energy. Therefore, the cor-
rect conservation law in this instance is that the 
total mass-energy of a system subject to no exter-
nal forces must be conserved. This follows from 
the theory of special relativity, but is not discussed 
further here. 
In any case the search for conservation laws has 
been and continues to be a fruitful field of activity 
for physicists. We conclude this section with a 
cautionary note. The "laws" of nature represent a 
synthesis of experimental evidence and cause-
effect relationships related to that evidence. As we 
noted in Section 9-1, our understanding of nature is 
subject to constant review (and revision when 
necessary). Hence, conservation laws obtained by 
formal manipulation of other "laws" of physics 
should be tested by careful experimentation to de-
termine the extent of their validity. In this context, 
the law of conservation of momentum appears to 
be universally valid, while some of the other con-
servation 
laws 
are more 
limited 
in 
their 
applicability. 
19-3 
E L A S T I C A N D I N E L A S T I C 
C O L L I S I O N S 
When two masses collide, their subsequent mo-
tion can be described by Eq. (19-3). If there are no 
external forces acting, the momentum of the sys-
tem is conserved and Eq. (19-8) applies. Example 2 
illustrates the fact that, for a two particle collision, 
it is not possible to determine the final momentum 
(and velocities) of each particle from a knowledge 
of the initial conditions and Eq. (19-8) only. How-
ever, it is possible to classify collisions. Specifying 
the type of collision is equivalent (for special cases) 
to providing an additional relationship between the 
particle velocities before and after the collision. 
A completely inelastic collision occurs when the 
two particles stick together and thus move with a 
common final velocity. When this is true, Eq. (19-8) 
becomes 
m^VA + maVB = (ΓΗΛ  + mB)V, 
(19-9) 
where V is the final velocity. Given the initial condi-
tions, V can now be calculated. 
For a one-dimensional 
elastic collision, the rela-
tive velocity of approach of two particles equals the 
negative of the relative velocity of separation. 
Using the notation of Eq. (19-8) for a one-
dimensional collision, this requirement becomes 
where 
and 
(vb-VA) 
= -(Vb-VA), 
(19-10) 
(vb-VA) 
= velocity of approach of Β 
relative to A 
ÍVb-VA) 
= velocity of separation of Β 
relative to A. 
An alternative form of the requirement that a colli-
sion is elastic is the following: 
I mAVA' +1 mBVB' = I 
VA' +1 m« VB' . 
(19-11) 
To see that Eq. (19-11) and Eq. (19-10) represent the 
same requirement, we combine Eqs. (19-11) and 
(19-8). Thus, dividing Eq. (19-11) 
mA(vA'-VA') 
= 
mB(VB'-VB') 
by Eq. (19-8) 
mAivA - VA) = mBiVB - Vb), 
yields 
+ VA = VB + Vb 
or Eq. (19-10) 
{Vb-VA)^-{Vb-VA) 
as asserted. 
The quantity imv^ is defined to be the kinetic 
energy of a particle of mass m and speed v, (The 
relationship between Newton's laws of motion and 
kinetic energy is discussed in Section 20-2.) There-
fore, Eq. (19-11) states that in an elastic collision 
the kinetic energy of the two particle system is 
conserved. By contrast, kinetic energy is not con-
served in an inelastic collision. For a one-
dimensional elastic collision, it is clear that Eqs. 
(19-8), and (19-10) or (19-11) are sufficient to deter-
mine both final velocities. For a two-dimensional 
elastic collision of two particles, additional infor-
mation will be required. This is because conserva-
tion of momentum and kinetic energy provide only 

Μ ο ί/ο η  of Systems with Variable Mass 
159 
three equations relating the velocities of the two 
particles, while each velocity has two components 
in a two-dimensional problem. As a result, there 
will be four unknown quantities, with only three 
equations relating them (an insoluble situation). 
Example 3. Indicate whether the following colli-
sions are elastic or inelastic. 
(a) A mass of 1 kg with an initial velocity of 
4 m/sec collides with an 8 kg mass, which is initially 
at rest. The final velocity of the 1 kg mass is 1 m/sec 
in the opposite direction. 
(b) A mass of 1 kg with an initial velocity of 
4 m/sec bounces off an initially stationary object 
of mass 3 kg. The 1 kg mass has a final velocity of 
2.83 m/sec in a direction 90° from its initial direc-
tion. 
SOLUTION 
(a) From Eq. (19-8), 
(1 kg)(4 m/sec) = (1 kg)(- 1 m/sec) + (8 kg)(i;2,) 
so that 
Ü2/ =|m/sec. 
For an elastic collision, Eq. (19-10) requires that 
1^10-1^20= 
V2f-Vif. 
In this case. 
? 1 . 
(4-0) m/sec = I-,-., 
m/sec = 1.625 m/sec. 
Therefore, the collision in (a) is not elastic. Since 
vifi" Vify it is not a completely inelastic collision 
either. 
(b) Applying Eq. (19-8) in this case; 
x-motion: 
(1 kg)(4 m/sec) = (1 kg)(0 m/sec) + (3 kg)(v2f.) 
y-motion: 
0 = (1 kg)(2.83 m/sec) + (3 kg)(i;2/,), 
from which 
V2fx = I m/sec 
V2fy = - 0.94 m/sec. 
If the collision is elastic, then Eq. (19-11) should be 
satisfied. In this case. 
^(1 kg)(4 m/sec)' + ^(3 kg)(0 m/sec)' = ^(1 kg) 
X (2.83 m/sec)'+ ^(3 kg) ( ! ) ' + (-0.94)' (m/sec)'; 
or, dropping the common factor of I and the energy 
units, 
16 = 8 + 3 
X 0.88) = 8 + 7.95 = 15.95. 
To the accuracy of the data the collision is 
elastic. 
19-4 
M O T I O N O F S Y S T E M S 
WITH 
V A R I A B L E 
M A S S 
Let us consider in more detail the impulse-
momentum relation. 
Γ ¥(t)dt = m v - m v o . 
We have dealt thus far with systems for which the 
mass does not change during the time t. When this 
is true, Eq. (19-3) is completely equivalent to New-
ton's second law of motion in the form 
F(i) = ^ ( m v ) . 
(19-12) 
Now, however, consider a system for which the 
mass is not constant. Experiment shows that Eq. 
(19-12) is a valid statement, while the form 
F(i) = m ^ = m a 
(19-13) 
is not. Since the rules of differentiation require that 
d(mv) 
d\ , dm 
we see that the momentum concept has led us to a 
generalized form of the second law, which applies 
whether or not the mass is constant. It is a matter of 
history that Newton himself recognized the impor-
tance of momentum and expressed the second law 
of motion in the form of Eq. (19-12). As an example 
of a system of variable mass that is of considerable 
current interest, we consider in the next example 
the launching of a rocket. Here, the mass of the 
rocket decreases with time (as fuel is consumed to 
provide launching thrust due to the escaping gases). 
Example 4. A rocket is to be launched vertically 
subject to the following conditions: at ί = 0, it is at 

160 
Momentum 
rest relative to the Earth; its mass (fuel plus rocket) 
is initially mo; the rate of mass decrease dmidt is 
assumed constant until a final mass m/ is attained. 
The velocity of the escaping gases relative to the 
rocket is v„ also assumed to be constant. Finally, 
we neglect air resistance, and assume that the 
"burn time" is short enough that the acceleration 
takes place under a constant acceleration due to 
surface gravity. Find the rocket velocity as a func-
tion of time during the "bum time." 
SOLUTION 
Consider the Earth to be an (approximate) iner-
tial reference system with ν  (the velocity of the 
rocket relative to the Earth) and ρ = mv (the cor-
responding momentum of the rocket) both at time i. 
In a small time interval dt, there is a mass decrease 
dm and a velocity increase d\ relative to the Earth. 
Now Vg, the velocity of the escaping gases relative 
to the Earth, is related to \ e and ν  by the equation 
Vg = Ve - V. 
(19-14) 
At time t + dt, the momentum of the system rela-
tive to the Earth, ρ + dp is in two parts: 
ρ + dp = (m - dm )(v + dv) - dm Vg 
Rocket 
gases 
or 
p + dp 
= mv 
- dm 
m d \ - dmd\- 
dmive 
-y). 
(19-15) 
Substituting Eq. (19-14) in Eq. (19-15), and neglect-
ing the small quantity dmdv gives 
p + d p = mv + m d v - d m Ve. 
(19-16) 
Using ρ = m V, and the fact that dp occurs in a time 
dt, we can write 
dp 
dv 
dm 
(19-17) 
This change in momentum of the system is related 
to F = mg, the external force (due to gravity) by 
Eq. (19-2), 
or 
mg = 
m-7-- dm 
dt 
dt 
(19-18) 
This can be re-arranged as follows: 
dv 
1 dm 
Now, both g and Ve are directed downward, while 
dv/di is upward (if launch is to occur!). Therefore, 
dv 
=-gdt-—dmVe. 
(19-19) 
m 
Equation (19-19) can now be integrated, noting that 
at Í = 0, r = 0, and m = mo, while at any later time 
(up to tp, the end of the "bum") the corresponding 
values are v ( t ) and m. Thus, 
Jo · 
Jo 
Jmo 
m 
which gives 
v(t) 
= - g t - \ \ n ( — ) ] v . . 
(19-20) 
Using the fact that m < mo, Eq. (19-20) becomes 
v(t) 
= -gt-^ve\n(^^y 
(19-21) 
As a numerical example, the following charac-
teristics apply to the Centaur rockett: 
mo «2.75x10'kg, 
mp^^ 2.45x10'kg, 
dm 
dt 
Since 
1300 kg/sec. 
dm. 
wily 
m — mo — mp — " ^ V Í F — uj. 
it follows that 
^ - = m Γ w T ' ^ ^ 2 3 1 s e c . 
(1300kg/sec) 
A reasonable estimate for the magnitude of the 
exhaust velocity for this rocket is 5 x 10^ m/sec. 
Thus, if i; = 0 at Í = 0, then the final velocity ι;(ίρ) 
will be, from Eq. (19-21), 
vitp) = (- 9.8)(231) + 5 x lOMn 1.11 
« 3000 m/sec, 
« 6700 mi/hr. 
tThe Centaur rocket is a [second-stage] rocket designed 
for use with the Saturn rockets used in the Apollo moon 
program. 

Problems 
161 
19-5 
FURTHER C O N S I D E R A T I O N S 
It was noted in Section 19-4 that Eq. (19-12) is a 
more generally valid form of Newton's second law 
of motion than the form given by Eq. (19-13). The 
example of a rocket launch with a time-dependent 
mass was used as an illustration of the general 
form. The reader should deduce that if d m / d i = 0 , 
Eqs. (19-12) and (19-13) become identical for typi-
cal types of motion. 
The phrase "typical types of motion" is meant to 
refer to situations where the speeds involved are 
small compared to that of light. We saw, in Chapter 
15, that the description of high speed kinematics re-
quires the modifications introduced by Einstein in 
special relativity theory. There are additional 
effects that must be considered, the most significant 
of which we present here without proof t. 
Suppose a particle of mass mo when at rest is 
caused to move at a speed ν  comparable to (but less 
than!) c. The theory of relativity requires, and ex-
perimental measurements confirm, that the mass mo 
must be replaced by a mass m related to mo by the 
expression 
m=-r 
?Vn75- 
(19-22) 
-(F) 
If the speed depends upon the time, m will also be a 
function of time. As a result, the momentum for 
relativistic speeds is given by the expression m v 
and n o t by the expression mot;. When this is done, 
it is found that Eq. (19-12) remains valid even for 
relativistic speeds. 
It was asserted in Section 19-2 that the law of 
conservation of momentum is universally valid. 
Historically, the firm conviction that momentum 
conservation was universal led Wolfgang Pauli to 
postulate a particle that was to have no charge and 
no mass (or a very small mass). It was needed in 
order to account for momentum and energy ap-
parently lost in a particular kind of nuclear decay 
process. The existence of this particle, called the 
neutrino ("Httle neutral one" in Italian) by Enrico 
Fermi, was not verified experimentally until 1958, 
about 25 years after its necessary properties were 
identified. 
In view of the several radical modifications of the 
classical concepts of mechanics that special rela-
tivity requires, it is remarkable that a quantity as 
fundamental as momentum retains its identity re-
gardless of the particle speed. This is not true for 
the kinetic energy. 
tSee any text on special relativity. 
P R O B L E M S 
1. An 8 gm bullet is fired horizontally into a 90 gm block of wood that is free to move. The velocity of the 
combined block and bullet after impact is 4000 cm/sec. What was the initial velocity of the bullet? 
2. If a bat is in contact with a ball for 0.02 sec, and it exerts an average force of 100 nt, what is the impulse 
given to the ball and the change in momentum of the bat? 
3. A 16 gm body traveling with a velocity of 30 cm/sec collides inelastically head on with a 4 gm body 
traveling in the opposite direction with a velocity of 50 cm/sec. What is the magnitude of the velocity of 
the 16 gm body after collision? 
4. A 3000 lb car traveling north and a 10,000 lb truck traveling east at 40 mph collide at an intersection; the 
combined pile of junk slides 30° north of east. Determine the speed of the car just before the collision. 
5. A 32 lb steel disc hangs at the end of a rope. A bullet moving horizontally with a speed of 1600 ft/sec 
strikes it and drops down. Find the speed with which the steel disc begins to move, if the weight of the 
bullet is 1 oz. 
6. A 2001b skater traveling with a velocity of 20 ft/sec eastward collides with a 1601b skater who is 
traveling northward at 15 ft/sec. 
(a) In what direction does the tangled heap of skaters slide? 
(b) What is the speed of the tangled skaters immediately after the collision? 
7. A bullet of mass 100 gm moves horizontally eastward at 10(X) m/sec, and has a perfectly inelastic colli-
sion with a block of mass 1 kg. The block is moving horizontally eastward at 10 m/sec at the moment of 
impact. If the coefficient of kinetic friction between the block and the plane is 0.51, how far does the 
block slide after collision? 

162 
Momentum 
8. A simple pendulum consists of an 800 gm bob hanging from a string. The length of the string is 130 cm. A 
100 gm projectile moving horizontally toward the center of the bob strikes the bob and is imbedded in it. 
After impact, the velocity of the bob plus projectile is 120 cm/sec. Compute: 
(a) the height through which the bob plus projectile rises and 
(b) the magnitude of the velocity of the projectile immediately before impact. 
9. An empty freight car weighing 10 tons rolls at 3 ft/sec along a level track and collides with a loaded car 
weighing 20 tons, standing at rest with the brakes released. If the two cars couple together, find: 
(a) their velocity after collision and 
(b) the impulse given to the loaded car. 
10. A 961b boy running 6 ft/sec north jumps into a 1121b boat moving 10 ft/sec in the opposite direction. 
Neglect friction between the boat and the water. Compute the velocity (magnitude and direction) of the 
boat after the boy arrives in it. 
11. A lead bullet has a weight of 2 oz and a velocity of 1280 ft/sec. It hits the center of a block of wood and 
remains imbedded, the mass of the block and bullet being 8 lb. Find: 
(a) the velocity with which the block (with the bullet imbedded) starts to move, 
(b) the acceleration and the retarding force if the block is on ice when fired at, and comes to rest in 10 
seconds, and 
(c) the coefficient of friction between the block and the ice. 
12. A golf ball weighing l|oz is driven from the tee with the initial velocity of 180 ft/sec. If the club is in 
contact with the ball for 0.0005 sec, compute the average force exerted on the ball. 
13. A 5 kg body with an initial velocity of 20 m/sec travels 6 m while acted on by a force of 500 nt in the 
direction of its initial velocity. Find the final velocity and the time during which the force acts. 
14. In accelerating a 0.50 lb hockey puck from rest to a speed of 80 ft/sec, the stick is in contact with the 
puck for 0.040 seconds. Compute: 
(a) the momentum of the puck as it leaves the stick and 
(b) the average force exerted on the puck by the stick. 
15. A 1501b man dives from a high springboard and reaches a maximum height above the surface of the 
water of 25 ft. After striking the water, he comes to rest in 5 sec. 
(a) What is the momentum of the diver at the instant he strikes the water? 
(b) What average force does the water exert on the diver? 

20 
Work and 
Mechanical Energy 
20-1 
Definition of Work 
163 
20-2 
Work and Kinetic Energy 
165 
20-3 
Conservative and Non-
Conservative Forces 
166 
20-4 
Potential Energy—Gravitational 
and Elastic 
169 
20-5 
Conservation of Mechanical 
Energy 
171 
20-6 
The Work-Energy Balance 
172 
20-1 
DEFINITION O F W O R K 
In this chapter, we develop another method for 
analyzing particle motion as an alternative to the 
direct application of Newton's laws of motion. It 
will be based upon these laws of motion but will 
involve the concept of work. It is first necessary to 
make clear what is meant by the term work in 
physics because it has so many meanings in our 
everyday language. 
In physics, a particle that experiences a displace-
ment As while subject to a constant force F that is 
directed at an angle φ  relative to As has, by defini-
tion, experienced an amount of work Δ W given by 
AW = |F||As| cos φ . 
(20-1) 
That is, the product of the magnitudes of the dis-
placement and the component of force parallel to 
the displacement during the application of the force 
gives the amount of work done on the particle. As 
Eq. (20-1) shows, work is a scalar quantity even 
though it is defined in terms of two vector quan-
tities. When two vectors are multiplied to yield a 
scalar quantity, the process of multiplication is 
called a scalar multiplication and yields a scalar 
product. This is indicated symbolically by the "dot" 
product notation 
A · Β  = |A||B| cos φ . 
(20-2) 
where φ  is the angle between A and B. (The scalar 
[dot] product is also discussed in the Appendix.) 
With this definition, we may rewrite Eq. (20-1): 
AW = F A s . 
(20-3) 
If the particle follows a winding path while subject 
to a force that is not constant but depends upon the 
position s along the path, then Eq. (20-3) must be 
replaced. By considering only an infinitesimal dis-
placement ds along the path and the force F(s) 
(which has a unique value when ds becomes 
vanishingly small), we can express the infinitesimal 
amount of work dW as 
dW = F ( s ) d s . 
(20-4) 
If the particle goes from point A to point Β along 
the path, the total work WA^B or WAB will be the 
sum of all the individual d W s for displacements 
from A to B. Such a sum is, of course, achieved by 
integration, indicated symbolically as 
WA 
s)'ds. 
(20-5) 
163 

164 
Work and Mechanical Energy 
This integral depends upon the form of the path 
from Λ  to β  and also involves a force vector 
whose magnitude and direction are functions of 
position along the path. Such an integral is called, a 
line integral to distinguish it from the simpler in-
tegrals encountered in determining ν  and r from a 
knowledge of the time dependence of a. (See Chap-
ter 14.) When the component of F(s) tangent to the 
path is known as a function of s for a given path, 
the line integral 
¡\(s)'ds 
becomes the simpler integral 
¡y^ds, 
where Ft{s) is the component of F(s) tangent to the 
path element. In the following sections, illustrations 
of the usefulness of this expression are given. 
Example 1. The graph of Figure 20-1 illustrates 
the variation of the tangential component of the 
force applied to a particle as a function of the par-
ticle displacement. Determine the work done as the 
particle moves from the origin to a point 50 m away. 
S O L U T I O N 
Since the graph illustrates Ft(s) versus s, Eq. 
(20-5) becomes 
, = 
j%Mds 
Ξ 50m 
in terms of this data. 
As we saw in Chapter 14, the value of such an 
integral is given by the area under the curve—here, 
Ft(s) versus s. In this case, the area is conveniently 
split into three separate regions for which the 
force-displacement relations are 
I F,(s) = 15s nt; 0 < s <30m. 
II 
=450nt; 3 0 < s <40m. 
Ill 
= [450 - 35(s - 40)] nt; 40 = 
= [1850-35s] nt 
From the graph, 
I W(^5o = |(30m)(450nt) 
II 
+(450nt)(10m) 
III 
+ ^(10 m)(350 nt) 
(100 nt)(10 m) 
= 14 X 10^ joules. 
^F(s) (newtons) 
500 -
400 
/ "1 
300 
/ 
\ 
200 
1 \ 
/ 
I 
π  ml 
100 
1 
· 
t
i
l
l 
C ) 
10 20 30 40 
50 
60 
70 80 
s (meters) 
Figure 
20-1 Force 
versus 
displacement 
for 
Example 1. 
As a check, notice that the analytic integration 
gives 
Wo-5o = I 15sds+ Í 
450ds 
Jo 
J30 
+ 450s 
15s^ 
- 3 5 ( - 4 0 s + y ) 
+ 450s 
+ 
4 5 0 s - 3 5 ( - 4 0 s + y ) 
= (450 nt)(15 m) + (450 nt)(10 m) 
H-(450nt)(50 - 40)m 
1, 
(-40(10))+ |(2500-1600) joules. 
^ 3 5 
= 14 X 10' joules, 
as before. 
Work is a derived physical quantity, and its units 
are determined by its definition. In MKS units, the 
product of force and displacement units yields the 
newton-meter, a quantity given a single label: the 
joule. In cgs units, the dyne-centimeter is called an 
erg; the reader can show that 10^ ergs = 1 joule.t In 
tAs an indication of the magnitude of an erg, it is 
approximately equal to the work required to lift a gnat 
1 cm on the Earth's surface. Thus, the reader has only to 
lift 10 million gnats 1 cm—or 100,000 gnats 1 m—to do a 
joule of work. 

Work and Kinetic Energy 
165 
the British engineering system of units the unit of 
work is called the ft-lb. (There is no single name for 
this unit,) As mentioned in the Introduction, the di-
mensions of work are m/'r'. By using the conver-
sions from meters and kilograms to feet and slugs, 
one finds that 1 joule = 0.7376 ft-lb. 
In many applications, it is desirable to know the 
amount of work done per unit time at any given in-
stant. The name given to this physical quantity is 
instantaneous power (P). It is defined by the rela-
tion 
(20-6) 
P = dt ' 
If Eq. (20-4) is combined with Eq. (20-6), we find 
that 
P = F . ^ = F.v. 
dt 
(20-7) 
The average power Ρ  during a time interval t is just 
thetotal work done W divided by the time interval 
t,P = Wit. From the point of view of an engineer, 
the rate at which work is done by a machine is of 
equal or greater significance than the total work 
done. The units of power are: 
M K S 
^ nt-m _ J joule _ ^ 
sec 
sec 
watt 
sec 
sec 
British Engineering: 1 ft-lb 
sec 
For historical reasons, English-speaking engineers 
adopted the horsepower (hp) as the British en-
gineering unit of power. It is related to other units 
by the relations 
1 hp = 550 — = 746 watts, 
sec 
With the present high demands for power in indus-
trialized nations, it is more common to speak of 
power in kilowatts (1 kw = 10^ watts) or even 
megawatts (1 Mw= 10* watts). Thus, for example, 
residential electricity meters record the total elec-
trical work delivered in kilowatt-hours (kwh), 
where 1 kwh = 3.6 x 10*^ joules. In the following sec-
tions, we will not be concerned further with power 
considerations in spite of their practical impor-
tance. 
20-2 
W O R K A N D K I N E TI C 
E N E R G Y 
Consider a particle of mass m having a velocity 
VA at point A and subject to a constant force F 
tangent to the path. The work due to this force will 
cause an increase in the velocity. To see this, first 
note that from Newton's second law; 
F = m a , 
so that 
WA 
Fds 
mads = mas 
AB, 
(20-8) 
where 
SAB 
is the magnitude of the displacement 
from A to B. [Since F, a, and ds are parallel, we 
need only consider magnitudes as indicated in Eq. 
(20-8).] Now, from the chain rule of differentiation, 
we could replace a by the expression 
_dv_dvds_ 
dv 
^~ dt~ ds 
dt'^ds' 
(20-9) 
and Eq. (20-8) becomes 
WAB = j 
mvdv = j 
mvdv 
1 
2 
1 
2 
= :^mvB - ^ ^ V A 
. 
(20-10) 
By combining Eqs. (20-8) and (20-10), we get the 
relation 
t;ß ' = t;A'-»-2a5Afl. 
(20-11) 
This is just the kinematical expression Eq. (14-15) 
relating velocity and displacement for a constant 
acceleration, now determined dynamically. 
We have a new relation in Eq. (20-10), however. 
It states that the result of applying a constant force 
tangent to the path traveled by a particle is to 
change the value of a quantity ΐ mvA^ to a value 
I mvB^, where VB is the magnitude of the velocity of 
the particle at point B, as indicated by Eq. (20-11). 
If we define the translational kinetic (or motional) 
energy (K.E.) of the particle by the relation 
K.E. = 
^mv\ 
(20-12) 
then Eq. (20-10) simply becomes 
WAB = ( K . E . ) B - ( K . E . ) A , 
(20-13) 
= Δ (Κ .Ε .), 

166 
Work and Mechanical Energy 
or, in words, the difference between the final and 
initial kinetic energy values of the particle is equal 
to the work done on it by the force F. 
It should be stressed that Eq. (20-10) is valid even 
when F (and, hence, a) is not constant [so that Eq. 
(20-8) is no longer applicable]. As an alternative to 
solving the vector relationships of Newton's laws 
of motion, it is now possible to apply the scalar re-
lationship between the work and the change in kine-
tic energy. 
Example 2. A 75 kg sled is made to move a 
horizontal distance of 60 m (starting from rest) by a 
constant horizontal force of 250 nt. 
(a) Calculate the work done by the force. 
(b) Determine the final velocity of the sled. 
SOLUTION 
(a) From Eq. (20-3), 
Wo^ = (250 nt)(60 m) = 15 x 10' joules. 
(b) Since the initial velocity is zero, Eq. (20-10) 
becomes 
(15xlO')iou,es = |(75kg)ri„ 
(m/sec)' = [4 X 10'](m/sec)', 
«60 = 20 m/sec. 
Notice also that 
a = Flm = 250 nt/75 kg = 10/3 
m/sec', and Eq. (14-15) yields Vf^ = (lasAB)"^ = 20 
m/sec, as it should. 
Example 3. An unknown force causes the veloc-
ity of a particle of mass 5 kg to increase from 
10 m/sec to 20 m/sec. Determine the work done by 
the force. 
SOLUTION 
From Eq. (20-10), 
WAS = | ( 5 kg)(20 m/sec)' -1(5 kg)(10 m/sec)' 
= 750 joules. 
20-3 
C O N S E R V A T I V E A N D N O N -
C O N S E R V A T I V E 
F O R C E S 
In work-energy considerations, forces are di-
vided into two classes—conservative and non-
conservative. What distinguishes the two classes is 
the fact that the work done on a particle by a 
conservative force does not depend upon the path 
traversed but only upon the initial and final loca-
tions of the particle. Thus, if the work done in going 
from A to β  by a given force is independent of the 
path, we can consider any two separate paths be-
tween A and Β which necessarily involve the same 
amount of work WAE. (See Figure 20-2.) Equation 
(20-5) requires that 
WA -Í: F d s , 
from which it follows that WBA, the work done by 
the force F in moving the particle from Β to A, will 
be given by 
WB 
(20-14) 
If the particle is moved from A to Β  on the path I, 
and returned to A by path II, the total work done 
by the conservative force involved will be 
W^O^^WABÍD-^WBAÍU). 
For a conservative force, Eq. (20-14) requires that 
WBA(ID 
= 
-WAB{D, 
and the total work done is, thus, 
W,OM=WAB(T)-WAB(1) 
= 0^ 
We can, therefore, give as an alternative definition 
of a conservative force the requirement that the 
work done by the force on a particle in moving 
around a closed path must be zero. This require-
ment is written in equation form as 
closed path 
F · ds = 0 (conservative force), 
(20-15) 
Figure 20-2 Alternate paths from A to B. 

Conservative and Non-Conservative Forces 
167 
where the symbol φ  denotes a line integral over a 
path beginning and ending at the same point. 
There are a number of conservative forces that 
are of considerable interest, including gravitational, 
elastic, and electrostatic forces. In the rest of this 
chapter, we shall consider the first two, leaving a 
discussion of the third to Chapter 29. 
Example 4. Figure 20-3 shows a particle of mass 
m initially located at a distance TÍ from a particle of 
mass M. The particle of mass m is to be carried at 
constant speed around the closed path consisting of 
the segments labeled 1, 2, 3, and 4, respectively. 
Show that the work done by the force of gravitation 
over this path is zero. 
SOLUTION 
In this case, the gravitational force exerted by Μ  
on m has a magnitude given by F = GMm/r^, and 
the direction of the force is radially inward from m 
toward M. (It is an attractive force, as discussed in 
Section 16-5.) Equation (20-15) can be written as 
Wclosed path =
^
1
+
^
2
+
^
3
+ 
W 4 , 
or, in words, the work done around the closed path 
is equal to the sum of the amounts of work done on 
the separate segments of the path. 
Paths 1 and 3 are arcs of concentric circles, so 
that on these paths the gravitational force and the 
displacement vector will be perpendicular. There-
fore, by Eq. (20-1) the work done on each of these 
segments will be zero. The work done along path 2 
is readily found. On this path. 
ds = dr 
and 
^ 
. 
GMm . 
F d s = 
2—dr. 
4 
l \ 
A S 
Μ  
2 
^ 
1 
(Since F is radially inward.) Therefore, 
Γ'» dr 
W2 = - GMm 
^ 
Jn r 
- ( - G M ™
) [ - ( Í - Í ) ; 
On path 4, we have 
ds = dr 
and 
F · ds = 
- GMm Ir^ dr. As a result, 
W4 = - GMm rdr 
Figure 20-3 Diagram for Example 4. 
= ( - G M m ) [ - ( i - i ) ; 
^-GMmi^-^). 
Since WA = - WI, we obtain the desired result, 
W e . o s e d p a t h = W . + 
W 2 + W 3 + W 4 
= 0 + ^ 2 + 0 - ^ 2 
= 0. 
Thus, the gravitational force is seen to be conserva-
tive, as asserted. Physically, what this analysis 
shows is that, in taking mass m along path 2, we 
must actually work against gravity, which acts in-
ward. Along path 4, the amount of work done by 
gravity is equal in magnitude, but opposite in sign to 
that done along path 2. In other words, we can think 
of work put in along part of the path and taken out 
along another part. For a conservative force, the 
net work done must be zero, leading to no change in 
kinetic energy for the closed path. For a non-
conservative force, Eq. (20-15) becomes 
Wciosed path = ^ F · ds # 0 
(nou-conscrvative force). 
(20-16) 
Alternatively, we can define non-conservative 
forces by saying that the work done by such forces 
is not independent of the path chosen. In moving a 
particle from point A to point Β in this case, the 
work done cannot be specified unless the path is 
also specified. The force of kinetic or sliding fric-
tion (discussed in Section 18-2) is an example of a 
non-conservative force. 
Example 5. An object of mass m is to be moved 
horizontally at constant speed from its initial posi-
tion A to a point Β a distance s along the horizon-
tal surface and back to A, as illustrated in Figure 

168 
Work and Mechanical Energy 
20-4. As the figure shows, there is a sliding friction 
force of magnitude μ κmg. Show that this force is 
non-conservative. 
Figure 20-4 Force diagram for Example 5. 
SOLUTION 
If 
the 
friction 
force 
is 
non-conservative, 
Wciosed path 5^0. To show that this is the case, con-
sider the closed path as composed of path 1 from A 
to Β and the path 2 from Β to A. For path 1, 
= 
F/ · ds = - 
iμ κmg)ds 
=- 
ß Kmgs, 
For path 2, 
Wi = 
Ff 'ds = j (μ κmg)ds 
= - μ κmgs. 
In words, for both segments of the closed path, the 
friction force acts in a direction opposite to the dis-
placement. Therefore, 
Wclosed path = 
" 
2μ κmgS, 
and we have shown that 
Wdosedpath 5*^ 0. 
Example 6. Figure 20-5 shows an object of mass 
m located initially at point A on the Earth's sur-
face. It is to be moved to a final position at point Β 
via two alternative paths. Along path 1, it slides 
horizontally a distance s on a surface for which a 
coefficient of sliding friction μ κ can be specified. It 
is then raised a vertical distance y to point Β 
against a gravitational force. Along path 2, it is 
raised a vertical distance y against the gravitational 
force, and then moved a horizontal distance s to 
the point B. Show that 
SOLUTION 
Along the horizontal portion of path 1, the forces 
acting are as shown in Figure 20-6. Since Ν  and mg 
are perpendicular to this part of the path, they do 
not contribute to the work done. Along the vertical 
portion of the path, only the gravitational force is 
involved. Therefore, 
W\ 
= 
^horizontal 
Wyertical 
= - 
μ κNds - 
mgdy 
= - μ κΝ 8 
- 
mgy. 
Since Ν  = mg, we can write 
Wi = - mg(μ κs + y). 
For path 2, the work done on the vertical path is 
again - mgy since only the gravitational force is 
involved, as before. In moving the horizontal dis-
tance 5 to point Β  to complete path 2, we now have 
a frictionless situation since the object is no longer 
in contact with the surface. Thus, no work is done 
on this horizontal path and 
W2 = - mgy. 
We see, therefore, that 
W2 
9^Wu 
as required. 
F, = μ κΝ  PI 
mg 
Figure 20-5 System diagram for Example 6. 
Figure 20-6 Force diagram for Example 6. 

Potential Energy— Gravitational and Elastic 
169 
20-4 
POTENTIAL 
E N E R G Y -
G R A V I T A T I O N A L A N D E L A S T I C 
In the last section, examples were presented in 
which conservative forces did work on particles 
without changing the speed of the particles. Since 
this means that there is no change of kinetic energy, 
the particles must have experienced a change in 
energy associated not with motion but with a 
change in configuration of the particles. This 
change of energy is called a potential energy 
change, abbreviated as Δ (P.E.) 
As an illustrative example, consider a particle of 
mass m subject to the gravitational force F g due to 
a mass M. If the particle is to move a distance s at 
constant speed, an applied force 
Fapp will be re-
quired while m is to move from ΓΛ  to ΓΒ, as shown 
in Figure 20-7. Then Eq. (20-5) gives 
WAS = 
(F.PP + F J · ds = Δ (Κ .Ε .) = 0. 
(20-17) 
This requires that 
Δ (Ρ .Ε .«) = 
Fapp · ds = - 
F, ' ds. 
(20-18) 
This change in potential energy, equal to the nega-
tive of the work done by the gravitational force and 
due to the action of the applied force, is called a 
gravitational potential energy change. Continuing, 
GMm\ . 
Δ ( Ρ . Ε . , ) = J \ a p p d s = 
-
£
(
-
= - G M m ( ^ - ^ ) = (P.E.,)B - (Ρ .Ε .,)Λ . 
(20-19) 
Since ΓΑ  < ΓΒ, Δ ( Ρ . Ε . β ) > O, which is consistent with 
having to do work on m to move it further from M. 
If the applied force is removed 
(Fapp = 0), m will 
move toward Μ  due to F g . This causes a change in 
kinetic energy Δ (Κ .Ε .) equal to the work done by 
gravity. Thus, 
Δ ( Κ . Ε . ) = J F , · ds = - 
dr, 
(20-20) 
Δ ( Κ . Ε . ) = - ΓGMm^= 
GMm(—-—). 
Therefore, we can see that 
(20-21) 
Δ (Κ .Ε .)Β .Λ  = ( Κ . Ε . ) Λ  - ( K . E . ) B = Δ ( Ρ . Ε . « ) Λ . Β . 
(20-22) 
Since 
Δ ( Ρ . Ε . , ) Λ . Β = ( Ρ . Ε . , ) Β - ( Ρ . Ε . , ) Λ , 
we can rearrange Eq. (20-22) to give 
(Κ .Ε .)Λ  + (Ρ .Ε .,)Λ  = (K.E.)B + (Ρ .Ε .,)Β . 
(20-23) 
Equation (20-23) shows that if there is no work 
done due to applied forces, the sum of the kinetic 
and gravitational potential energies at A equals the 
sum of the same quantities at B, In other words, the 
total energy (kinetic plus potential) of the particle 
cannot change when no work is done on it by ap-
plied forces. This statement, which is called the 
principle of conservation of mechanical energy, is 
one of the keystones of classical physics. More will 
be said about it in Section 20-5. 
From Eq. (20-19), we obtained 
( P . E . J B ^ - ^ ^ ' " 
( Ρ . Ε . , ) Λ = -
rB 
GMm 
TA 
(20-24) 
Figure 20-7 Force diagram for a particle under-
going motion at constant speed. 
The minus sign in Eq. (20-24) indicates that when m 
and Μ  are at rest and separated by a finite distance, 
work must be supplied by an external applied force 
in order to cause a larger separation. At infinite sep-
aration (rB 
on the other hand, the gravitational 
potential energy goes to zero. 
Example 7. A rocket is launched from the Earth 
and reaches a maximum height of 8 x 10* m above 
the Earth. 
(a) What was its change of potential energy be-
tween launch and this maximum separation, and 
(b) what must have been the launch speed if its 
speed is zero at the height of the flight? 

170 
Work and Mechanical Energy 
SOLUTION 
(a) From Eq. (20-19), 
GhÍEm 
GhÍEm 
IRE+AR 
RE J 
Δ (Ρ .Ε .,) = -
GMEIti 
RE 
1 
- 1 
= -I-
GMETtiAR 
R.' (-0 
In this example. 
and 
AR^ 
(8xlO')m ^ 1 0 
RE 
(6.4xl0')m 
8 ' 
^ = g = 9.8 m/sec' 
R E 
from Section 16-5. Therefore, 
A . D i 7 
. 
mgAR 
(mkg)(9.8m/sec')[(8xl0')m] 
1 + RE 
1 + 8m 
= [3.49x 10'm] joules, 
(b) From Eq. (20-23), 
(K.E.)o + (Ρ .Ε .,)ο  = (K.E.)p + (P.E.Jp 
or 
(K.E.)o = 0 + (P.E.,)p - (Ρ .Ε .,)ο  = Δ (Ρ .Ε .,). 
Therefore, 
I mü o' = [3.49 X 10'm] joules 
2(3.49 X 10'm) joules 
m kg 
= 8.35 X 10' m/sec. 
The reader should note that these calculations are 
unrealistic in the sense that the mass m of the 
rocket is not constant as assumed, since fuel (and, 
hence, mass) is expended to achieve the rocket 
flight. However, if it is reasonably assumed that the 
rocket "burn" is over in a distance small compared 
to RE, then the results will not be greatly in error. 
A particle subject to an elastic restoring force ex-
periences a force that is linearly dependent upon 
the displacement of the particle. That is. 
F = -/c(r-ro). 
(20-25) 
The minus sign indicates that, when the particle is 
displaced from its equilibrium (original) position, 
the elastic force is directed back toward the equilib-
rium position. The elastic constant (or spring con-
stant) k has the dimensions of force/length. If one 
measures the restoring force as a function of dis-
placement, k can be found as the slope of the force 
versus displacement curve {if the curve is linear). 
As an example, a mass attached to a light spring 
and resting on a frictionless surface will experience 
no force if the spring is neither stretched nor com-
pressed. Either stretching the spring or compres-
sing it by an amount χ  will result in a force on the 
mass that is directed toward χ  = 0, the original 
equilibrium position. The work involved in moving 
the mass from position Xi to position Xi is, by Eqs. 
(20-5) and (20-25): (Note that F · ds = kxdx.) 
Ww2 = 
kxdx 
= |fcx.'-|ibc,'. 
(20-26) 
If the displacement occurs at constant speed, we 
can write—as in Eq. (20-18)—that 
VV,.2 = A(P.E.)e,= 
ibcdx=(P.E.)2-(P.E.),. 
(20-27) 
Therefore, elastic potential energy satisfies the rela-
tion 
(P.E.)e, = ^kx'. 
(20-28) 
As in the gravitational case, Eq. (20-23) is still valid 
if no external forces do work on the spring-mass 
system. Thus, a mass attached to a spring, dis-
placed from X = 0 to X = A, and held at rest, has a 
total energy \kA\ If it is then released, Eq. (20-23) 
requires that 
total energy = | k A ' = ^foc' + ^mi;', 
(20-29) 
for any other displacement x. We see, therefore, 
that the (P.E.) is a maximum (IfcA') when the (K.E.) 
is zero, and the (K.E.) is a maximum (íkA') when 
the (P.E.) is zero (corresponding to χ  = 0). As the 
motion proceeds, there is a periodic change from 
maximum displacement to maximum speed as re-
quired by Eq. (20-29). The details of this periodic or 
oscillatory motion are presented in Chapter 22. 
In concluding this section, it is worth emphasiz-
ing that potential energy is related to work done 
against conservative forces. By working against 

Conservation of Mectianical Energy 
171 
conservative forces, a change in the configuration 
of the system is accomplished and energy is in 
effect stored in the system and is potentially avail-
able for subsequent conversion to kinetic energy or 
to be dissipated as energy lost due to frictional or 
non-conservative forces. 
20-5 
C O N S E R V A T I O N O F M E C H A N I C A L 
E N E R G Y 
In the discussion of Eq. (20-23), the sum of the 
kinetic and potential energies of a particle was 
called the total mechanical energy. Henceforth, we 
shall use the symbol Ε for this quantity. Thus, 
Ε 
= ( K . E . ) + ( P . E . ) . 
(20-30) 
In a situation where there are no non-conservative 
forces acting, the value of Ε will remain constant 
even though the kinetic and potential energies 
change. This information can be advantageously 
displayed graphically. Figure 20-8 shows the poten-
tial energy [P.E.(x)] as a function of position χ  for 
a particle executing one-dimensional motion. Sev-
eral horizontal lines, Ei, E2, and E3, on the same 
graph represent different (constant) total mechani-
cal energy states. As indicated by the ordinate 
scale. El < E2 < E3. With such a potential energy 
curve and Eq. (20-30), it is possible to immediately 
construct the [K.E.(x)] versus χ  curve since 
[K.E.(x)] = E-[P.E.(x)]. 
Notice that the total energy curves are shown 
dashed wherever E<[P.E.(x)]. If E<[P.E.(x)], 
Eq. (20-30) requires that [K.E.(x)]<0. Since this 
would require ¡mv^ < 0, either the mass would have 
to be negative or 
would be negative (v must then 
be a purely imaginary number in the algebraic 
sense). Either possibility does not correspond to 
Figure 20-8 Schematic graph of potential energy as a 
function of position. 
the physical world, so that we require that (for clas-
sical physics at least) Ε >[P.E.(x)]. 
For E l , there are two points (XIL and x^) where 
El = [P.E.(x)] and (K.E.) = 0. Therefore, the veloc-
ity of the particle is zero at these points. Between 
these points ( K . E . ) > 0 (since E > [ P . E . ( x ) ] ) and, 
since (K.E.) = |m(± ι;)^ we can deduce that for 
total energy E i the particle can move back and 
forth along the x-axis between the turning points 
XiL, XiR, where t? = 0. At XIL, there must be a force 
acting to the right which slows down a particle ap-
proaching from the right and which speeds up a 
particle receding from XIL to the right. Conversely, 
at XiR the force must act to the left to stop a particle 
approaching from the left or to speed up a particle 
receding to the left from x^. When we calculate the 
potential energy change associated with the work 
done against a conservative force, we use the rela-
tion (for one-dimensional motion) 
d[P.E.(x)] = Δ ( Ρ . Ε . ) = - 
F(x)dx, 
(20-31) 
This integral relationship is equivalent to the differ-
ential relationship 
Ρ Μ  
= - ^ ί ψ ^ . 
(20-32) 
With 
this relation, it is possible 
to 
sketch 
(qualitatively at least) the graph of F(x) versus χ  
if [P.E.(x)] versus χ  is known. Thus, at x,/, 
- d[P.E.(x)]/dx is positive (convince yourself this 
is so!) so that indeed F(x) is directed to the right as 
we have already deduced on physical grounds. 
Similarly, we find F(x) is directed to the left at X,R. 
In sununary, given [P.E.(x)] versus χ  and the value 
of Ε for a conservative system, the motion of the 
system is completely determined. Not only the 
kinetic energy but also the force acting (and the 
consequent acceleration) for all allowed positions 
can be found, frequently by simple graphical 
analysis. 
Notice that for total mechanical energy E2 there 
are two possible regions in which oscillatory mo-
tion can occur, with a "forbidden" region between 
them. Where the particle actually moves will de-
pend upon the initial location of the particle, which 
would have to be specified if the motion is not to be 
ambiguous. It is left for the reader to deduce 
[K.E.(x)] versus χ  and F(x) versus χ  curves for 
the energy states E i , E2, and E3 of Figure 20-8. 

172 
Work ar)d Mechanical Energy 
20-6 
T H E W O R K - E N E R G Y 
B A L A N C E 
There are many non-conservative forces that can 
act upon particles in a physical situation. Thus, it is 
necessary to develop an expression more general 
than Eq. (20-30), but one which reduces to Eq. 
(20-30) when the vector sum of non-conservative 
forces is equal to zero. We have already associated 
the work due to conservative forces with a change 
in the potential energy so that (see Section 20-4) 
Δ Ε = Δ ( Κ . Ε . ) + Δ ( Ρ . Ε . ) = 0. 
(20-33) 
It seems reasonable, therefore, to assume that any 
work done by non-conservative forces must be re-
flected in a change in the total energy Δ Ε . That is, 
using WN to denote work due to non-conservative 
forces, 
^ ν = Δ Ε = Δ ( Κ . Ε . ) + Δ ( Ρ . Ε . ) 
or 
(Κ .Ε .)Λ  + ( Ρ . Ε . ) Λ  + 
= (K.E.)B + ( P . E . ) B . 
(20-34) 
When WN is positive (as, for example, when a 
proton experiences an increase in speed with each 
orbit in a cyclotron), the ñnal energy will exceed the 
initial energy. For dissipative forces (friction), 
WN < 0 , and the final energy will be less than the 
original energy. Note that Eq. (20-34) reduces to 
Eq. (20-30) when WN = 0. Since the conclusions ob-
tained from Eq. (20-34) are physically acceptable, 
we accept it and refer to it as a work-energy bal-
ance. This emphasizes the fact that analyzing 
dynamical situations via the scalar work-energy 
balance reduces the problem essentially to a debit-
credit type of bookkeeping problem. This is in con-
trast to the equally valid but usually more tedious 
analysis required in a direct application of the 
(vector) laws of motion. 
Example 8. A 3 kg block initially at rest is re-
leased to slide along a surface of non-simple shape. 
When it is 2 m below its original position, its speed 
is observed to be 4 m/sec. (See Figure 20-9.) Find 
the work done by friction. 
AÍ Ί 
VA 
= 0 
2m 
\ . 
, 
Ve = 4 m/sec 
Β  
Figure 20-9 Diagram for Example 7. 
SOLUTION 
Using Eq. (20-34), 
Wf= 
(K.E.)B - ( Κ . Ε . ) Λ  + ( P . E . ) B - ( Ρ . Ε . ) Λ  
= ^(3 kg)(4 m/sec)'-0 + 0 
- ( 3 kg)(9.8 m/sec')(2m) 
= (24 - 58.8) joules = - 34.8 joules. 
The negative sign indicated that EB < EA, a result 
consistent with our experience with frictional sys-
tems. Notice that this problem could not be solved 
using Newton's laws of motion unless additional in-
formation were provided. Using Eq. (20-34), how-
ever, the solution was straightforward. 
P R O B L E M S 
1. A 50 lb box is pulled 10 ft up a rough plane (μ κ = 0.10) inclined 25° to the horizontal by a force of 100 lb 
acting parallel to the plane. Find: 
(a) the work done by the 100 lb force, 
(b) the work done against gravity, 
(c) the work done against friction, and 
(d) the increase in kinetic energy of the body. 

Problems 
173 
2. A 40 lb box is given an initial velocity of 30 ft/sec up a surface inclined 30° with the horizontal. Consider 
the first 10 ft of motion up the incline. Determine: 
(a) the initial kinetic energy of the box, 
(b) the minimum possible change in kinetic energy (for 10 ft), and 
(c) the maximum possible final kinetic energy (at 10 ft). 
3. A 10 lb block slides down the frictionless arc of radius 10 ft, starting from rest at the position shown in 
Figure 20-10. When the block reaches the horizontal portion of the track, it is brought to rest after 
traveling 20 ft. 
(a) What is the kinetic energy of the block just 
as it reaches the horizontal portion of the 
track? 
(b) What is the coefficient of friction between 
the block and the horizontal portion of the 
track? 
Figure 20-10 
4. A force of 230 lb parallel to a plane is required to pull a 300 lb box at constant speed up a plane inclined 
30° above the horizontal. The box is pulled a distance of 16 ft along the plane. Determine: 
(a) the total work done in moving the box, 
(b) the work done against friction, and 
(c) the efficiency of this inclined plane. 
5. A gun at the edge of a cliff 100 m above sea level fires a 2 kg shell towards the sea at an angle of 53° 
above the horizontal with an initial velocity of 1000 m/sec. Find: 
(a) the initial kinetic energy and potential energy of the shell, 
(b) its kinetic and potential energy at the highest point of its path, and 
(c) its kinetic and potential energy as its strikes the sea. 
6. One force continuously pulls 40 nt horizontally north on a 98 kg block, initially at rest on a frictionless, 
horizontal surface, while another force pulls 30 nt horizontally east on the same block. Compute the 
kinetic energy of the block when it has moved 20 m. 
7. (a) A walking horse can pull a load equal to one-half his weight. If the horse weights 1600 lb and walks 
at the rate of 0.50 mi/hr, what horsepower does he develop? 
(b) Calculate the power output for a motor that pulls a car on a horizontal track at a constant velocity of 
15 m/sec against a frictional force of 20 nt. 
8. A 60 lb body is being pushed a distance 20 ft up an inclined plane by a force Ρ  of 50 lb parallel to the 
plane. The plane rises 2 ft for every 5 ft of incline, and the force of friction between the body and the 
plane is 181b. Compute: 
(a) the work done on the body by P, 
(b) the amount of energy transformed into heat, 
(c) the increase in potential energy of the body, and 
(d) the change in kinetic energy of the body. 
9. A box weighing 120 lb is pulled 30 ft up a rough plane inclined 30° to the horizontal by a force of 90 lb 
parallel to the plane. The coefficient of friction between the box and the plane is 0.20. Find: 
(a) the total work done by the 90 lb force, 
(b) the work done against gravity, 
(c) the work done against friction, and 
(d) the increase in kinetic energy of the box. 

174 
Work and Mechanical Energy 
10. A 100 lb box starting from rest at the top of a plane inclined 30° with the horizontal slides down the plane 
until it attains a velocity of 20 ft/sec and has then covered a distance of 22 ft measured along the plane. 
Find: 
(a) the kinetic energy gained by the box, 
(b) the potential energy lost by the box, and 
(c) the work done against friction while the box was sliding down the plane. 
11. A tractor hitched to a load of hay hauls it from the ground to the loft of a bam in 1 minute. If the load 
weighs 500 lb and the height through which it is lifted is 20 ft, find the power developed in horsepower 
and watts. 
12. A 20 kg block on a plane inclined at 37° with the horizontal is pulled 10 m up the plane by a force Ρ  of 
250 nt parallel to the plane. Friction between the block and the plane is 40 nt. 
(a) Find the total work done by P. 
(b) Find the work done against friction. 
(c) Find the increase in potential energy of the block. 
(d) Without finding the velocity, find the increase in the kinetic energy of the block. 
(e) Find the velocity of the block when its kinetic energy is 1000 joules. 
13. A 2 lb hammer moving with a velocity of 20 ft/sec strikes a nail and drives it in a plank a depth of 1 in. 
(a) Find the average force exerted upon the nail. 
(b) If the operation lasted only ^ second, what was the horsepower of the blow? 
14. On a plane making an angle of 37° with the horizontal, the point Β is further down the plane than A by a 
distance of 6 m measured along the plane. A 100 kg block starting at A slides down past B. The 
coefficient of kinetic friction between the block and the plane is 0.250. Compute: 
(a) the decrease in potential energy of the block from A to B, 
(b) the work done against friction by the block from A to B, and 
(c) the kinetic energy of the block at B. 
15. A plank that is 12 ft long has one end resting on the ground and the other end 4 ft higher. On the plank is 
a trunk weighing 125 lb, which is just kept from sliding by friction. 
(a) How great must the force of friction be? 
(b) If the trunk is pulled up the plane a distance of 10 ft, how much work is done? 
(c) If the previous amount of work was done in 10 seconds, how much horsepower was required? 

21 
Rotation 
21-1 
Moment of Force and Center of 
Gravity 
175 
21-2 
Rotational Motion of an Extended 
Object 
179 
21-3 
Kinetic Energy of a Rotating Body 
179 
21-4 
Moments of Inertia 
181 
21-5 
Newton's Laws of Motion for 
Rotating Bodies 
183 
21-6 
Angular Momentum and Its 
Conservation 
184 
21-7 
Rolling Motion 
185 
21-1 
M O M E N T OF F O R C E t A N D 
C E N T E R OF GRAVITY 
It is appropriate to remind the reader that our dis-
cussion of particle motion in the previous seven 
chapters was limited at the outset to ideal or point-
particles (see Section 14-1). As a result, any system 
of forces applied to such a particle is necessarily a 
concurrent one—each force acts at the same point, 
which is the location of the (point-) particle. When 
the resultant force is tangent to the direction of 
motion, straight line, rectilinear (or translational) 
motion results (Chapters 17 and 18). When the re-
tThe term moment of force, like the term moment of 
momentum, involves in its definition the concept of a 
'^moment arm." When the meaning of "moment arm" is 
clear, terms such as moment of force are quite descriptive 
and practical. From the standpoint of the algebra of vec-
tor quantities, a "moment arm" is directly related to a 
cross (or vector) product. In most modern literature, the 
term moment of force is replaced by the term torque. 
Similarly, moment of momentum is replaced by angular 
momentum. Although we mention the early terms for 
their descriptiveness, we shall conform to the modern 
usage in what follows. 
sultant force is instead perpendicular to the direc-
tion of motion, uniform circular motion results 
(Sections 18-4 to 18-6). In both cases, the point-
particle concept allowed us to ignore the possibility 
of any intrinsic rotational motion (such as the rota-
tion of the Earth on its axis as it executes orbital 
motion about the Sun). 
If we now consider rigid objects (objects whose 
non-negligible size and shape does not change) we 
encounter the possibility that the system of applied 
forces may not all act at the same point (be concur-
rent). The resulting motion can then be more com-
plex in that it may involve a combination of transla-
tion and rotation. This is the subject of this chapter. 
Consider a point-particle of mass m attached to a 
weightless arm of length r, which is in turn attached 
at right angles to a long slender rod that can rotate 
freely about an axis along the rod, as illustrated in 
Figure 21-1. A force F is applied to the point mass 
m, in order to cause rotation about the rod axis. 
Now, any component of the force that is parallel to 
the rod axis or parallel to the supporting arm cannot 
cause rotation about the axis. On the other hand, an 
applied force will have maximum rotational effec-
tiveness if it is applied perpendicular to both the 
supporting arm and the rod axis. Thus, rotation de-
175 

176 
Rotation 
Figure 21-1 Diagram illustrating parameters involved 
in rotation about an axis. 
pends not only on the force magnitude but also on 
the orientation of the line of action of the force. As 
a familiar example, the reader may consider the 
most effective way of causing a door to swing on its 
hinges. A force applied perpendicular to the plane 
of the door and to an axis through the hinges will be 
best. 
This example is also useful in pointing out that 
the effectiveness of a given applied force is in-
creased if the distance between the axis of rotation 
and the point of application of the force is in-
creased. (The skeptic can test this by comparing the 
results of pushing normally on the doorknob and 
then pushing with a normal force of the same mag-
nitude along the inner or hinge edge of the door.) 
Still another facet of this situation is the fact that 
reversing the direction of the applied force simply 
reverses the direction of the rotation around the 
same axis of rotation. Our problem then is to incor-
porate all these observations in a relationship that 
makes possible quantitative predictions for analo-
gous systems and that is compatible with the 
mathematical requirements the vectors (force and 
particle displacement) must satisfy. By referring to 
Figure 21-1, it is seen that only the component of F 
that is perpendicular to the arm can cause rotation 
about the axis. If we call this component F x , it 
follows from the figure that 
F x = F sin a 
(21-1) 
Relative to the rod axis, the point of application 
(the location of m) of the force F has a displace-
ment r of magnitude r, directed outward along the 
arm from the rod axis to the mass m. Now we 
introduce the quantity torque (τ ) to denote quan-
titatively the effectiveness of a force F in causing 
rotation about an axis. In order to display all the 
necessary properties, the definition of torque can 
be written as 
τ  = r X F. 
(21-2) 
The expression r x F is a vector (or cross) product 
in contrast to the scalar (or dot) product we have 
encountered earlier. As is indicated in the Appen-
dix, the vector product magnitude r x F is given by 
Ir X FI = Irl IFI sin θ = Irl sin θ |F|. 
(21-3) 
(The quantity |r| sin θ is referred to in the older lit-
erature as the moment arm of the force F. Hence, 
the magnitude of the torque is just the product of 
the moment arm and the magnitude of the force. 
Thus, a force applied at the axis has a zero moment 
arm, yielding a zero torque as required.) 
The direction of r x F is perpendicular to the 
plane containing r and F according to the conven-
tion that if the fingers of the right hand curl so that r 
swings toward F, then the thumb points along the 
direction of r x F. It follows from this that reversing 
the direction of F reverses the direction of r x F 
along the rod axis. Thus, r x F is a product of vec-
tors which yields a third vector perpendicular to the 
two vectors. We intend that this vector product is 
to relate to rotational motion. It is therefore clear 
that for torque to have a unique meaning it cannot 
lie in the plane containing r and F, for, once rotation 
begins, the directions of r and F can both change, 
while the sense of the rotational motion remains 
the same. In addition, when two or more different 
torques are applied to the mass m, one might expect 
that the resulting rotational motion is related to the 
vector sum of the applied torques, just as the resul-
tant translational motion of an object is related to 
the vector sum of the applied forces. In fact, it is 
our intent to show in this chapter that (with suitably 
defined quantities pertinent to rotational motion) 
Nevi^on's laws of motion are also adequate to de-
scribe rotational motion. 
Next, we must introduce the quantities necessary 
to describe rotational kinematics. Not surprisingly, 
they are (analogous to rectilinear kinematical 
terms): angular displacement (Θ ), angular velocity 
(ω ), and angular acceleration (a). If the angular 
displacement changes with time, the following rela-
tions are valid: 
a . = f . 
(21-4) 

Moment of Force and Center of Gravity 
177 
a 
ά ω  
d'e 
dt 
(21-5) 
It should be stressed that the vectors ft a>, a are 
directed parallel to the axis of rotation about which 
the particle moves, rather than in the plane of mo-
tion. 
Suppose the force F of Figure 21-1 (perpen-
dicular to the axis of rotation and the supporting 
arm) remains constant in magnitude and always 
directed tangent to the circular path of radius r 
traversed by the mass m. This is shown in Figure 
21-2. At any given instant, the tangential accelera-
tion a and the force F are related by Newton's sec-
ond law, 
F = m a . 
(21-6) 
Figure 21-2 Circular motion of a particle about an 
axis due to a tangential force F. 
For circular motion of fixed radius, the following 
scalar relations hold (the reader should verify them 
from the geometry of Figure 21-2): 
s= rS. 
de 
τω . 
a = r d^e 
dω  
(21-7) 
(21-8) 
(21-9) 
It is essential in these relations that the angu-
lar 
displacement 
be 
measured 
in 
radians-
abbreviated, rad—with analogous units for ω  and 
a. Often, however, a rotational displacement may 
be given in degrees—abbreviated, °. From the 
geometry of a circle: 
c = Ittt, 
where c is the circumference and r is the radius in 
the same units. Since 27Γ is the number of radians in 
one complete revolution—equivalent to 360°— we 
obtain the conversion relation 
Ö(rad) = Ö(°)x 27r(rad) 
360 (°) * 
Thus, if the magnitude of F is constant, the mass m 
will experience tangential acceleration, which is 
also constant in magnitude. From Eq. (21-9), the an-
gular acceleration is also constant in magnitude. 
Let us now determine the torque τ  due to F. Using 
Eqs. (21-2) and (21-6), 
τ  = r x F = r X ( m a ) = m r x a . 
(21-10) 
The right-hand rule for the cross product shows 
that Γ X a, like τ , is perpendicular to the plane con-
taining r and F. Equation (21-9), in addition, re-
quires that the magnitude of the torque is given by 
T = mr^a, 
(21-11) 
Equation (21-11) expresses the fact that for a given 
mass m at fixed radial distance r from an axis of 
rotation, the angular acceleration and the applied 
torque are proportional. This is the desired analogy 
(for translational or rectilinear motion of a given 
mass m, the acceleration and the applied force are 
proportional). For rotational motion, the inertial 
quantity is seen to be dependent upon both the 
mass and its radial distance from the axis of rota-
tion. If we introduce the term moment of inertia (I) 
with the relation 
I = mr\ 
(21-12) 
then the rotational form of Newton's second law of 
motion becomes 
τ  = la, 
(21-13) 
If the net applied torque is zero, it follows that the 
angular acceleration is also zero. Then the angular 
velocity is constant (or zero as a special case), 
which is equivalent to saying that the particle is 
experiencing uniform circular motion (or else it is at 
rest). 
The reader may recall that Section 17-1 discussed 
static and dynamic equilibrium of point-particles. 
We see now that for equilibrium of a real object to 
exist, we must require not only that the net applied 
force be zero but also that the net applied torque 
vanish. 
Now consider a real object of weight W, which is 
to be supported at rest at the Earth's surface by a 

178 
Rotation 
Figure 21-3 Diagrams illustrating the location of the 
X- and /-coordinates of the center of gravity of a body. 
single force F. The problem is to locate where the 
force should be applied. It is clear from the discus-
sion of Chapter 17 that F must be equal in mag-
nitude but directed opposite to the weight W. For 
complete equilibrium, we must also require that F 
and W lie along the same line of action. If this were 
not the case, rotation about some axis would result, 
contrary to the conditions given. (Everyone has at 
one time or another experimentally solved this 
problem—for example, in the form of stacking 
blocks or balancing on a narrow rail.) A material 
object of total weight W on the Earth's surface can 
be regarded as a point-particle of weight W located 
at a point known as the center of gravity of the 
object. This point is located by determining the sum 
of torques due to the weight elements making up 
the body. Thus [see Figure 21-3(a)], we must re-
quire that the χ -coordinate of the center of gravity 
X satisfy the relation 
Wx=2(:c.AW), 
where Ψ =Σ  
(AW¡) = total weight of the object, 
i = l 
and XiAWi 
is the torque of the ith element of the 
object about the z-axis. Similarly, Figure 21-3(b) 
illustrates the requirement satisfied by the y-
coordinate of the center of gravity: 
Wy=E(y.AW). 
(21-15) 
Thus, if the object is supported by a force equal in 
magnitude to the weight passing through the center 
of gravity, it will remain at rest. Stated differently, 
Eqs. (21-14) and (21-15) simply require that the sum 
of clockwise torques balance the counterclockwise 
torques about the center of gravity in any orienta-
tion of the object. 
If the weight of equal volume elements varies 
continuously throughout the object, Eqs. (21-14) 
and (21-15) become 
c= jxdW, 
(21-16) 
with 
Wx 
W= 
ydW, 
dW, 
(21-17) 
(21-18) 
Example 1. A student has a quantity of laundry 
to be done in a machine which takes loads of 8 lb 
each. Although he does not own a set of bathroom 
scales, he possesses a uniform metal rod 1 m in 
length weighing 4 lb and a pair of shoes weighing 
3 lb. To insure that his machine loads are the max-
imum weight, he attaches the laundry at one end 
of the rod, his pair of shoes at the other, and sup-
ports the rod at a point such that the rod rests in a 
horizontal position. How far from the laundry load 
end is this position? 
SOLUTION 
Since the rod is uniform, the center of gravity is 
at the center (5 m from each end). When the rod is 
supported at a position χ  meters from the 8 lb load 
(see Figure 21-4), it remains horizontal. In equilib-
rium, the counterclockwise torques must balance 
the clockwise torques. Thus, 
(81b)(x) = (41b) _g-x)m + (31b)[(l-x)m] 
or 
(21-14) 
Therefore, 
(151b)(x) = 51b-m. 
X = ^m from 8 lb load. 
Figure 21-4 Force diagram for Example 1. 

Kinetic Energy of a Rotating Body 
179 
21-2 
ROTATIONAL 
M O T I O N O F A N 
E X T E N D E D 
O B J E C T 
In the previous section, we analyzed the rota-
tional motion of a point-particle about a fixed axis 
of rotation at a definite radial distance from the 
axis. If instead we wish to consider the rotation of 
an extended rigid object (for example, a thin flat 
plate), as in Figure 21-5, it is only necessary to 
subdivide the plate into small elements of mass 
(dm) and apply the previous analysis to each ele-
ment. Thus, the magnitude of the torque 
re-
quired to produce an angular acceleration a for the 
mass dm a distance r from the axis of rotation 
through 0 is given by 
dr = dm r^a. 
(21-19) 
The total torque required is then just the sum (an 
integral for continuous distributions of mass) of the 
individual contributions; 
= j d T = 
(^jdmr^a 
la. 
(21-20) 
where the integration extends over the entire plate, 
yielding the moment of inertia of the plate. The sig-
nificance of Eq. (21-20) is that, for a rigid body, the 
inertial and dynamical parts of the problem have 
been separated. The moment of inertia calculation 
is primarily a geometric question (for uniform den-
sity). When it is known, the torque required to 
produce a given angular acceleration (or vice versa) 
follows immediately. 
Figure 21-5 Rotational motion of a thin flat plate 
about an axis i to the plate. 
Example 2. Figure 21-6 represents a uniform 
slender rod of length L meters, total mass M, and 
cross-sectional area A. If the rod is to rotate about 
the axis as indicated with an angular acceleration a, 
determine the magnitude of the torque required. 
Figure 21-6 Diagram for Example 2. 
SOLUTION 
From Eq. (21-20) and the discussion preceding it, 
it is clear that we need only determine /, the mo-
ment of inertia, to obtain the desired result. The 
moment of inertia is given by 
r'dm. 
From Figure 21-6, 
Since the integral must include all elements of mass 
from r = 0 to r = L, we have 
ML' 
L 3 
ML' 
Equation (21-23) then gives τ  =-=f^a. 
21-3 
KINETIC E N E R G Y O F A 
ROTATING 
B O D Y 
To continue the study of rotational motion, we 
next consider the motional or kinetic energy as-
sociated with rotation. Our starting point is the 

180 
Rotation 
point-particle discussed in Section 21-1. Since the 
tangential velocity at a given instant has a mag-
nitude V, the kinetic energy at the same instant is 
given by the rectilinear expression of Eq. 20-12: 
K.E, = 
jmv\ 
or, equivalently in rotational quantities (since 
V = Γω ), 
| m ( r í ü ) ' = | m r ' ω ' = | Ι ω '. 
(21-21) 
For the point-particle, the kinetic energy for both 
the rectilinear and rotational forms is thus seen to 
be one-half the product of an inertial factor and the 
square of the relevant velocity magnitude. 
From Eq. (20-13), we can write 
F d s , 
where ds is an element of the circular path 
traversed by m due to the applied force F. If F is 
tangent to ds, 
¥ ' ds = F ds = mads 
= (mra)(rd0) = (mr'a)(í¿0) 
= (Ia)de = Tde = r'de. 
(21-22) 
For the point-particle of Section 21-1, we now can 
present a table of analogues for describing dynamic 
and kinematic quantities. Table 21-1 is such a table. 
It contains additional analogues than will be discus-
sed in Section 21-6. 
The reader should be able to extend the discus-
sion of rigid bodies in Section 21-2 to see that the 
quantities and analogues of Table 21-1 remain valid 
for the case of extended rigid bodies. Thus, the 
reader who has mastered 
the mechanics of 
translational motion may now obtain twice as much 
utility from this knowledge by using the table of 
analogues. 
Example 3. The moment of inertia of a wheel is 
400 nt m\ At a given instant, its angular velocity is 
10 rad/sec. A constant torque is then applied which 
increases the angular velocity to 30 rad/sec after 
the wheel has rotated through an angular displace-
ment of 100 rad. Calculate the magnitude of the 
applied torque and the work produced by it. As-
sume no frictional loss of energy. 
SOLUTION 
For an object of fixed moment of inertia subject 
to a constant torque, Eq. (21-13) requires that the 
angular acceleration be constant. Therefore, we 
must first determine the magnitude of α if τ  is to be 
found. For constant a, Eqs. (21-4) and (21-5) can be 
used to find, in complete analogy with Eq. (14-15), 
that 
ω ' = ω ο ' + 2αΔ 0. 
(21-23) 
Table 21-1 A Table of Translational and Rotational Analogues 
Physical quantity 
Translational 
Rotational 
Displacement 
r (meters) 
θ (radians) 
Velocity 
v = ^(m/sec) 
ω  = ^(rad/sec) 
Acceleration 
a= ^^2(m/sec^) 
a=^(rad/sec^) 
Inertia 
m(Kg) 
/ = mr'(kg m') 
Law of motion 
F= ma(nt) 
τ  = /«(nt-m) 
= rXF 
Kinetic energy 
K.E. = imv'(joules) 
K.E. = i/ω ^ü oules) 
Work-kinetic energy 
relation 
Δ Κ .Ε.= | 
F d s 
ΔΚ .Ε. = 1 
τ de 
Momentum 
mv (kg m/sec) 
Ιω  (kg mVsec) 
Impulse-momentum relation 
I'^Fdi =A(mv) 
Ι'τ Λ  
=Δ (/ω ) 
Equilibrium condition 
Σ  F' = 0 
and 

Moments of Inertia 
181 
Therefore, 
ω '-ω ο ' 
(30)^-(10)^ 
A r . A U . . ^ 
« = ^ Δ β - = 
2(10') 
=4rad/sec. 
and 
T = Ia 
= 1.6 X 10^ nt-m. 
Now the work-energy relation: 
Thus, 
Δ Κ .Ε . = work due to τ  = ί \ 
- dft 
Je, 
work = I Ι ω ' -^Ιω ο ' 
= 1.6 x 10' joules. 
The reader may already have noted that this result 
also follows from 
work 
Je, T'de 
= 
τ Α Θ  
(since τ  is constant) 
= 1.6 X 10' X 10' = 1.6 X 10' joules, 
as before. 
21-4 
M O M E N T S O F INERTIA 
As indicated in Eq. (21-20), the determination of 
the moment of inertia of an extended rigid body is a 
calculus problem (assuming the variation of density 
with position in the body is known). Such calcula-
tions are in fact frequently used as practical ex-
amples of integration in an elementary calculus 
course. Thus, in principle, one could proceed in this 
fashion whenever necessary. However, the world 
of interest to scientists and engineers is one pos-
sessing many kinds of symmetry—cars have cir-
cular wheels, axles roll on spherical bearings, 
diatomic molecules can be viewed approximately 
as two spherically uniform masses connected by a 
massless rod (an ideal "dumbbell"), etc. The physi-
cist has usually found that exploiting these symmet-
ries can reduce or even eliminate the integration 
necessary to obtain a given moment of inertia. 
Thus, in Example 2, we found the value of / for 
the slender rod of Figure 21-6 to be I ML' with re-
spect to an axis perpendicular to the length of the 
rod and at one end of the rod. Suppose, instead, we 
have been required to determine the value of 
ICM 
(for an axis parallel to the axis at the end but pass-
ing through the center of mass CM, of the rod nor-
mal to the length of the rod). We could, of course, 
do the same integration except that the limits of 
integration would now be from - L/2 to L/2 in-
stead of from 0 to L. The reader can show that 
IcM=j^ML\ 
Rather than repeating the first calculation with dif-
ferent limits, we proceed as follows. Recognize 
that a rod of length L with an axis through the 
center of mass normal to the rod length is the same 
as two rods of length L' = L/2 with an axis at one 
end and normal to the length of the rod. Using the 
result already obtained. 
as before. 
Now suppose that for a given body we had first 
calculated ICM and then wished to determine I for 
an axis parallel to the one through the center of 
mass but located a distance h from it. We will show 
that the result is obtained by means of the 
"parallel-axis theorem," 
I = IcM+Mh\ 
(21-24) 
First, however let us test it for the example just 
presented. In that case, h = L/2; it is indeed true 
that 
The proof is not difficult. Figure 21-7 is a cross-
sectional view of an arbitrarily shaped object of 
mass Μ  with the center of mass at point C. If the 
object rotates with angular speed ω  about an axis 
through point P, its kinetic energy by Eq. (21-21), is 
Κ .Ε . = | ΐ ω ' , 
(21-25) 
Figure 21-7 Cross-sectional view of an arbitrary ob-
ject used to illustrate the parallel-axis theorem. 

182 
Rotation 
where / is the moment of inertia with respect to the 
axis through P. 
Alternatively, we may view the kinetic energy of 
the rotating object as composed of two parts: (1) 
the kinetic energy of a particle containing the total 
mass Μ  and moving with the center of mass (for 
which vcM = Η ω ) about point P, and (2) the rota-
tional kinetic energy computed as though the body 
were in pure rotation about the center of mass. 
For (1), we have 
r'dm 
for (2), 
(K.E.),=|M(ha))^ 
(K.E.)2 = ^/CMCÜ\ 
(21-26) 
(21-27) 
Equating Eq. (21-25) with the sum of Eqs. (21-26) 
and (21-27) yields 
or 
I = Mh' + IcM, 
as stated. The reader may wish to execute the proof 
replacing sums by the appropriate integrals. 
'1 
' 
• 
^ 
Figure 21-8 The flat plate used to illustrate the 
perpendlcular-axis theorem. 
Still another useful device to avoid integration is 
available for any rigid body whose thickness is neg-
ligible compared to its area. (Such bodies are called 
thin plates or laminar bodies.) The situation is 
shown in Figure 21-8. The problem is to determine 
the moment of inertia with respect to an axis (the 
ζ -axis) perpendicular to the plane of the body (the 
x-y plane). Because the body is thin (all Zi «0), 
= | ( x ' + y')dm 
= J x'dm + J y'dm 
(21-28) 
In words, this is a "perpendicular-axis" theorem in 
that, for laminar bodies, the moment of inertia with 
respect to an axis perpendicular to the plane of the 
body equals the sum of two moments of inertia in 
the plane of the body that are at right angles to each 
other. 
Example 4. Calculate the moment of inertia of a 
thin uniform circular disk of radius R and thickness 
t with respect to an axis lying along a diameter of 
the disk and in the plane of the disk. 
Figure 21-9 Diagram for Example 4. 
SOLUTION 
Figure 21-9 illustrates the problem. Here h = 
ly = ID, where ID is the desired result. Using Eq. 
(21-28), 
L=2ID 
or 
ID=\L 
To calculate L, divide the disk into circular seg-
ments of thickness t, width dr and volume lirrtdr 
at a radial distance r from the ζ  axis. Then 
^^2nrtdr 
2Mrdr 

Newton's Laws of Motion for Rotating Bodies 
183 
therefore, 
L = 
- ^ l r d m = ^ l r d r = - j ^ = - ^ M R . 
Finally, 
I O = \ M R \ 
(21-29) 
The reader may wish to compare this calculation 
with a direct calculation of /D. The required choice 
of elements for dm is reasonably straightforward, 
but the integration involved should convince the 
skeptic of the practical value of Eq. (21-28). 
In summary, it is possible to calculate many com-
mon moments of inertia using Eqs. (21-24) and 
(21-28) together with a limited number of results 
calculated directly from the integral definition. 
Table 21-2 is a modest list of such results. 
Finally, just as it is often useful to replace an 
extended rigid body conceptually by a point-
particle of mass Μ  at the center of mass, so it is 
also useful to locate all the mass at a single distance 
from the axis of rotation such that the moment of 
inertia for this arrangement equals that of the ex-
tended body. This distance is called the radius of 
gyration k, which satisfies the defining equation 
I = 
Mk\ 
(21-30) 
tem also involves translational motion. It is then 
necessary to first subdivide the problem (in free-
body diagrams) into its rotational part and the 
translational part. The translational part is analyzed 
using Newton's laws (Chapter 18). The rotational 
part, by analogy, is analyzed using Eq. (21-13). It 
should be stressed here that Eq. (21-13) is valid only 
with respect to an axis through the origin of an 
inertial reference frame.t Again this is analogous to 
translational dynamics in that the laws of motion 
are valid for inertial frames of references only as 
discussed in Chapter 18. 
Consider the system shown in Figure 21-10, com-
posed of a uniform circular cylinder of mass Μ  and 
radius R , free to rotate about an axis through its 
center parallel to the axis of the cylinder. Two mas-
ses, mi and m2, connected by an inextensible string 
of negligible mass are hung over the cylinder, and 
the system is released from rest. Assume that mi > 
m2, the string does not slip on the cylinder, and the 
bearings supporting the rotation axis are friction-
less. When m, has dropped a distance h below its 
original position, what are the linear speeds of mi 
and m2 and the angular velocity of Μ ? 
As indicated in the figure, there are three free-
body diagrams: two translational and one rota-
21-5 
N E W T O N ' S L A W S OF M O T I O N FOR 
ROTATING B O D I E S 
We are now ready to discuss the dynamics of a 
system involving a rotating body. Suppose the sys-
tln rotational motion this is equivalent to rotation about 
an axis fixed in space or an axis through the center of 
mass. When the axis of rotation is in translational motion 
(rolling motion) the analysis is somewhat more involved. 
We will discuss this case in Section 21-7. 
Table 21-2 Moments of Inertia and Radii of Gyration for Some Simple Bodies of Uniform Density. 
Body 
ζ -axis 
Slender rod, length L 
1 rod at center 
12 
Thin hoop, radius R 
Thin circular ring, 
radii Rx, R2 
1 plane of hoop, 
at center 
_L plane of ring, 
at center 
MR' 
R^ 
|(ΐ?,' + ι?Λ  
Sphere, radius R 
Thin rectangular plate, 
sides W, L 
Thin rectangular plate, 
sides W, L 
through center 
||L, at center 
1 plate, at center 
\mr' 
12 
12 
Right circular cylinder, 
radius 
length L 
1 L, through center 
3Ä' + L' 
12 

184 
Rotation 
Finally, using either 
Figure 21-10 Free-body diagrams for a system show-
ing both translational and rotational motion. 
tional. Since m i > m 2 , nti accelerates downward. 
Because the string cannot stretch, it follows that 
the magnitude of the acceleration of nti (upward) 
must be the same as that of mi. The angular accel-
eration of the cylinder is similarly required to be 
counterclockwise, as shown. Using Newton's laws 
and Eq. (21-13), we obtain (the reader should verify 
this!): 
m,g -Ti 
= m,a, 
T2 - rriig = 
mia, 
TiR 
- T2R =Ia=^MR'a= 
^MRa. 
Solving for a , we obtain the result 
^ 
( m , - m . ) g 
(21-31) 
(21-32) 
(21-33) 
(21-34) 
Since the forces (and torques) are constant, we can 
write (since υ,ο  = υ» = 0, ω ο  = 0) 
VI 
or 
and 
Vi = -
t?2 = 
2(mi - 
m2)gh 
r2(mi - m2)gfi 
(21-35) 
(21-36) 
or 
yields 
V = Ρ ω  
ω '^Ιαθ 
( α = | ^ and Ö = | - ) , 
-il/2 
2(mi-m2)gfe 
(m,4-m2 + y)l?' 
(21-37) 
This shows that when mi = m2, then Vi = V2 = 0 and 
ω = 0 (equilibrium exists). It shows further that the 
more massive the cylinder when 
η ΐχ >η ΐ2, 
the 
slower will be the motion—certainly a reasonable 
result. 
It is left as an exercise for the reader to solve this 
problem using the conservation of energy and to 
show that the same result is obtained. 
21-6 
A N G U L A R M O M E N T U M A N D 
ITS C O N S E R V A T I O N 
We saw in Chapter 19 that in some situations an 
analysis based on Newton's laws of motion was 
less convenient than an analysis making use of the 
impulse-momentum theorem. As indicated in Table 
21-1, a rotational impulse-momentum theorem can 
be derived from the rotational second law, Eq. 
(21-13). Thus, 
= Δ (Ι ω ) = (Jcö )2 - (Ι ω ),. 
(21-38) 
Here, τ  = Ι α is replaced with the more fundamen-
tal relation 
_ ά {Ι ω ) 
in analogy with the translational second law discus-
sion of Chapter 19. 
Continuing the analogy, Eq. (21-38) states that, in 
the absence of any external torque on a system un-
dergoing rotational motion, there can be no change 
in angular momentum. Thus, if there is a change in 
the moment of inertia of the system, it must be 
accompanied by an appropriate change in the angu-
lar velocity. 
Example 5. A lecturer demonstrating the conser-
vation of angular momentum stands upon a rotat-

Rolling Motion 
185 
able platform holding a 16 lb weight in each hand at 
his sides. The platform is set in rotation with an 
angular speed of 1 rev/sec. He then extends his 
arms horizontally, putting the weights at a larger ra-
dial distance from the axis of rotation. Assume that 
frictional effects can be ignored and that the lectur-
er's moment of inertia is 5 slug-ft' independent of 
the position of his arms. In addition, let the original 
radial distance of the weights be i ft and their final 
distance 2 ft from the axis of rotation. What will be 
the final angular speed of the lecturer? 
SOLUTION 
There are no external torques acting; the forces 
involved in shifting the weights are radial and con-
tribute no torques to the system. Therefore, from 
Eq. (21-38), 
( Ι ω ) , = (Ι ω )2 
or 
Ι ιω ι = 
Ιιω ι. 
Initially, 
11 = Jlecturer + 2(mbaliri') 
= 5 + 2(^xQy) = 5.25slug-ft.' 
FinaUy, 
II = /lecturer + 2(mbalir2') 
= 5 + 2(^x(2)') = 9slug-ft'. 
Therefore, 
ω2 = - ^ ω ι = - ^ X 1 
« 0.58 rev/sec. 
A more glamorous version of this example is the 
graceful pirouette executed by an ice skater who in-
itiates a rotational motion with arms outstretched, 
then clasps her arms tightly to her body, and im-
mediately experiences a much more rapid rate of 
rotation due to her lower moment of inertia. A 
similar analysis can be made of the way a diver exe-
cutes a one turn (or multiple turn) somersault from 
a diving board. The reader can supply the details in 
this case. 
21-7 
R O L L I N G M O T I O N 
To conclude this chapter, we now consider rota-
tion about an axis that is in translational motion, or 
what is conunonly called rolling motion. It is neces-
sary to use both the translational and the rotational 
relationships simultaneously. Either type of motion 
can be analyzed from the point of view of the laws 
of motion, the impulse-momentum theorem, or 
energy considerations. It should not be surprising, 
therefore, that rolling motion can be analyzed in 
several alternative ways. To illustrate some of the 
possibilities, consider the motion of a uniform cyl-
inder of mass Μ  and radius R , released from rest, 
to roll without slipping down an inclined plane of 
incline angle Θ , as in Figure 21-11. What will the 
acceleration of the center of mass of the cylinder be 
when it has moved a distance s down the incline? 
Figure 21-11 Force diagram for a system exhibiting 
rolling motion. 
Since there is no energy lost due to friction (no 
slipping), it is possible to apply the conservation of 
energy principle. Thus, at the starting position, 
y CM = y I and Etotai = Mgyi. At the final position, 
ycM 
= y2, 
and 
Etotai = Mgy2 + jMv CM+1Ιω  \ 
Since 
energy is conserved, 
Mg(y,-y2) = ^Mi;'cM + | ( | M l ? v ) . 
(21-39) 
The reader can show from Figure 21-11 that 
y2-yi = 5sinö  
(21-40) 
and 
VcM = Rω . 
(21-41) 
By combining Eqs. (21-39), (21-40), and (21-41), we 
obtain 
i;'cM = 2 ( | g s i n 0 ) s . 
(21-42) 
This relation is in a familiar form, which relates an 
initial velocity of the center of mass (vcmo = 0 in this 
case), the final velocity of the center of mass, the 

186 
Rotation 
displacement of the center of mass, and the acceler-
ation down the incline of the center of mass. That 
is. 
from which we conclude that 
OcM =^g sin Θ . 
(Notice that had the cylinder slid without rolling, 
the acceleration would have been g sin Θ , a larger 
result.) 
Next, consider the same situation from the con-
text of the laws of motion. Considering rotation 
about the center of mass, Eq. (21-13) requires that 
FfR = 
la 
(21-44) 
(we have chosen clockwise torques as positive). 
For the translational motion, we have: 
perpendicular to the incline the sum of forces is 
Ν  - Mg cos θ =0, 
(21-45) 
parallel to the incline the sum of forces is 
Mg sin e-Ff 
= MocM, 
(21-46) 
with down the incline taken as positive. By dif-
ferentiating Eq. (21-41) with respect to i, we obtain 
Ra. 
(21-47) 
(21-43) 
Combining Eqs. (21-44), (21-46), and (21-47), 
Mg sin β  - ^ 
= MocM 
(21-48) 
or 
Mg sin θ 
Mg sin θ 
= | g sino, 
as before. 
Notice in this calculation that no use was made of 
Eq. (21-45). It is not possible, for example, to calcu-
late the coefficient of static friction since this would 
imply that slipping was about to occur and there is 
no evidence either favorable or unfavorable to sup-
port such an assumption. 
P R O B L E M S 
1. A wheel of 10 cm in radius turns through an angle of 500 rad in 10 seconds starting from rest. The 
acceleration is constant. Calculate: 
(a) the average angular velocity, 
(b) the angular acceleration, 
(c) the final angular velocity after 10 seconds, 
(d) the tangential acceleration, 
(e) the tangential velocity of a particle on the rim after 10 seconds, and 
(f) the centripetal acceleration of the particle after 10 seconds. 
2. A car goes around a curve of 121 ft radius at a speed of 30 mi/hr. What must be the coefficient of friction 
between the wheels and the level pavement in order that the car does not skid? 
3. A string is wound around the rim of a cylindrical grindstone of mass 2 slugs and radius 8 in. By pulling 
on the end of the string, a man exerts a constant tangential force of 3 lb. Assume the bearings are 
frictionless. 
(a) Find the magnitude of the torque applied to the grindstone. 
(b) Find the magnitude of the angular acceleration of the grindstone. 
(c) Find the angular velocity of the grindstone 6 seconds after it starts from rest. 
(d) What is the kinetic energy of the grindstone 6 seconds after it starts from rest? 
(e) How far does the man pull the free end of the string in 6 seconds starting from rest? 
4. (a) On a horizontal frictionless surface, a 500 gm body revolves in a circle whose radius is 90 cm. Find 
the magnitude of the centripetal force on the body if it makes 1 rev/sec. 
(b) If the body in (a) should accelerate 0.50 rad/sec^ from the given velocity, what would its angular 
velocity be at the end of 10 seconds? 

Problems 
187 
5. A figure skater with arms outstretched starts to rotate with angular velocity of 1 rad/sec and then 
inunediately brings in her arms. Given that her initial and final radii of gyration are 20 cm and 5 cm, 
respectively, what is her resulting angular velocity? 
6. On a horizontal turntable a 100 gm lead block lies whose center of gravity is 40 cm from the axis. The 
angular velocity of the turntable is very slowly increased until it becomes 0.50 rev/sec, at which instant 
the block begins to slip. Find: 
(a) the radial acceleration of the block at this time, 
QD) the centripetal force on the block at this time, and 
(c) the coefficient of static friction between the block and the turntable. 
7. A car starts from rest on a circular race track of radius 1000 ft and increases its speed at the rate of 
4 ft/sec'. 
(a) How long a time will be required to attain the speed at which the tangential and radial components of 
the car's acceleration are equal? 
(b) How far does the car move along the track before the radial and tangential components of its 
acceleration are equal? 
8. A 40 gm ball on the end of a string revolves in a horizontal circle of 50 cm radius on smooth ice at a 
uniform speed of 60 cm/sec. 
(a) Find the angular velocity of the ball. 
Qoi) Compute the tension in the string. 
(c) If the ball should decelerate uniformly at 0.03 rad/sec', through how many radians would it travel 
before it stopped? 
9. A light rigid rod has point masses of 5 kg and 10 kg at 3 m and 1 m, respectively, from the supporting 
frictionless pin Ρ  about which it is free to rotate in a vertical plane. 
(a) If the rod is released with negligible angular 
velocity from the position shown in Figure 
21-12, find the angular velocity with which 
the 5 kg mass passes the horizontal line 
through P. 
(b) Calculate the angular acceleration of the 
rod at the moment that the rod becomes 
horizontal. 
Figure 21-12 
10. A wheel 5 ft in diameter starts from rest and acquires an angular speed of 720 rev/min in 12 seconds. A 
rivet on the rim weighs i lb. Compute: 
(a) the angular acceleration, 
(b) the linear acceleration of the rivet, 
(c) the instantaneous tangential velocity of the rivet 12 seconds from the start, 
(d) the radial acceleration of the wheel 12 seconds from the start, and 
(e) the centripetal force exerted on the rivet. 
11. (a) An airplane in turning should not be subjected to an acceleration greater than 7 g. Find the radius of 
the smallest possible circle when flying at 350 mi/hr. 

188 
Rotation 
(b) A car with wheels of 2 ft diameter is traveling 20 ft/sec. Find the centripetal force necessary to 
prevent 1 lb of mud from leaving the rim of the wheel. 
12. Disks A and Β are mounted on the same shaft so that they may rotate at different speeds or may be 
connected together by a clutch to rotate at the same speed. Initially, A has an angular velocity 
ω ο  rad/sec and a kinetic energy of 400 ft-lb, while Β is stationary. Then they are coupled together, and 
heat is produced in the clutch. The moment of inertia of B is 3 times the moment of inertia of A. 
(a) Find the ratio of the final angular velocity of the two disks to the initial angular velocity of A. 
(b) Compute in ft-lb the heat produced in the clutch. 
13. A stationary disk of mass 0.10 kg and radius 0.10 m, free to rotate about an axis through its center, has a 
little projection on its circumference. A 0.02 kg bit of putty with a velocity of 5 m/sec strikes the 
projection tangentially and sticks to it. What is the angular velocity given to the disk? 
14. In Figure 21-13, a weight W is attached to a light 
string wrapped around a solid cylinder of mass 
M, mounted on a frictionless axle at 0. If the 
weight starts from rest and falls a distance h, 
show that its tangential speed is given by 
V = 
Figure 21-13 
15. Starting from rest, a small car of mass m 
rolls down the incline of the "loop-the-loop" 
shown in Figure 21-14. It starts from a height Η  
of 32 ft, where 1? = 8 ft is the radius of the loop. 
Find the magnitude of the velocity of the car as 
it passes point A on the top of the loop. Neglect 
friction. 
Figure 21-14 
16. (a) Find the moment of inertia about an axis perpendicular to the axis of a slender rod of uniform 
cross-sectional area A and length L, if the density varies according to the relation 
ρ = a-^br, 
where a and b are constants, and r is the distance along the rod from one end. 
(b) A flywheel 3 ft in diameter is pivoted on a horizontal axis. A rope is wrapped around the outside of 
the fli^heel and a steady pull of 101b is exerted on the rope. It is found that 24 ft of rope are 
unwound in 4 seconds. What was the moment of inertia of the flywheel? 

Problems 
189 
17. Consider a circular ring of outer radius Ru inner radius R2, with a thickness t, and of uniform density p. As 
indicated in Table 21-2, the moment of inertia about an axis 1 to the plane of the ring and through the 
center is given by \M(Ri^ +1?2^). Show that this result can be obtained without integration by noting (as in 
Example 4) that for a solid circular disk (R2 = 0), I = ΙΜ Κ Λ  and considering the ring as a solid disk of 
radius Ri from which another soHd disk of radius R2 has been removed. That is, jR¡ng = hARi) 
- IdiskiRi)-
Note: the masses of the two rings must be calculated from the expression Μ  = pV = ρττΑ ^ί. 
18. Use the perpendicular-axis theorem to verify the expression for the moment of inertia of a thin 
rectangular plate of sides W and L about an axis 1 to the plate and through the center, given the validity of 
the preceding result in Table 21-2. 
19. The moment of inertia of a sphere of radius R, mass M, about an axis through the center, is given by 
icM =lMR^. Show that the moment of inertia for the sphere about an axis tangent to the sphere is 
I = lMR\ 

22 
Harmonie 
Motion 
22-1 
Periodic Motion 
191 
22-2 
Simple Harmonie Motion 
192 
22-3 
Examples of Simple Harmonie 
Motion 
192 
22-4 
Energy Relationships 
193 
22-5 
Damped Oscillations 
194 
22-6 
Forced Oscillations 
196 
22-1 
P E R I O D I C M O T I O N 
One of the physical concepts discussed in a 
number of earlier chapters (beginning with Chapter 
10) had to do with repetitive phenomena. Thus, a 
transverse sinusoidal wave propagating along a 
string has the principal feature that any given par-
ticle of the string will take on a continuous se-
quence of displacements that lie between two ex-
treme values symmetrically located on either side 
of the equilibrium position. When the particle has 
gone through the entire sequence of displacement 
values, it repeats the sequence or cycle of events 
again and again, as long as the same wave motion is 
maintained. As we have seen, the extreme displace-
ment value is called the amplitude of the wave mo-
tion and the time required for one cycle of the mo-
tion is the period. 
In the present chapter, we will discuss the nature 
of forces that can produce such repetitive motion, 
which is generally referred to as periodic motion. 
We shall be concerned primarily with sinusoidal 
motion, but the reader should understand that a 
more general periodic motion such as the "saw-
tooth" wave form illustrated in Figure 22-1 can be 
discussed similarly (although the mathematics is 
necessarily more involved). This can be partially 
understood if it is recalled from Section 10-2 that 
any periodic disturbance can be expressed as an 
infinite series of sinusoidal disturbances. Such 
Figure 22-1 Displacement versus time for a "saw-tooth" wave. 
191 

192 
Harmonic Motion 
series are called Fourier series, and they are ex-
tremely useful in the analysis of general periodic 
motion. 
22-2 
S I M P L E H A R M O N I C 
MOTION 
A particle is said to experience simple harmonic 
motion (SHM) if the displacement-time relationship 
is a sinusoidal one. For example, one-dimensional 
SHM is given by 
x=Asm(^-l·φ j, 
(22-1) 
where φ  is the phase angle and is dependent upon 
initial conditions. Since Τ  = 1//, where / is the fre-
quency, and 27Γ / = ω , Eq. (22-1) can also be written 
X = A sin (ω ί + φ ). 
(22-2) 
We wish to know the kind of force that could cause 
a particle of mass m to experience the SHM of Eq. 
(22-2). This is accomplished by recalling that the 
velocity and acceleration of the particle are given 
by dxidt and dhldt\ 
respectively. That is. 
and 
or 
ν ,=^=ω Α 
cos (ω ί + φ ) 
(22-3) 
α . = ^ = ^ = - ω ' Α 8 ΐ η ( ω ί + φ ), (22-4) 
ü x = - ω  χ . 
(22-5) 
where Eq. (22-5) follows from Eqs. (22-2) and 
(22-4). The force F„ which is responsible for the 
motion described by Eq. (22-2), can be determined 
from Newton's second law: 
Fx = mux = - mω 'x. 
(22-6) 
Since mω Ms a constant, we can rewrite Eq. (22-6) 
as 
Fx=-kx, 
(22-7) 
where 
k = mω \ 
(22-8) 
The meaning of Eq. (22-7) is clear: a particle ex-
periencing SHM does so because of the existence 
of a force F which is proportional, but directed op-
posite, to the displacement of the particle. That is, 
the direction of Fx is always toward the equilibrium 
position, and is greater in magnitude the further 
from equilibrium the particle is located. This kind 
of force is known as a linear restoring force or a 
Hooke's law force because the magnitude of the 
force is a linear function of the displacement and 
because Robert Hooke first stated the relationship. 
Very few physical systems exhibit true SHM. (Fric-
tional effects in most systems cause the oscillations 
to die out.) However, the simple harmonic oscil-
lator has proved to be a reasonable starting approx-
imation for a wide variety of situations, from a 
simple pendulum to the small oscillations of one 
atom in a diatomic molecule due to the force 
exerted on it by the other atom. 
22-3 
E X A M P L E S O F S I M P L E 
H A R M O N I C 
M O T I O N 
One of the simplest examples of SHM that can be 
studied experimentally consists of a mass m con-
nected to a helically coiled spring (of negligible 
mass), fastened to a rigid support as in Figure 22-2. 
(See Section 20-4.) It is assumed that the surface on 
which the mass moves is frictionless, and that the 
amplitude of the motion of the mass is small, so that 
the restoring force due to the spring is not altered 
by distortion of the spring. (For large amplitude, a 
coiled spring will suffer a permanent stretch, as a 
result of which the restoring force is also changed.) 
At equilibrium, the position of the mass relative to 
the origin at the wall at left is Xo, so that the dis-
placement from equilibrium is given by x' = χ  - Xo, 
where χ  is the displacement of m from the origin. If 
x' = A at ί = 0, then the motion is described by 
x' = A cos ω ί. 
(22-9) 
where ω  is given by Eq. (22-8). For a spring, the 
constant k is known as the force constant of the 
spring, or simply the spring constant. Experimen-
tally, it is determined by measuring the force Ρ  
necessary to cause a given stretch x' of the spring. 
Figure 22-2 An example of SHM. 

Energy Relationships 
193 
since the applied force is equal in magnitude but 
opposite in direction to the restoring force due to 
the coiled spring. It follows from Eq. (22-7) that 
k = 
-Fx 
x'' 
(22-10) 
An experimental test for the suitability of SHM 
conditions can be made by comparing the value of k 
determined statically by Eq. (22-10) and the value 
obtained dynamically by finding the period τ  of the 
motion and using ω  = Ιττ/τ in Eq. (22-8). 
As another example, consider the simple pen-
dulum shown in Figure 22-3. For a simple pen-
dulum, it is understood that the dimensions of the 
mass m are small compared to the length / of the 
supporting cord, which has a negligible mass. Ap-
plying Newton's second law along the string and 
perpendicular to it, we have 
Τ
- m g cos θ = 0, 
- mg sin θ = Fs = 
mas. 
(22-11) 
For motion along the arc of the circular path, we 
can write 
s = 16, 
(22-12) 
from which 
,ά 'θ 
(22-13) 
since / is a constant. Combining Eqs. (22-11) and 
(22-13) gives 
,ά 'θ 
or 
m / - ^ = - mg sin θ 
(22-14) 
/ / / / / / / / / / / / / 
1 
γ 
/ 1 
\ 
/ 
! 
y
^
"
"
^ 
-^^/^ \ 
^ \ 
mg sin Ö = - F , 
This equation is complicated and leads to equally 
complicated solutions without further assumptions. 
Therefore, let us consider oscillations for which θ 
is small enough to permit use of the approximation 
sin β  « ft If this is done, Eq. (22-14) becomes 
ά 'θ 
(22-15) 
which is identical in form to Eq. (22-5). It follows, 
therefore, that the solution to Eq. (22-15) is 
e = em sm(ω t-l·φ ), 
(22-16) 
where dm is the amplitude of this rotational SHM, 
and 
- 
fi 
(22-17) 
by comparison with Eqs. (22-5) and (22-8). [The 
skeptical reader is invited to show for himself by 
substitution that Eq. (22-16) is the solution of Eq. 
(22-15).] Rearranging Eq. (22-17) leads to the result 
that 
Τ  = 27ΓΛ Ρ . 
(22-18) 
^ g 
We conclude, therefore, that the period of a simple 
pendulum is independent of the amplitude of the 
oscillations for small oscillations. A more sophisti-
cated analysis shows that the solution to Eq. (22-14) 
can be written in the form 
= 2 7 Γ ^ { ΐ + ^ sin' θη , + ¿ sin^ θ „  
. (22-19) 
Figure 22-3 Force diagram of a simple pendulum. 
It is left to the reader to verify that for amplitudes 
up to θη , <25° , Eq. (22-18) is in error by less than 
1 % . Because the period of the simple pendulum is 
independent of the amplitude of osciUation, it is a 
convenient device for keeping time. Alternatively, 
by means of an auxiliary time-keeping device, the 
simple pendulum can be used to determine the 
value of g , the acceleration due to gravity. 
22-4 
E N E R G Y R E L A T I O N S H I P S 
Throughout the discussions above, it has been as-
sumed that there are no frictional (dissipative) 
forces. When this is true, it is clear from Chapter 20 
that the principle of conservation of mechanical 
energy can be applied. For the examples of the last 
section, this means that the total energy of a system 
exhibiting SHM must remain a constant. For the 
mass attached to the spring of Figure 22-2, the total 

194 
Harmonie Motion 
energy is partly kinetic energy and partly potential 
energy stored in the coiled spring, in amounts that 
are sinusoidal functions of time. 
Thus, the kinetic energy is given by 
K.E. = ^mi;' 
= 2 ^ ω ' Λ ' sin' ω t 
= ^mvj 
sin' ω t = (K.E.)„.ax sin' ω ί 
(22-20) 
from Eq. (22-9). In the last expression, ν „  is the 
maximum speed of the mass m. 
From Eq. (20-28), the elastic potential energy of a 
stretched spring is given by 
P.E. = |fcx". 
(22-21) 
Therefore, from Eqs. (22-8) and (22-9), we obtain 
P.E. = |fcA' cos' ω ί = | m ω ' Λ ' cos' ω ί 
= (P.E.)„ax cos' ω ί. 
(22-22) 
The total energy Ε for the system is the sum of 
kinetic and potential energy expressions, or 
Ε = K.E. + P.E. 
= ^mω Ά '(sin' ω ί + cos' ω ί) 
: ^ Γ η ω ' Λ ' = (Κ .Ε .).3χ  = (Ρ .Ε .). 
(22-23) 
which is a constant for fixed m and k. 
It is clear, then, that the transformation of energy 
from a kinetic form to a potential form is beauti-
fully synunetric in cases of SHM. For maximum 
displacement from equilibrium, the system is at rest 
and the total energy is potential in form. At the 
equilibrium position, there is no restoring force so 
the potential energy vanishes whue the speed of the 
system as it goes through the equilibrium position 
becomes a maximum, giving maximum kinetic 
energy. At any other position, the total energy is 
distributed in the two forms. Alternatively, we may 
say that the kinetic and potential energies differ in 
phase by 7Γ/2 or 90°. 
Example 1. A mass m = 0.1 kg is attached to a 
helical spring of spring constant k=4nt/m 
as in 
Figure 22-2. If the mass is pulled out 0.10 m from 
the equilibrium position and released, determine: 
(a) the period of the resulting SHM, 
(b) the total energy Ε of the system, 
(c) the earliest time for which the kinetic energy 
equals the potential energy, and 
(d) the displacement from equilibrium for which 
the kinetic energy is one-half as large as the poten-
tial energy. 
SOLUTION 
(a) From Eq. (22-8), 
2τΓ ^ 
[m ^ 
/0.1kg 
, 
(b) E=ikA' 
= K4nt/m)(0.1 m)' = 0.02joule. 
(c) K.E. = P.E. = EI2, when sin' ω ί = cos' tut = 5 
or ω ί = IT 14. Therefore, f = IT «0.12 sec. 
(d) If K.E. = |P.E., then 
E = | p . E . + P.E. = | p . E . 
As a result, -2kA^ = Mkx\ Therefore, 
or 
x=± 
A =«±0.82 A = ± 0.082 m, 
since A = 0.1 m. 
22-5 
D A M P E D O S C I L L A T I O N S 
Let us now turn to the effect of friction forces on 
a system subject to a linear restoring force. Since 
most freely oscillating systems do not continue to 
oscillate indefinitely, it is clear that some dissipa-
tive mechanism must be included to account for the 
damping (steady reduction of oscillation amplitude) 
of the motion. The exact nature of the friction force 
is a complex problem depending to a great extent 
on the specific situation involved. Thus, a mass sus-
pended from a helical spring and immersed in a 
beaker of oil will not experience the same frictional 
forces as a mass fastened to a similar helical spring 
and moving horizontally along a surface that is 
rough enough to provide a drag on the mass. How-
ever, it is an experimental fact that in many cases 
one can obtain a satisfactory analysis of the motion 
by assuming that the friction force is proportional 
to the speed of the oscillating object and is opposite 
in direction to the motion of the object. Therefore, 

Damped Oscillations 
195 
we express the frictional force F/ in the form 
(22-24) 
where X is a proportionality constant that is 
strongly dependent upon the system being consid-
ered, and dridt is the velocity of the particle. For 
motion in one dimension, we can omit the vector 
notation and write, for example. 
(22-25) 
Applying Newton's second law to this situation 
gives the equation 
d'x 
, 
^dx 
m-TT = -kx- 
X^r. 
'dt' 
'dt' 
(22-26) 
The solution of this equation is a straightforward 
exercise in differential equations, but we prefer to 
use physical arguments to infer the form of the pos-
sible solution(s) to it. Thus, let us suppose that the 
linear restoring force could somehow be neglected 
in comparison with the friction force. This could be 
done, for example, if the system involved a mass 
suspended from a helical spring with a weak spring 
constant and immersed in a highly resisting fluid 
such as molasses. If this is true, it is not difficult to 
deduce that the particle will approach the equilib-
rium position exponentially with time. That is, 
χ = Α β Λ  
(22-27) 
where c is a constant, is a solution to the equation 
d'x 
m dt' 
On the other hand, we have seen that SHM results 
if the frictional force is neglected. It is tempting to 
suppose that a solution to Eq. (22-26) would be a 
trigonometric expression (for the SHM) multiplied 
by a decreasing exponential (due to the damping 
force). Let us, therefore, try the expression 
X = Ae"'' 
s i n ( í t í í + δ ) 
(22-28) 
as a possible solution. Substituting in Eq. (22-26), 
and equating the coefficients of sin(ω í + δ ) and 
cos (ω ί -I- δ ) on both sides of the equation, shows 
that if 
(22-29) 
c = 
Κ  
2m 
and 
ω  = 
then Eq. (22-28) is indeed a possible solution. It 
should be noted that if Κ  = 0, Eq. (22-30) reduces 
to Eq. (22-8), and Eq. (22-28) becomes Eq. (22-2), as 
they should. We have obtained much more, how-
ever. Thus, Eq. (22-30) predicts that for damped 
oscillatory motion the frequency is a constant de-
pending upon k, Ky and m. The maximum fre-
quency occurs for the undamped (K = 0) situation. 
For X ^ O , the frequency is reduced to a lower 
value, but the motion is still oscillatory. This case is 
called underdamped 
motion. If {KI2mf 
= klm, 
ω  = 0 and the motion ceases to be oscillatory, be-
coming exponential instead—a situation referred to 
as critically damped motion. For 
(KI2mf>klm, 
the motion is still non-oscillatory (since ω  cannot 
be negative), but the exponential decay proceeds 
much more slowly, giving rise to overdamped mo-
tion. If (KI2mf>klm, 
Eqs. (22-28) to (22-30) no 
longer apply, but the motion is still exponential in 
form. We will not pursue this point. Figure 22-4 is a 
graph of χ  versus t for the three cases. Finally, we 
remark that the exponential term can be rewritten 
in the form e'"\ 
where 
τ  = 
2m 
X 
(22-31) 
is called the decay time for the system. That is, in a 
time equal to the decay time the amplitude of the 
motion decreases to He of its original value. The 
reader should find physical arguments for the decay 
time increasing with increasing mass m or decreas-
ing frictional effects (decreasing X). 
A 
\ 
X 
f\ 
Ό  
l/ 
' 
(a) underdamped 
(b) critically damped 
(c) overdamped 
Figure 22-4 Displacement versus time for damped 
oscillatory motion. 

196 
Harmonie Motion 
22-6 
F O R C E D 
O S C I L L A T I O N S 
If the oscillations of a damped system are to 
remain constant in amplitude, it is clear from the 
last section that energy must be supplied by some 
external agency to compensate for the energy lost 
because of the frictional force. This can be done by 
applying an external force to the system. To be 
effective, however, it is necessary that the force 
applied be an oscillatory function of time. This is 
because a constant force would resist the motion of 
the system whenever the system was moving in the 
dh-ection opposite to the applied force. Further-
more, it is not difficult to see that an applied force 
with a frequency equal to that of the oscillatory 
system will be more effective than a force with a 
different frequency. An example of this is a child 
being pushed in a swing. Only when the swing is 
moving in the direction of the applied force will that 
force have its maximum effect. 
The mathematics of a driven, damped oscillator 
is substantially more complex than the damped free 
oscillator. To begin with, when the driving force is 
initially applied, there will be an initial motion de-
scribed by an equation of the form of Eq. (22-28). 
This motion dies out rather quickly (that is, for 
t ä  4T), and is therefore referred to as the transient 
part of the solution. The more important part of the 
solution is the steady-state 
part, which describes 
the motion for all time once the transient part has 
damped out. In the analysis that follows, we will 
discuss only the steady-state solution for the case 
of an applied force varying sinusoidally with time. 
That is, we assume a driving force given by the 
equation 
Fd = Fo sin cü dí, 
(22-32) 
where ωd is the angular frequency of the applied 
force. In this case, 2 F = ma leads to 
m § = Fo sin Wät-kx- 
X ^ , 
(22-33) 
'dt^ 
where the various quantities retain the meanings as-
signed to them in the previous section. 
Since we are interested in the steady-state mo-
tion, it is certainly reasonable to assume that the 
time dependence will be sinusoidal with the fre-
quency of the driving force, 6>d. On the other hand, 
ii ω ά φ  ω  =(k/my'^, 
the natural frequency of the 
undamped oscillator, we should expect that the 
amplitude of the oscillations will not be a maximum 
and will be frequency dependent. To test these as-
sumptions, we substitute a solution of the form 
X = A (ú>d) sin (ωdí -h δ ) 
(22-34) 
in Eq. (22-33). The result is 
- mω dΆ (ω d) sin (wdt + δ ) 
= Fo sin ωdí - kA(ω d) sin (ωdí + δ ) 
- Κ ω ά Α  
(ω ά ) cos (a>dí + δ ) 
or 
ö )d')A(wd) sin (wdt + δ ) 
= —sin Cüdí -^^^A(ö >d)cos (a)dí + δ ). 
m 
m 
(22-35) 
This can be simplified by using the identities 
sin (a + β ) = sin α cos β  + cos a sin β  
and 
cos (a + β ) = cos α cos j3 - sin a sin β , 
and rearranging terms to give 
A(wd) ( ω ' - ω d ' ) c o s δ + ^ s i n δ  
m 
m j sin <üdí 
- ( ω ' - ω d ' ) s i n δ - ^ c o s δ  
m 
A(a>d) cos ω dt, 
(22-36) 
where ω ^ = k|m, from Eq. (22-8). Since cos Wdi^ 
sin a>di, Eq. (22-36) is satisfied only if the coeffi-
cients of sin ωdí and cos Wdt vanish independently. 
We then have 
ί ο  
m 
A(ω d) = 
and 
{ ( ω ' - ö>d') COS δ  + 
m sin 
in δ } 
Kωd 
(22-37) 
tan δ  
sin δ  
cos δ  
( ω ' - íü d')* 
From Eq. (22-38), it follows that 
_ Ka>d 
sin δ  = 
m 
(22-38) 
m 
cos δ  = 
175. 
Ta' 

Problems 
197 
When these expressions are substituted into Eq. 
(22-37), we obtain 
772. (22-39) 
These somewhat laborious manipulations show 
that, as we had anticipated, the steady-state does 
indeed possess the frequency of the driving force. 
The amplitude of the oscillations depend not only 
upon the driving frequency and the natural fre-
quency of the undamped oscillator but also upon 
the friction force constant X. The displacement of 
the oscillating object differs in phase from the ap-
plied force by the phase angle δ , given by Eq. 
(22-38). 
We can draw additional conclusions from our 
solution. As asserted above, A (ω ά ) will be a max-
imum when ω α = ω . In fact, for an undamped oscil-
lator (K = 0), Eq. (22-39) indicates that the am-
plitude increases without limit as ωd 
ω. This is a 
situation known as resonance. At resonance an os-
cillating system absorbs a maximum amount of 
energy from the driving force mechanism during 
each cycle of the motion, and the amplitude in-
(a) solid curve: K=0 
(b) dashed curve: Κ ψ ^Ο  
Figure 22-5 Amplitude versus driving frequency. 
creases until some portion of the system breaks 
down. The torsional oscillations induced in the 
Tacoma Narrows suspension bridge by winds of 
resonant frequency resulted in gigantic oscillations, 
which ultimately led to a collapse of the bridge. Fig-
ure 22-5 is a plot of amplitude versus frequency for 
undamped and damped driven oscillations. 
P R O B L E M S 
1. A body of mass 0.10 kg hangs from a long spiral spring. When pulled down 0.10 m below its equilibrium 
position and released, it vibrates with a period of 2 seconds. 
(a) What is its velocity as it passes through the equilibrium position? 
(b) What is its acceleration when it is 0.05 m above the equilibrium position? 
(c) What is the force constant of the spring? 
2. A spiral spring of negligible mass hangs vertically with a 200 gm lead ball fastened to its lower end. The 
ball is then pulled down 4 cm and released. In 60 seconds it completes 150 vibrations or cycles. For the 
vibrating ball, compute: 
(a) its maximum speed and 
(b) the magnitude of its maximum acceleration. 
When the ball is 2 cm below its equilibrium position, compute the magnitude of: 
(c) its speed and 
(d) its acceleration. 
3. A 3 kg mass is attached to a spring and set into oscillation. The mass executes 9 oscillations in 40 
seconds. The total energy associated with the simple harmonic motion is 2 joules. Determine: 
(a) the maximum velocity of the mass and 
(b) the amplitude of the motion. 
4. For a body undergoing simple harmonic motion, 
(a) when the displacement is one-half the amplitude, what fraction of the energy is kinetic and what 
fraction is potential? 
(b) At what displacement is the energy half kinetic and half potential? 

198 
Harmonie Motion 
5. A fifty cent coin is placed at the end of a rough plank that is vibrating horizontally with simple harmonic 
motion of an amplitude of 10 cm and a period of 2 seconds. Compute the coefficient of friction between 
the coin and the plank if the coin is just on the verge of slipping. 
6. A body oscillates with simple harmonic motion according to the equation 
10 sin 
(swt-h^m. 
where t is in seconds. Find: 
(a) the displacement, velocity, and acceleration when ί = 2 seconds and 
(b) the amplitude, frequency, and period of the motion. 
7. A 0.070 kg mass is attached to a spring and placed on a horizontal frictionless plane as shown in Figure 
22-6. The mass is displaced 0.12 m from equilibrium and released. The force constant of the spring is 
8.47 nt/m. Determine the following quantities: 
(a) the frequency of oscillation, 
(b) the total energy of motion, 
(c) the maximum speed of the mass, and 
(d) the maximum acceleration of the mass. 
Figure 22-6 
8. For a mass that is oscillating harmonically, plot the kinetic energy as a function of time and the potential 
energy as a function of time. Assume that the mass is displaced an amount A and released from rest at 
ί =0. 
9. (a) A particle oscillates with a frequency of 1000 cycles/sec if the amplitude is 0.010 cm and its mass is 
O.lOgms. What is the maximum restoring force on the particle? 
(b) Write the equation for a sinusoidal transverse wave of wavelength 2 m, traveling in the positive 
X-direction with a period of 0.05 seconds. 
10. A body is vibrating with simple harmonic motion of amplitude 10 cm and frequency 3 vibrations/sec. 
Compute: 
(a) the maximum values of acceleration and velocity, 
(b) the acceleration and velocity when the displacement is 5 cm, and 
(c) the time required for the body to move from the equilibrium point to a point 8 cm distant from it. 
11. A point at the end of one prong of a tuning fork makes 128 vibrations/sec, and the amplitude is 0.10 cm. 
Calculate the maximum velocity and maximum acceleration of this point. 
12. In what ratio will the frequency, maximum potential energy, and maximum kinetic energy of simple 
harmonic motion in a spiral spring be altered if the load mass is quadrupled? If the amplitude is doubled? 

23 
Properties 
of Matter 
23-1 
Elasticity 
199 
23-2 
Stress, Strain, and Elastic Moduli 
23-3 
Interrelationships of the Elastic 
Moduli 
202 
200 
23-4 
Inelastic Properties of Solids 
204 
23-5 
Elastic Waves 
205 
23-1 
ELASTICITY 
When a substance is described as being rather 
elastic, one conunonly pictures something like a 
rubber band or even a steel band which will, if 
deformed by stretching or bending, "snap back" to 
its original configuration. This qualitative behavior 
is similar to the Hooke's law situation of the last 
chapter. Therefore, if we wish to develop a more 
quantitative means of classifying substances that 
exhibit elastic behavior, it seems reasonable to pro-
ceed by discussing them in terms of SHM. 
No real substance is completely elastic, if this 
means that it exhibits perfect SHM or is completely 
restored to its undisturbed state with no increase or 
decrease in energy content. Thus, the helical spring 
was idealized to permit the discussion in Chapter 
22. It is easily demonstrated that such a system, if 
set in motion, will eventually come to rest because 
of internal dissipative forces that cannot entirely be 
eliminated. Furthermore, if such a spring is ex-
tended beyond a particular limiting extension, it 
becomes permanently deformed and assumes new 
characteristics. From these observations one can 
postulate a number of things: 
1. Within rather restricted limits, many sub-
stances may exhibit properties that approxi-
mate those related to SHM. 
2. Beyond those limits, the properties exhibited 
could vary markedly in a manner dependent 
upon both the material and the extent to which 
limiting values have been exceeded. 
3. It might be useful to classify materials in gen-
eral in terms of Hooke's law and to give some 
indication of the manner in which they fail to 
satisfy the ideal behavior requirements. 
Ultimately, one would prefer a microscopic or 
detailed understanding of the interatomic or inter-
molecular forces that make it possible for a mater-
ial to appear to behave like a harmonic oscillator. 
To attempt such a program here is not desirable for 
several reasons, although it does make an interest-
ing study. Instead, we present a macroscopic de-
scription relating the application of external forces 
to changes in the size or shape of isotropic material 
and the forces produced by these changes. 
Consider a metal wire, which is subjected to a 
tensile or stretching force parallel to the length of 
the wire. Measuring the applied force and the re-
sulting elongation or extension of the wire for a 
wide range of values, we can illustrate the results 
graphically. It is customary to speak of the tensile 
stress, which is the applied force divided by the 
original cross-sectional area of the wire upon which 
it acts normally. The strain is given by the elonga-
1£ 

200 
Properties of Matter 
Stress 
A
' 
/ 
/ 
Rupture 
/ 
/ 
point 
/ 
/ 
Strain 
• 
Figure 23-1 Stress-strain relation for a metal wire. 
tion divided by the original length. Since the origi-
nal area and the original length are constants, it is 
clear that an equivalent statement to Hooke's law is 
that the tensile stress is proportional to the strain. If 
the relevant data is plotted in this fashion, the re-
sults are as shown in Figure 23-1. Point A, the pro-
portional limit, indicates where non-linearity begins 
on the curve. For metals, this corresponds to a 
strain of less than \%. Above A, at point B, a 
permanent "set" or deformation takes place, so 
that reducing the stress to zero yields the dashed 
curve instead of the original curve. Beyond point B, 
a very slight increase in stress produces a marked 
increase in the strain. This region of the curve is 
referred to as the plastic region. The rupture point, 
as the name implies, is the point representing the 
stress that causes the sample to fail. 
If we limit our discussion to stresses falling in the 
proportional region, we can expect that the material 
will behave in a manner approximating the SHM of 
Chapter 22. In the next section, the constant of 
proportionality in Hooke's law (in stress-strain 
form) will be identified for the common types of 
deformation that can occur at low stress. In Section 
23-3, we discuss the relationships that exist be-
tween these elastic moduli, and present a table of 
moduli values for some conunon materials. Section 
23-4 gives a brief indication of phenomena for 
which the SHM approximation is not a satisfactory 
description. The propagation of a longitudinal elas-
tic wave along a metallic rod is the subject of Sec-
tion 23-5. 
23-2 
S T R E S S , S T R A I N , A N D E L A S T I C 
M O D U L I 
If Hooke's law is to be written in the form dis-
cussed above (stress « strain), our attention should 
be directed to the evaluation of the proportionality 
constant that is required if Hooke's law is to be an 
equation. This proportionality constant is called an 
elastic modulus and is equal to the ratio of stress to 
strain, provided that the material has not been sub-
jected to stresses exceeding the proportional limit. 
It is important to distinguish between types of 
deformation that can occur and to properly define 
the stress and strain related to these types. Three 
cases are presented here: the tensile case discussed 
above, tangential deformation or shear, and bulk or 
volume deformation. Figure 23-2 illustrates the 
three cases, with solid lines indicating the initial 
geometry and dashed lines indicating the geometry 
resulting from the application of the appropriate 
stress. 
As indicated in Figure 23-2(a), the tensile force Ft 
is applied perpendicular to the cross-sectional area 
Area of cross-section A 
(a) Tensile stress 
F. 
1 
1 
~ / ~ 
yy 
(b) Shear stress 
. F x 
Fx 
/ 
l ^ w — I 
(c) Bulk stress 
Figure 23-2 Stress-strain geometry for tensile stress, shear stress, and bulk stress. 

stress, Strain, and Elastic Moduli 
201 
A . The tensile stress or normal stress is thus defined 
to be 
tensile stress = E / A , 
(23-1) 
and the related strain is given by 
tensile strain = γ^. 
to 
(23-2) 
The elastic modulus for this case is called Young's 
modulus and is indicated by Y. Thus, Hooke's law 
for this case becomes 
A 
' 
lo' 
so that 
YA Δ / 
(23-3) 
(23-4) 
can be compared with 
F = IcM, 
(23-5) 
which follows from Eq. (22-8) for the extension of a 
material obeying Hooke's law. We see, therefore, 
that by determining Young's modulus it is possible 
to calculate a force constant for the material. This 
constant represents an estimate of the force per 
unit separation exerted by one atom on another in a 
metal sample. Typical values of k obtained in this 
manner are of the order of 10 nt/m. One would ex-
pect, therefore, that any microscopic theory of the 
mechanical behavior of metal wires should predict 
values of this magnitude. 
For the shear case. Figure 23-2(b) indicates that 
by applying the force Fs tangentially to the surface 
area A = Iw, the solid is deformed linearly through 
an angle Αφ . For this case, the shear stress is de-
fined by the relation 
shear stress = FslA= 
Fsllw, 
(23-6) 
and the shear strain is taken to be 
Δ w 
shear strain = 
= tan Αφ . 
(23-7) 
Since we are concerned with strains of less than 1% 
(<0.01), we can use the relation 
so that 
tan Αφ  « Αφ  
(in radians). 
shear strain = ^ 
« Δ φ . 
(23-8) 
the symbol S, Hooke's law becomes 
(23-9) 
or 
5 = § / Δ Φ
, 
Fsh 
IwAw' 
for the geometry of the figure. 
Figure 23-2(c) illustrates a rectangular block of 
material subject to forces that are all normal to the 
surface to which they are applied. To avoid motion 
of the sample as a whole, it is clear that the forces 
applied to the surfaces must be equal in magnitude. 
Like the tensile case, it is natural to define the 
stress in this case as the ratio of the force applied 
normally to the surface divided by the area of the 
surface. The strain is the sum of the strains. There-
fore, we write 
bulk strain = ^ 
+ ψ  + ^, 
(23-10) 
where w = total change in the width due to the bulk 
stress, etc. Since the volume of the solid is given by 
V = lwK 
(23-11) 
Therefore, the bulk strain is equal to the fractional 
change in volume produced by the bulk stress. The 
elastic modulus for this situation is given the sym-
bol B, and we define it by the relation 
bulk stress 
(23-13) 
If the elastic modulus in the case of shear is given 
where the minus sign is introduced to insure that 
the bulk modulus (B) will be a positive number. An 
increase in applied force causes a decrease in 
volume. 
In order to determine the bulk stress, we first in-
troduce the concept of fluid pressure. An extended 
discussion of fluids is given in Chapter 24, so that 
we merely note here that a fluid is not capable of 
resisting shear or tangential stresses. When a gas or 
a liquid is subjected to a shear stress, slip motion 
occurs freely until the removal of the stress. Now 
consider a fluid enclosed in a container equipped 
with a tightly fitting piston. If the fluid is subjected 
to a compressive force by moving the piston, this 

202 
Properties of Matter 
force will be transmitted uniformly throughout the 
fluid if static equilibrium is to be maintained. Figure 
2 3 - 3 shows a wedge-shaped portion of the fluid that 
has been isolated from the remaining fluid in the 
piston. Since the fluid is in equilibrium and shear 
forces do not exist, the forces F i , F2, and F3 on the 
surfaces of the wedge due to the surrounding fluid 
must act normal to the respective surfaces of area 
A i , A2, and A3. This means that F i must be equal in 
magnitude to F2:c, and F3 is equal in magnitude to 
Fly (see the decomposition of force F2 into compo-
nent forces Fix and Fiy in the figure). As a result, 
the magnitude of the stresses 
and 
F, 
Fl. 
Fl sin θ 
Fl 
A, 
A, 
Al sin θ 
Al 
F, 
F2, 
Fl cos θ 
Fl 
A, 
A, 
Al cos θ 
Al 
We see, therefore, that the magnitudes of the 
stresses are equal for an arbitrary wedge angle θ; 
the stress must, therefore, be independent of orien-
tation of the fluid wedge. This compressive stress is 
called the fluid pressure Δ ρ (or hydrostatic pres-
sure in the case of a liquid), and it follows that the 
pressure is the same on all the surfaces of the 
wedge. 
Returning to Eq. ( 2 3 - 1 3 ) , we can now write 
bulk stress = Δ ρ 
and, therefore. 
Β 
( 2 3 - 1 4 ) 
( 2 3 - 1 5 ) 
In the case of solids and liquids, the fractional 
change due to a given pressure increase is very 
A. 5 
/ 
1 
/ 
1 
J,. 
Fi 
A
y 
Fa 
Figure 23-3 Segment of a fluid subject to a compres-
small, so that the value of Β is truly constant. For 
gases, however, the volume changes markedly for 
rather small changes in pressure. In this case, it is 
useful to use the relation 
Β 
( 2 3 - 1 6 ) 
sive force. 
together with the gas law relating pressure and vol-
ume (and temperature). In fact, it is possible to 
obtain an infinite number of values for J3, depend-
ing upon how the temperature varies during the var-
iation of pressure that produces the volume change. 
To avoid confusion or ambiguity, it is customary to 
quote only values of the adiabatic bulk modulus. 
Bad, and the isothermal bulk modulus, Βτ, B^ is the 
value obtained when the pressure volume change 
takes place in an insulated situation, so that no heat 
energy can enter or leave the system. Equivalent 
results are obtained if the change takes place so 
rapidly that heat energy does not have time to enter 
or leave the system before the change is complete. 
It is this modulus that is required in discussing the 
propagation of a sound wave in a gas (Chapter 2 8 ) , 
since the periodic variations of pressure due to the 
sound wave take place too rapidly for heat energy 
to enter or leave a given region of the gas during the 
variation. The isothermal modulus on the other 
hand is the value obtained when the temperature is 
held fixed during the change in pressure. 
Finally, it should be noted that some tables pre-
sent values not of the bulk modulus but rather of 
the compressibility /c, where 
Thus, the compressibility gives the fractional de-
crease in volume per unit increase in pressure, and 
is usually quoted in reciprocal atmospheres or other 
(pressure units)'. As an example a one atmosphere 
increase in pressure on a volume of water will pro-
duce a decrease in volume of less than 0 . 0 1 % . 
23-3 
I N T E R R E L A T I O N S H I P S O F T H E 
E L A S T I C M O D U L I 
It was assumed in our discussions of Section 2 3 - 2 
that no changes occur in the areas upon which the 
deforming forces are applied. It is known experi-
mentally, however, that in the tensile case an in-
crease in length is accompanied by a decrease in 
cross-sectional area at right angles to the length in-

Interrelationships 
of the Elastic Moduli 
203 
crease. If the substance is isotropic (as we have as-
sumed in this chapter), the connection between the 
length strain and the linear strains associated with 
the fractional area change can be written 
we have 
/ 
(23-18) 
where σ  is called Poisson's ratio, and the cross-
sectional area A = wh. Since an increase in / is 
accompanied by decreases in w and h, this defini-
tion assures that σ  is positive. Experimentally, σ  is 
found to be about 0.3. Theory shows that it cannot 
be negative and must be less than I for isotropic ma-
terials. Furthermore, it is possible to show that for 
homogeneous isotropic materials a knowledge of σ  
and Y will provide the values of S and B. The rela-
tionships connecting them are: 
Υ  = 25(1 + σ ) 
y = 3Β (1-2σ ). 
(23-19) 
(23-20) 
Example 1. Derive Eq. (23-20). 
SOLUTION 
Consider a rectangular solid of length /, and 
transverse dimensions w and h, which is placed in a 
pressure tank and subjected to a uniform hydro-
static pressure Δ ρ on all surfaces. As a result, com-
pressional forces will produce corresponding com-
pressional strains in all three directions. From the 
definition of Poisson's ratio (σ ), however, it is clear 
that a compressional strain in the (w) (or h) direc-
tions will be related to a tensile strain in the / 
direction. Let us therefore determine the total 
strain in the / direction and relate this to the volume 
strain. 
First of all, the compressional stress in the (/) 
direction (Δ ρ) produces a strain given by 
or 
Υ ' 
For the (w) direction, we can write 
Since 
/Δ >ν \ 
_Δ ρ 
\w ) 
Y' 
(ί) 
(ii) 
(iii) 
(iv) 
Similarly, in the (h) direction we obtain 
(ν ) 
(vi) 
Now the total strain in the (/) direction becomes 
Since the material is isotropic and the applied 
pressure is uniform, the same result is obtained if 
we calculate the total strain in the (w) or (fi) direc-
tions. That is, 
Δ / 
Aw 
Ah 
Δ ρ.. 
^ . 
. .... 
j = - 
= - ^ = - f ( 1 - 2 σ ) . 
(vm) 
Notice now that the volume of the solid is 
V = IwK 
(ix) 
so that the fractional change in volume due to the 
applied pressure is just the volume strain; 
Δ ν ^Δ / 
Δ η ; ΔΛ  
V 
ι 
w 
h' 
Substituting (viii) in (x), we obtain 
or 
( 1 - 2 σ ) 
(X) 
(xi) 
(xii) 
Equation (23-13) then leads to the desired result 
Υ  = 3Β (1-2σ ). 
(xiii) 
It should be noted that Eq. (xi) requires that 
σ  < I, since a greater value would result in an ex-
pansion when the solid is subjected to a compres-
sion, which is contrary to experiment. 
It should bé emphasized that the assumption of a 
homogeneous isotropic material is an idealization. 
This is because no real material is microscopically 
homogeneous (there is empty space between the 
atoms constituting the material). Furthermore, 
most materials are not isotropic. Their properties in 
fact vary markedly with direction in the material, 
due to the forces holding the material together. 
In Table 23-1, approximate values of the moduli 

204 
Properties of Matter 
and Poisson's ratio are given for some materials. 
The reader can determine readily the extent to 
which Eqs. (23-19) and (23-20) are satisfied. Even 
where the equations fail, the lack of validity is not 
generally a gross failure. As a result, these equa-
tions can be used for estimation purposes when it is 
not desirable to measure all the moduli for materi-
als whose values are not tabulated. 
Table 23-1 Approximate Elastic Moduli and Pois-
son's Ratio for Selected Materials. 
Sub-
stance 
y ( . o - » ^ ) 
5 ( , o . » ^ ) 
B(.O-S) 
Aluminum 
7.0 
2.8 
7.0 
0.34 
Copper 
12.0 
4.2 
12.0 
0.34 
Iron 
20.0 
8.2 
16.0 
0.30 
Lead 
1.16 
0.5 
3.3 
0.40 
Sü ver 
8.0 
8.0 
10.0 
0.38 
23-4 
I N E L A S T I C P R O P E R T I E S O F S O L I D S 
It is possible to demonstrate experimentally that 
the elastic properties discussed in the preceding 
sections provide a satisfactory description of real 
materials. For example, consider a mass m sus-
pended from a wire and set into small amplitude 
vertical oscillatory motion. The observed fre-
quency of vibration is consistent with the value 
ω  ={klmY' 
calculated from a knowledge of the 
mass m and the value of k obtained from a com-
parison of Eqs. (23-4) and (23-5). Other examples of 
such tests are given in the problems. In the next 
section, the propagation of longitudinal elastic 
waves in solids is developed in terms of elastic 
moduli. The results can similarly be shown to be in 
accord with physical reality (within the limits im-
posed on the development at the outset). 
As a result, it is tempting to speculate that the 
world around us is primarily governed by linear re-
storing forces which produce oscillatory motions 
that are essentially symmetric about the equilib-
rium position. Providing the amplitudes of the mo-
tions are suitably limited, this speculation is in fact 
essentially correct. On the other hand, some situa-
tions cannot be described by an elastic model for 
solids. When this is the case, the force law govern-
ing the motion is found to have a more complicated 
form, a principal feature of which is a departure 
from the symmetry of Hooke's law. For example, 
when the displacement Δ / of a solid rod is negative. 
the restoring force is substantially larger than the 
restoring force for a corresponding positive dis-
placement, and the disparity becomes more pro-
nounced at larger displacement magnitudes. This 
behavior is not unreasonable from a microscopic 
point of view. As the atoms or molecules of a solid 
are brought closer together, the restoring forces 
might be expected to increase sharply, especially 
when the atoms are essentially in contact with one 
another. Conversely, as they become more widely 
separated, one would expect the restoring forces to 
diminish steadily. 
Physical evidence favoring this picture of inelas-
tic or asynunetric forces is not hard to find, al-
though a complete discussion of the various phe-
nomena belongs in a study of the solid state. We 
shall now discuss qualitatively the situation in-
volved in the thermal expansion of solids and 
merely list some other inelastic properties. For ad-
ditional details, the reader can consult any intro-
ductory text devoted to the solid state. 
In Chapter 3, the length of a solid rod as a 
function of temperature was given by Eq. (3-2), 
where 
= /ο (1 + αΔ ί), 
It = length at temperature i, 
/o = length at temperature ίο  (usually 0° C), 
α = coefficient of linear expansion, 
Δ ί = ί - Í0 = change in temperature. 
This relationship can also be written in the form of 
Eq. (3-1): 
a 
= 
or 
Δ /= α/ο Δ ί . 
(23-21) 
This latter form can be interpreted as a linear rela-
tionship between the displacement ( Δ / ) from the 
equilibrium configuration (h) and the temperature 
(measured from the reference temperature ίο ). If 
the solid is viewed microscopically as a collection 
of atoms each oscillating about an equilibrium posi-
tion, it is reasonable to assume that the higher 
the temperature the greater will be the range of 
displacements from the equilibrium positions. 
Furthermore, it is also reasonable to consider the 
displacement ( Δ / ) in Eq. (23-21) as the average dis-
placement resulting from the oscillations of the 
atoms of the solid. For an atom oscillating in SHM, 
the symmetry of the motion would produce an av-

Elastic Waves 
205 
erage displacement of zero magnitude. In other 
words, the elastic model predicts no thermal expan-
sion. On the other hand, an inelastic model predicts 
that the oscillating atom would vibrate to larger 
positive displacements than negative displacements 
from the equilibrium position for a given total 
energy because of the lack of symmetry of the re-
storing forces, as mentioned earlier. As a result, the 
average displacement of each atom will be positive, 
corresponding to an increased separation of the 
atoms. If the total energy is regarded as a linear 
function of temperature, it is plausible to assert that 
the inelastic model predicts a thermal expansion 
that increases linearly with temperature, in agree-
ment with Table 3-1. From the values indicated in 
that table for a , it is clear that the linear expansion 
is very small for small changes in temperature 
(Δ ί «0), giving further support to the assumption 
of elastic forces for small displacements. 
Finally, other properties requiring an inelastic 
force model are: (1) the dependence of the elastic 
constants of a solid on temperature and pressure; 
(2) the temperature dependence of the thermal con-
ductivity of a solid; and (3) the temperature depen-
dence of the heat capacity of a solid at high temper-
atures. 
23-5 
E L A S T I C 
W A V E S t 
The discussion of SHM in Chapter 22 considered 
the motion of a mass under the influence of an 
elastic or linear restoring force. Such a force can be 
provided by a helical spring or, as mentioned in the 
previous section, by suspending the mass from a 
slender wire, which is given an initial extension, 
and then releasing it. In these two situations, it was 
tacitly assumed that the mass undergoing the mo-
tion was large enough to permit neglecting the mass 
of the elastic material. In this section, we consider 
the motion of an elastic medium itself. 
As a specific example, we will consider the prop-
agation of longitudinal waves in an isotropic solid in 
the form of a long slender rod of uniform cross 
section. For simplicity, the cross-sectional area is 
assumed to be small enough that transverse vibra-
tional effects (related to Poisson's ratio considera-
tions) are negligible. It is not difficult to visualize 
such a system. For example, by striking the end of 
tThis section is not essential in the remainder of the 
text and may be omitted if desired.) 
the rod periodically with a hammer, a wave (or train 
of pulses) will be produced that travels along the 
rod. At any given thin section of the rod, the dis-
placement from its equilibrium position depends 
upon time and upon position along the rod. Figure 
23-4 illustrates the situation. We now write New-
ton's second law of motion as it applies to the thin 
section whose volume is A dx. For a uniform den-
sity p, the mass of this section is pAdx. If m, the 
displacement of the segment from equilibrium, is 
small, then we may neglect the small change in size 
of the section without appreciable error. As indi-
cated in the diagram, the force on the section di-
rected to the right is then given byt 
F.diFjA)dx 
A 
dx 
A = F , + dFdx 
dx ' 
while the force on the section directed to the left is 
As a result, the net force becomes 
and the second law becomes 
-dx=pAdx^. 
But from Hooke's law, we have 
Ft^ydu 
A 
^ dx' 
(23-22) 
(23-23) 
where duldx is the strain. Note again that the par-
tial derivative notation is required, since u also de-
pends upon both χ  and t. From Eq. (23-23), we see 
that 
dF\ 
dx 
Combining Eqs. (23-22) and (23-24), we get 
d'u 
Yd'u 
dt"" ρ dx 
(23-25) 
ÍThe notation dPJdx 
indicates a partial derivative. 
That is, since F, depends upon both position and time, 
dF,/dx indicates that the derivative with respect to χ  is to 
be determined, while the dependence on time is not al-
lowed to vary. Similarly, dF,/dt indicates a differentiation 
with respect to time, while the χ  dependence is held fixed. 
For example, suppose F, =4xí^ Then dE/dx =4í^ and 
dFjdt=Sxt. 

206 
Properties of Matter 
Figure 23-4 Longitudinal stresses acting upon a thin section of a long rod. 
Equation (23-25) is a one-dimensional wave equa-
tion. Its solution is clearly a rather special kind of 
function in that differentiating it twice with respect 
to time must be equal to a constant times the result 
of differentiating twice with respect to position. 
Rather than considering this problem from a 
strictly mathematical point of view, it is more 
appropriate to be guided by the physical considera-
tions presented in Chapter 10, where we first dis-
cussed traveling waves. Thus, let us assume in anal-
ogy with Eq. (10-5) that the solution we seek can be 
written in the form 
u=AsmY(x-vtl 
(23-26) 
Equation (23-26) leads to 
du 
(2τ τ \ 
. 
2 7 Γ . 
^ 
= - ( - ) A s . n - ( x - « 0 
= - { ψ ) \ . 
(23-27) 
du 
(Itr \ . 
2tr, 
Substituting Eqs. (23-27) and (23-28) in Eq. (23-25), 
we obtain 
or 
(23-29) 
We see, therefore, that our assumed solution, Eq. 
(23-26), is satisfactory provided that the velocity of 
propagation is related to the elastic and mechanical 
properties of the rod by Eq. (23-29). It is left for the 
reader to show that the expression (Υ ΙρΫ " has the 
dimensions of velocity. Since y«10"nt/m^ and 
ρ « 10^ kg/m\ ϋ « 3 X 10^ m/sec for a solid rod. 
Although we shall not do so here, it is not difficult 
to show by similar reasoning that similar wave 
equations result for transverse waves upon a string, 
longitudinal waves in gases, or longitudinal and 
transverse waves in solids of unlimited extent. In 
every case, it is found that the velocity of propaga-
tion is proportional to the square root of the ap-
propriate elastic modulus divided by the density of 
the medium. 

Problems 
207 
P R O B L E M S 
1. A 100 kg weight W is attached by means of a 
hook to a horizontal rod 2 m long, suspended by 
two wires of equal length, as shown in Figure 
23-5. Wire A has a cross-sectional area of 
1 π ml^ and wire Β has a cross-sectional area of 
3 π un^ The weight of the rod may be neglected. 
At what distance from A should the weight be 
placed to give equal stresses in wires A and Β ? 
Figure 23-5 
/¿y/////////////////////////////// 
A 
s 
A 
s 
( 
2. A wire of 0.5 nun in diameter and 10 m long hangs from the ceiling, and an added weight of 2 kg causes 
an extension of 1 mm. Find the Young's modulus for the wire. 
3. Find the extension caused in a steel wire 1 nun in diameter and 5 m in length when an extra weight of 
900 gm is hung on it. Ysteei = 22 x 10'' nt/m'. 
4. When a 5 kg block is attached to the end of a spring hanging vertically, the spring experiences an 
elongation of 10 cm. Find the force constant of the spring. 
5. A hollow cylindrical iron colunm 12 ft high, 10 in. external diameter, and 8 in. internal diameter is 
anchored in a vertical position to a rigid base. The colunm is then subjected to a load of 220 tons placed 
at its upper end. Yiron = 2x10^ lb/in' for this specimen. 
(a) Find the amount by which the column is shortened. 
(b) Find the work done in shortening the colunm. 
6. A 9.8 kg weight fastened to the end of a steel wire of unstretched length 1 m is whirled in a vertical circle 
with an angular velocity of 1 rev/sec at the bottom of the circle. The cross section of the wire is 1 π un^ 
Calculate the elongation of the wire when the weight is at the lowest point of its path. 
y s t c e i = 
22xl0''nt/m'. 
7. A mass of 5 kg attached to a spring causes it to stretch 1 cm. What is the work done in stretching the 
spring? 
8. A rod of elastic material is elongated 2% by a stress of 10' nt/m\ assuming this to be within the elastic 
limit. What is the Young's modulus of the material? 
9. The force constant of a certain spring is 1 nt/cm. What is the potential energy of the spring when it is 
extended 10 cm? 
10. A spiral spring with a force constant of 0.50 lb/ft is 3 ft long when unstretched. What is its elastic 
potential energy when it is stretched to a length of 4.20 ft? 
11. How much force is necessary to stretch a steel wire 2 mm' in cross-sectional area and 2 m long a 
distance of 0.50 nun? Ysteci = 22 x 10'° nt/m'. 
12. A high pressure apparatus can produce pressures as high as 2 x 10' nt/m'. What change in volume will a 
cube of quartz 1 cm on a side undergo if subjected to this pressure? Β  quam = 2.7 x 10'° nt/m'. 
13. Suppose a rubber ball 4 in. in diameter is immersed in water until the pressure on its surface is 20 lb/in'. 
What is the change in volume of the ball? Β  rubber = 1 x 10^ nt/m'. 
14. What is the shear modulus of a rod 1 m long and 2 mm in diameter if the torsion constant of the rod is 
5000 m-nt/rad? 

208 
Properties of flatter 
15. The tension in a rod, whose cross-sectional 
area is 25in^ is 10001b. What is the shearing 
stress on the inclined section of the rod shown 
in Figure 23-6? 
Figure 23-6 

QA 
Fluid 
Mechanics 
24-1 
Ideal and Real Fluids 
209 
24-2 
Fluid Statics 
209 
24-3 
Archimedes' Principle 
211 
24-4 
Applications of Fluid Statics 
211 
24-5 
Fluid Dynamics—Bernoulli's 
Equation 
213 
24-6 
Applications of Bernoulli's 
Equation 
215 
24-1 
IDEAL A N D R E A L FLUIDS 
A fluid is a substance that has the ability to flow 
or alter its shape to conform more closely to that of 
its container. Thus, solids are non-fluid, but both 
gases and liquids are classiñed as fluids. Although 
glass exhibits an extremely slow flow, it behaves in 
many respects more like a solid and is not usually 
considered to be a fluid. Another way of describing 
fluids is to say that they have little or no rigidity. 
From Chapter 23, a rigid substance is one that can 
resist a change in shape when acted upon by 
tangential or shearing forces. All real substances 
must exhibit some tendency toward rigidity.t This 
is due to the existence of the forces that the 
molecules of the substance exert upon each other 
and upon the molecules of any containing vessel. 
For solids, these forces are quite large, but for 
fluids they are much weaker and in some cir-
cumstances can be neglected. 
We can distinguish between a real fluid and the 
concept of an ideal fluid in terms of this rigidity. A 
real fluid exhibits some rigidity while an ideal fluid 
does not. This absence of rigidity simplifies the dis-
tWe do not include superñuids in this discussion. Their 
properties are discussed in the June 1958 issue of Scien-
tific American. 
cussion of fluids in static equilibrium (Sections 24-2 
and 24-3) and in motion (Sections 24-4 and 24-5). 
Because of the mathematical complexities in-
volved, we will not discuss modifications that are 
required in dealing with the motion of real fluids. 
24-2 
FLUID S T A T I C S 
If a given body of fluid is at rest, Newton's first 
law of motion requires that at any point in the fluid 
the resultant force at the point is zero. Further-
more, for any small surface area element about a 
point in the fluid, any force acting on that surface 
must be acting in a direction perpendicular to the 
surface. This assertion follows from our assump-
tion that the fluid is ideal and so cannot respond to 
tangential or shearing forces except by flowing 
freely, which would not be a static situation. Even 
for real fluids, there can be no tangential forces if 
static equilibrium exists. It simplifies the applica-
tion of Newton's laws of motion to ideal fluids if the 
concept of pressure is used. Pressure is a scalar 
quantity, which is equal to the magnitude of the 
normal force per unit surface area acting on a given 
element of surface area. If we consider a small sur-
face area A in the fluid, a total force F due to the 
fluid on one side of A will be exerted on the fluid on 
the other side of A (and, conversely, by Newton's 
209 

210 
Fluid Mechanics 
third law). In terms of these quantities, the average 
pressure ρ is given by 
(24-1) 
_ 
F 
The pressure at a given point in the fluid is given by 
the ratio of the force to the surface area as the 
surface area (bounding the point in question) be-
comes vanishingly small. That is. 
ρ = hm 
-Γ. 
(24-2) 
Additional simplification becomes possible by 
using the density of the fluid at the point in question 
instead of the mass of a given part of the fluid. Den-
sity is defined as the mass per unit volume of the 
substance, 
P = V . 
(24-3) 
As we have already seen in Chapter 3, a change in 
temperature can alter the dimensions (and thus the 
volume) of a substance. Since the mass of the sub-
stance will not be altered by temperature (unless 
the substance somehow disintegrates and disperses 
or else undergoes a chemical reaction), it is reason-
able (and correct!) to assume that the density of a 
substance depends upon the temperature. There are 
other factors that can influence the density of a 
substance which vary in importance for the differ-
ent phases of matter. Because of their very low 
compressibilities, solids and liquids show little 
change in density as the pressure varies, in marked 
contrast to the situation for gases. Densities of vari-
ous substances are given in Table 24-1. 
Let us now apply the condition of static equilib-
rium to determine the variation of pressure with 
depth in a fluid assumed to be at rest. Consider a 
small volume of fluid at a depth y below the sur-
face, as in Figure 24-1. The element of fluid has a 
horizontal surface area A and a thickness dy. For 
Reference 
level 
^=0 
# 
·• 
i ί 
dy 
Τ  
^lp+dp)A 
(pAdy)g 
Figure 24-1 Pressure variation with depth in a fluid. 
fluid density p, the mass of fluid in the volume is 
ρ A dy, and its weight is therefore (pA dy )g. Since the 
fluid is at rest, the sum of all forces must be zero. 
The horizontal forces on the volume element must 
cancel one another in pairs. Since the horizontal 
forces are due to the pressure exerted by the sur-
rounding fluid, symmetry requires that the pressure 
be the same at all points in a given horizontal plane. 
In equilibrium, the vertical forces must also vanish. 
From Figure 24-1, it is clear that 
or 
Σ  Fy = 0 = (p + dp)A -pA- 
(pAdy)g, 
dp = pgdy 
dy 
pg' 
(24-4) 
(24-5) 
From Eq. (24-5), we see that there is a downward 
pressure gradient; that is, the pressure increases 
with increasing depth below the surface. To deter-
mine the pressure difference between two points, it 
is necessary to integrate Eq. (24-4). Thus, if ρ i is the 
pressure at depth yi, and pi is the pressure at depth 
y 2 , we have 
pgdy 
Tabie 24-1 Approximate Densities of Solids and Liquids (values are given for 20°C except when otherwise stated). 
Liquids (10' kg/m') 
Metallic solids (10' kg/m') 
Non-metallic solids (lO'kg/m') 
Ethyl alcohol 
0.789 
Steel 
7.80 
Ice 
0.992 (0°C) 
Carbon tetrachloride 
1.59 
Copper 
8.90 
Concrete 
2.30 
Sea water 
1.03 
Brass 
8.70 
Glass 
2.60 
Pure water 
1.000 (4°C) 
Gold 
19.3 
Cork 
0.240 
Mercury 
13.6 (0Ό  
Lead 
11.3 
Balsa wood 
0.130 
Benzene 
0.880 
Aluminum 
2.70 
Oak 
0.720 
Ebony 
1.20 

Applications of Fluid Statics 
211 
or 
P2 - ρ , = Δ ρ = ( ^ pgdy. 
(24-6) 
The integral on the right-hand side of Eq. (24-6) 
cannot be performed unless the relationship be-
tween density and depth is known. Furthermore, 
one must be sure that there is no variation of the 
value of g in the depth interval involved. For liq-
uids on the Earth's surface, this is true, but in the 
Earth's atmosphere there is a non-negligible varia-
tion of density with height (negative depth) above 
the Earth's surface, and, for sufficiently large differ-
ences in height, g varies also. 
Example 1. Find the relationship between pres-
sure and depth below the surface of water. 
SOLUTION 
At the surface (yi = 0), the pressure pi = po is that 
due to the atmosphere, at y i = d, the pressure is 
P2 = Po + Δ ρ . For an incompressible fluid and as-
suming no variation in g, Eq. (24-6) gives 
P2 - Po = Po + Δ ρ - Po = pg (yi - y i) = 
pgd. 
or, alternatively. 
Δ ρ = 
pgd 
P2 = Po + 
pgd. 
(24-7) 
The weight density (pg) of fresh water is about 
62.4 lb/ft'. Equation (24-7), therefore, indicates that 
the pressure increases by 62.4 Ib/ft^ for every foot 
of depth below the surface in a lake. Since atmos-
pheric pressure at sea level is about 151b/in^ the 
reader can show that an increase in pressure equal 
to atmospheric pressure corresponds to a depth of 
approximately 35 ft in a fresh water lake. 
24-3 
A R C H I M E D E S ' P R I N C I P L E 
Over 2000 years ago, Archimedes deduced that a 
body inunersed in a fluid should experience a lifting 
(or buoyant) force equal to the weight of fluid dis-
placed by the body and acting in a vertical direction 
through the original center of gravity of the dis-
placed fluid. To see that this must be so, consider a 
volume of the undisturbed fluid having the same 
shape as the solid to be immersed in the fluid. From 
the discussion of the previous section, the net force 
exerted on the volume of fluid by the surrounding 
fluid must be just equal in magnitude and opposite 
in direction to the weight of that volume of fluid for 
equilibrium to exist. In addition, that force must act 
through the center of gravity, as does the weight of 
the volume of fluid involved. If the volume of fluid 
is replaced by a body with the same shape and vol-
ume, the remaining fluid will still exert the same 
force as it did on the volume of fluid, since no 
change has been made in the remaining fluid. If the 
body remains at rest when inunersed, it follows that 
its average density must correspond to that of the 
displaced fluid. 
When the buoyant force is greater than that of 
the weight of the body (density of the fluid greater 
than density of the body), the body floats in a liquid 
or gains altitude in a gas. Conversely, when the 
buoyant force is less than the weight of the body 
(density of the fluid less than density of the body), 
the body sinks in a liquid or loses altitude in a gas. 
For example, if a gas-filled ballon is released in still 
air, it will rise to a height at which the density of the 
supporting air equals the average density of the bal-
loon and the gas inside. At that height, the buoyant 
force will equal the weight of the balloon and its 
contents; no further increase in height will occur. 
Buoyant forces must be taken into account if one 
makes precision "weighings" with a sensitive equal 
arm analytical balance when the density of the ob-
ject being measured differs appreciably from the 
density of the standard "weights" used with the 
balance. For an object of volume and density Vx 
and px, and using "weights" of density pw and 
balancing volume Vw in air of density pa, a balanced 
situation requires that 
( p . V, - paVx)g 
= ( p . V. - p . VJg. 
(24-8) 
The desired quantity is the actual mass (pxVx) of the 
object, which from Eq. (24-8) is 
Pχ Vχ =p^V^-l·pΛ Vχ -VΛ  
(24-9) 
If px differs greatly from pw, then 
and Vw will 
also differ markedly, and the buoyant force exerted 
by the air cannot be neglected. 
24-4 
A P P L I C A T I O N S O F FLUID 
S T A T I C S 
The mercury barometer can be used to measure 
atmospheric pressure. As shown in Figure 24-2, it 
consists of a glass tube filled with mercury. The 
tube is then inverted in an open dish of mercury. 
The mercury in the tube then drops, until the weight 
of the remaining column of mercury exerts the 
same pressure on a unit cross section of surface of 
mercury at point 2 as does the atmosphere outside 

212 
Fluid Mechanics 
Figure 24-2 The mercury barometer. 
the tube on a unit cross section of the surface at 
point 3. (The small amount of mercury vapor in the 
evacuated space above the column is negligible at 
ordinary temperatures.) By Eq. (24-7), 
P2 = Pl + 
pHggh 
= PHggd. 
Since points 2 and 3 lie in the same horizontal 
plane, symmetry requires that the two points are at 
the same pressure. Therefore, 
Patm = PHgg^. 
From Table 24-1, the density of mercury is 13.6 x 
10^ kg/m\ 
At 0°C at sea level, where g = 
9.81 m/sec', a pressure of one atmosphere is taken 
to be equal to exactly 0.76 m of mercury. Thus, one 
atmosphere is equivalent to 
Patm = (13.6 X 10' kg/m')(9.81 m/sec')(0.76 m) 
= 1.013x10'nt/m' 
= 14.70 lb/in'. 
= 1 atm. 
It is common to quote pressures in centimeters or 
inches of mercury, but it should be understood that 
this is done for convenience. The example above 
indicates the calculation that is required to convert 
centimeters of mercury to actual units of force per 
unit area. If a liquid of density px (and negligible 
vapor pressure) is used in the barometer instead of 
mercury, it should be clear that equilibrium with at-
mospheric pressure requires that 
or 
Pxhx —  Phs^Hs 
Example 2. What is the height of a colunm of 
water in a water barometer on a day when atmos-
pheric pressure is one atmosphere? 
SOLUTION 
From Eq. (24-10), 
. 
_ (13.60 X 1 0 ' k g / m ' ) . ^ - . . 
(IxlO'kg/m') 
^^'^^"^^ 
= 10.34 m of water 
= 33.9 ft of water. 
As another example of the application of the con-
ditions of static equilibrium to fluids at rest, con-
sider the situation shown schematically in Figure 
24-3. An upright gate of width L and height d, 
hinged along its base, serves as part of a dam. If the 
water surface is level with the top of the gate: 
(a) what is the total horizontal force acting on the 
gate, and (b) what is the torque about the hihge due 
to the body of water? 
To begin with, we can neglect atmospheric pres-
sure since it acts on both sides of the gate, directly on 
the one side and as an increase in hydrostatic pres-
sure on the upstream side. From Eq. (24-7), neglect-
ing PAtm, the pressure ρ at depth y is 
Ρ  = P^y, 
so that the resulting horizontal force on the element 
of area 
dA==Ldy 
is 
dF = pdA = pgLydy. 
The total force will be 
F = 
dF = [ 
pgLydy 
(24-10) 
t 
*4 
1 
d-y 
τ  
- 
Τ  1 
Figure 24-3 A hinged vertical gate in a dam. 

Fluid Dynamics—Bernoulli's Equation 
213 
The torque acting about the hinge due to the 
pressure ρ on the area element d A is 
dT=ipdA){d-y) 
= 
pgLy(d-y)dy, 
and the total torque about the hinge will be 
= ί dT=¡ 
pgLy{d-y)dy 
Jo 
Jo 
4pgLd^ 
(24-11) 
Let us examine more closely the assertion in the 
example above that the effect of the external 
(atmospheric) pressure on the surface of the water 
is to increase the total pressure at any point by ex-
actly the same amount. This follows directly from 
Eq. (24-7), and is known as Pascal's principle since 
it was stated by Blaise Pascal in the mid-
seventeenth century. A major application of the 
principle can be found in the operation of a hy-
draulic jack or hydraulic press. As indicated in Fig-
ure 24-4, a small force F , is applied to a piston of 
cross-sectional area A„ which thus exerts a pres-
sure ρ = FslAs 
on the liquid in the device (oil, for 
example). By Pascal's principle, this pressure is 
transmitted undiminished to a larger piston of area 
AL. Since the increase in pressure is the same at 
both pistons, the force FL exerted on the larger pis-
ton because of the applied force is found from the 
relation 
F , 
FL 
or 
Thus, the hydraulic press is a force multiplying de-
vice that depends upon the ratio of the piston areas 
for its multiplying factor. 
Figure 24-4 The hydraulic press. 
24-5 
FLUID D Y N A M I C S — B E R N O U L L I ' S 
E Q U A T I O N 
We turn now to fluid dynamics, the study of 
fluids in motion. To begin our study, we continue to 
discuss only ideal fluids which are free from dis-
sipative forces. For simplicity, it is also assumed 
that the flow pattern is static, which means that at 
any given point in the fluid the velocity remains 
constant even though the particles of the fluid are 
always in motion from point to point. When this is 
true, one can sketch representative paths followed 
by the particles. Such paths are called streamlines, 
and the velocity vectors of the various particles 
along the streamline are tangent to the streamline. 
This latter requirement means that streamlines can-
not cross, since to do so would involve points of 
intersection at which the velocity vector would 
have to be double-valued in order to be simultane-
ously tangent to both streamlines. In considering 
the fluid confined between two surfaces formed by 
sets of adjacent streamlines, the absence of inter-
secting streamlines leads to a view of the fluid flow 
as occurring in sheets or layers between the two 
streamline surfaces. For this reason, laminar flow is 
an alternative description of streamline flow. 
A tube of flow exists if the boundary of a given 
portion of flowing fluid is a set of adjacent stream-
lines. The assumption of an ideal fluid means that a 
tube of flow and the flow pattern in the containing 
pipe should be the same at any given cross section 
of the pipe, since there are no dissipative forces be-
tween the pipe and fluid or between adjacent layers 
of fluid. For the moment, let us also assume the 
fluid is compressible and that at point 1 all fluid par-
ticles have a velocity of magnitude Vi directed per-
pendicular to the cross-sectional area A i of the 
pipe, while at point 2 the particle velocities are all 
of magnitude Vi directed perpendicular to the 
cross-sectional area Ai of the pipe, as in Figure 
24-5. In a short time interval Δ ί , the particles that 
were crossing A i will move a distance ϋ ιΔ ί and 
those that were crossing Az will move a distance 
ϋ2Δί. If the density of the fluid at point 1 is pi, the 
mass of fluid crossing A i in Δ ί will be approxi-
mately given by 
Am, = ρ ιΑ ιΓ ιΔ ί ,, 
and the mass of fluid crossing A2 in Δ ί will similarly 
be 
Am2 = 
piAiViAt. 

214 
Fluid Mechanics 
Figure 24-5 Streamline flow in a pipe of varying cross 
section. 
If the time interval becomes infinitesimally small 
(Δ ί ^ 0 ) , the mass flux at 1 and 2 become 
dmi 
dt 
= 
PiAiVi 
and 
dt • = P2A2V2. 
In the pipe between points 1 and 2, there are no 
locations where mass can be removed from the flow 
(sinks) nor are there locations where mass can be 
introduced (sources). As a result, the mass flux at 
any point in the pipe must be identical to that at any 
other point in the pipe. Therefore, we can write 
dmi _ 
dm2 
dt ~ 
dt' 
This result is known as the equation of continuity, 
and is expressed in the form 
PiAiVi 
= p2A2t;2. 
(24-12) 
For incompressible fluids, p, = p2, and the equation 
of continuity simplifies to the form 
AiVi 
= A2V2. 
(24-13) 
For liquids then, Eq. (24-13) indicates that where 
the cross section increases the velocity must de-
crease, and conversely. In terms of the streamlines 
of Figure 24-5, it is clear that in regions where the 
streamlines are close together the fluid velocity is 
greater than in regions of wider spacing. 
Now let us consider the conservation of energy 
for an ideal liquid in streamline or laminar flow. In-
stead of the pipe of Figure 24-5, we shall use the 
modification shown in Figure 24-6. In a short time 
interval Δ ί, the net work done on the fluid will be 
W = Fll;,Δ í - Ε 2ΐ;2Δ ί = (ρ ιΑ ,υ, -ρ2Α 2ΐ ;2)Δ ί. 
(24-14) 
Figure 24-6 Streamline flow in a pipe of varying cross 
section and elevation. 
The work-energy theorem states that 
W= (K.E.)p - (K.E.)/ + (P.E.)p - (P.E.), 
= Δ (Κ .Ε .) + Δ (Ρ .Ε .). 
(24-15) 
For the system of Figure 24-6, 
Δ (Κ .Ε .) = ^pA2V2'At -^ρΑ ,ι;,'Δ ί 
:|ρ(Α 2ΐ ;2'-Α ,ι;.')Δ ί 
and 
Δ (Ρ .Ε .) = pA2i;2Aígy2 - pA,ü iAígy, 
= pg(A2t;2y2-Aiü iyOAí. 
Using Eq. (24-13), 
Δ (Κ .Ε .) = ^ρ(ι;2'-ι;.')(Α ,ι;,Δ ί), 
Δ (Ρ .Ε .) = pg(y2- y.)(A,ι;,Δ í). 
(24-16) 
Combining Eqs. (24-14), (24-15), and (24-16), we 
obtain 
P. - P2 = \piv2' - Vi') + pg(y2 - y.) (24-17) 
or 
ρ -l·^pυ'-l· pgy = constant. 
Equation (24-17) is known as Bernoulli's equa-
tion for the streamline flow of an incompressible 
ideal fluid. From the form of our derivation, it is 
clear that it is just the statement of conservation of 
energy for a moving ideal fluid. It might also be 
noted that Eq. (24-17) reduces to Eq. (24-7) when 
Ü1 = t?2 = 0, as it should for energy conservation in a 
static situation. 

Applications of Bernoulli's Equation 
215 
24-6 
A P P L I C A T I O N S O F B E R N O U L L I ' S 
EQUATION 
There are a variety of devices for determining 
fluid speeds by means of pressure measurements 
and Eq. (24-17). One such device is the Venturi 
meter, which consists of a manometer tube (simply 
a U-tube) located so that it can measure the pres-
sure difference between a point in the main body of 
a pipe and a point in a constricted portion of the 
pipe (called the throat) as shown in Figure 24-7. The 
Pm=manometer 
fluid density 
Figure 24-7 A Venturi meter. 
value of ν  is to be found with the device. Since the 
pipe is horizontal, yi = y i in Eq. (24-17), measuring 
from the axis of the pipe. As a result, we can write 
Pi-^\pfv' 
= 
P2-^^PfV2' 
Now 
(24-18) 
(24-19) 
Pl-P2 
= (Pm 
-Pf)gh. 
This is because 
P i + Pfgih 
+ y ) = P2 + Pmgh 
+ 
pfgy 
is the pressure at the level of the top of the left-
hand column of manometer fluid. From Eq. (24-13), 
we find 
A, 
Λ 2 
(24-20) 
Combining Eqs. (24-18), (24-19), and (24-20), 
(Pm -Pi)gh 
=^Pf 
and, therefore. 
Example 3. Water flows in a pipe of cross-sec-
tional areas A l = 200cm^ A2 = 50cml The man-
ometer fluid is mercury (pm = 13.6 gm/cm\ pf = 
gm/cm'), and h is found to be 30.4 cm. Find: 
(a) vi and 
(b) Q = AiV, the volume flux in the pipe. 
SOLUTION 
(a) Using Eq. (24-21), 
2(12.6 gm/cm^)(980 cm/sec^)(30.4 cm) 
(1 gm/cm^) 7200 cm^ 
LV15— 
cm )-] 
= 224 cm/sec. 
(b) 
Q = (200 cm')(224 cm/sec) = 4.48 x 10' cmVsec. 
Notice that continuity requires that Q = A2V2 = 
AiUi. 
Although our derivation of Bernoulli's equation 
was for an ideal liquid, it can be used to give a 
qualitative explanation for the curving of a pitched 
spinning baseball in air, a compressible fluid. The 
air flow past a stationary spinning ball, as indicated 
in Figure 24-8, will be the same as that for a spin-
ning ball moving through stationary air. Due to the 
roughness of the ball cover, some air will be carried 
in rotation with the ball. As a result, the net velocity 
on either side of the ball will be the vector sum of 
the flow and rotational air velocities. In Figure 24-8, 
the velocity at the right is a difference, while the 
velocity at the left is a sum of the two velocities. 
Figure 24-8 
ball. 
Top view—air flow around a spinning 
Therefore, Umet > uinet. Since y i = yz, it follows that 
Pi < P2, and the pressure difference results in a side 
thrust to the left that creates the curved path. 
In concluding this section, two warnings are ap-
propriate. Many devices measure not the total or 

216 
Fluid Mechanics 
absolute pressure at a point but the gauge pressure, 
which is the difference between the absolute pres-
sure and atmospheric pressure. Bernoulli's equa-
tion requires absolute pressures in all calculations. 
Furthermore, it is necessary to check carefully to 
be certain that all quantities involved in Bernoulli's 
equation are expressed in a single system of units. 
Otherwise all efforts will be futile. 
P R O B L E M S 
1. What is the pressure at the bottom of a vessel 76 cm deep when filled with water? 
2. Rain falls into a boat pulled up on the beach; can it be emptied by a siphon? If the boat were floating on 
the water, could it be emptied? A siphon is a pipe or tube bent to form two legs of unequal length by 
which a liquid can be transferred to a lower level over an intermediate elevation. 
3. In a hydraulic press, the large piston has cross-section area A i = 200cm', and the small piston has 
cross-section area A2 = 5 cm'. If a force F2 = 25 nt is applied to the small piston, what is the force on the 
large piston? 
4. A large tank contains water to a depth of 4 m, and a round hole 1 cm' in area is opened 1 m from the 
bottom. Find the rate of flow in liters/min. 
5 . If the velocity of water in a pipe 10 cm' in area is 2 m/sec, what is the velocity in a pipe 5 cm' in area that 
connects with it, both pipes flowing full? 
6. If a baseball 4 inches in diameter is thrown 
through the air with a velocity of 20 ft/sec and is 
rotating once per second, as shown in Figure 
24-9, compare the velocities of the air at the top 
side and bottom side of the ball. 
Figure 24-9 
7. A glass ball of mass 25 gm has an apparent mass of 15 gm in water and 13 gm in brine. Find the density 
of the glass and the brine. 
8. Because the density of ah- ( p « ) is so small, the mass of an object ( p x V J will not differ greatly from the 
mass of the "weights" ( p w V w ) . Show that, to a good approximation, Eq. (24-9) can be written 
Using 
this can be written 
pxVx 
- p w V w 
+ p e 
ntx = pxVx, 
mw = Pw V . 
v.. 
ntx —  ntw 
Pw \Px 
/ J 
9. A block of oak ( p , = 0.720 g/cm') is balanced by brass weights (pw = 8.70 gm/cm') of mass 
= 10 gm. 
Taking the density of air to be pa = 1.3 x 10' gm/cm', find: 
(a) the true mass nix of the oak and 
(b) the percentage error involved if the buoyant effect of air is neglected. 
10. What is the approximate volume of a floating log of specific gravity 0.60 which just supports a 160 lb man? 

Problems 
217 
11. The pipe illustrated in Figure 24-10 carries 
water in streamline flow. The water discharges 
into the atmosphere with a velocity of 60 ft/sec 
at point B. The cross-sectional area of the pipe 
is 0.03 ft' at point Β and 0.09 ft' at point A. The 
elevation of point Β is 15 ft above the level of 
A. Determine the gauge pressure at point A. 
Figure 24-10 
12. A frictionless fluid flows through the horizontal 
system shown in Figure 24-11. Points 1,2, and 3 
are on the axis of the system, and point 3 is at the 
wall. At point 1, the pipe has a cross-sectional 
area of 80 cm'; at point 2, the area is 20 cm', and 
the two side tubes each have areas of 10 cm'. The 
fluid flowing from the left discharges from the 
system at a rate of 0.02 mVsec. 
(a) Determine the speed of the fluid at points 
1, 2, and 3. 
(b) Determine the pressure of the fluid at points 
1, 2, and 3. 
13. The volume flow rate of water in the pipe shown 
in Figure 24-12 is 12 ftVsec. The cross-sectional 
area at points 1, 2, and 3 are 0.4 ft', 0.3 ft', and 
0.3 ft', respectively. The water discharges from 
the right end of the pipe into the atmosphere. 
(a) Find the velocities of flow at points 1 and 2. 
(b) Find the heights hi and hi to which the 
water wiD rise in the two vertical tubes. 
Figure 24-12 
Figure 24-11 
1-· J 
—  
2 » - t 
Τ  
- - 3 - · 
14. The section of pipe shown in Figure 24-13 car-
ries water in streamline flow. At point A, the 
water discharges into the atmosphere at the rate 
of lOftVsec. The cross-sectional area of the 
pipe at point A is 0.25 ft' and 1 ft' at point B, 
Point Β  is 8 ft above point A. What is the gauge 
pressure at point B? 
Figure 24-13 

218 
Fluid Mechanics 
15. The section of pipe shown in Figure 24-14 car-
ries water in streamline flow. At point A, the 
water discharges into the atmosphere at the rate 
of 9 ftVsec. The cross-sectional area of the pipe 
at point A is 0.75 ft' and 0.5 ft' at point B. Point 
Β is 12 ft below point A. What is the gauge 
pressure at point B? 
Figure 24-14 

25 
The First Law of 
Thermodynamics and 
Thermodynamic 
Processes 
25-1 
Work 
219 
25-2 
Heat and Internal Energy 
220 
25-3 The Mechanical Equivalent of Heat 
221 
25-4 The First Law of Thermodynamics 
221 
25-5 
Thermodynamic Processes 
222 
25-6 
Specific Heat of an Ideal Gas 
224 
25-7 
Adiabatic Process for an Ideal Gas 
224 
Thermodynamics is a subject that involves the per-
formance of work, the flow of heat, and the change 
in energy of a system. The laws of thermodynamics 
are important to many branches of engineering, 
physics, and chemistry. In this chapter, we will dis-
cuss work, heat and internal energy, and the 
mechanical equivalent of heat. After a discussion 
of the first law of thermodynamics and thermo-
dynamic processes, we will consider the specific 
heat of an ideal gas and the special features of an 
adiabatic process for an ideal gas. 
25-1 
W O R K 
Here we shall discuss work in terms of it being 
done on or by a system. This is called external 
work. Internal work, such as the work done by one 
part of a system on another or by atoms or 
molecules on one another, is not considered in ther-
modynamics. 
We begin by showing that the work done on or by 
a system is equal to the pressure exerted on or by 
the system times the change in volume of the sys-
tem. It will further be shown that this work depends 
upon how the process is carried out. 
Suppose we consider a gas contained in a cylin-
der of cross-sectional area A fitted with a movable 
piston subjected to an external force F. If the pis-
ton is moved a small distance Ad by the force F, as 
illustrated in Figure 25-1, an amount of work W 
equal to 
W = FM 
(25-1) 
is done on the system. Since force is equal to the 
product of pressure and area, Eq. (25-1) may be 
written as 
W = pAAd. 
(25-2) 
Figure 25-1 Work done on a gas by a force F. 
219 

220 
The First Law of Thermodynamics and Thermodynamic Processes 
But 
(25-3) 
where Δ V is the change in the volume occupied by 
the gas. Therefore, the work done on the system is 
given by 
W = pAV. 
(25-4) 
Example 1. One kilogram of water of volume 
1 m' is all changed to steam at a volume 1671 m' 
when boiled at atmospheric pressure of 1.01 x 
10^ nt/m'. What is the external work done by the 
system? 
SOLUTION 
External Work = ρΔ V 
= (1.01 X 10'nt/m')[(1671-l)m'] 
= 1.69x10'joules. 
So far, we have considered work done under con-
stant pressure. If, however, the pressure changes as 
the volume changes, as in the case of the compres-
sion or expansion of a gas, this knowledge can be 
obtained by plotting the pressure versus the volume 
of the gas as work is done on or by the gas. The 
graph obtained should be similar to Figure 25-2, and 
the work done will be equal to the area under the 
curve. 
Figure 25-2 Work done when a gas changes its 
volume. 
The area under the curve can be obtained as fol-
lows: 
W = 
pdV, 
(25-5) 
where Vo is the initial volume and V} is the final 
volume. Since the pressure changes as the volume 
changes, the integration cannot be performed 
analytically unless the pressure can be expressed as 
a function of the volume. If the gas can be treated 
as an ideal gas, the curve can be fitted by Boyle's 
law or, in other words, the pressure-volume relation 
obeys Boyle's law. Here the pressure at any point is 
given by 
^ P o V o 
and 
W -I" 
PoVc 
^ - d V = p o V o l n ^ . 
(25-6) 
It is obvious that the work depends on the path 
followed. If the curve of Figure 25-2 bowed upward 
rather than downward, the work done would be 
greater since the area under it would be greater. We 
can see, therefore, that the work done on or by a 
system depends not only on the initial and final 
states but also on the intermediate states, that is, on 
the path followed from the initial to the final state. 
25-2 
HEAT A N D I N T E R N A L E N E R G Y 
We discussed these two concepts briefly at the 
beginning of Chapter 1. However, it seems a good 
idea to repeat what we said in addition to making 
another point regarding both concepts. 
Heat is energy transferred between two or more 
material substances or from one portion of the sub-
stance to another on a macroscopic scale. In other 
words, heat is energy in transit. Heat added to or 
removed from a system, like work done on or by a 
system, depends on the path followed from one 
state of the system to another. Again, suppose we 
have a gas contained in a cylinder fitted with a 
weighted movable piston, which can be clamped. 
Now, consider the two different processes. First we 
place the cylinder on a hot stove with the piston 
clamped. Then we place the same cylinder on the 
hot stove with the piston undamped, so that it can 
rise as the gas expands. The amount of heat trans-
fer required to increase the temperature of the gas 
over the same interval will be different for the two 
processes. In other words, the heat flow into the 

The First Law of Thermodynamics 
221 
system depends on how we go from one state of the 
system to another. 
If work is done on or by the system, and/or heat 
is added to or removed from the system, we have a 
change in what is called the internal energy of the 
system. From the molecular point of view, the in-
ternal energy of a system is the sum of the total of 
the kinetic and potential energies of its molecules. 
Consider a system that undergoes a change from 
one state to another by means of several alternative 
paths. Each path involves a definite heat transfer 
and a definite amount of work done on or by the 
system. An important observation will be that for 
all the paths there will be the same change in inter-
nal energy of the system equal to the difference be-
tween the heat added and the work done by the 
system. It follows, therefore, that the change in 
internal energy of a system is independent of the 
path. 
25-3 
THE M E C H A N I C A L EQUIVALENT OF 
HEAT 
Heat energy can be converted into mechanical 
energy and vice versa. Rumford, in 1798, was the 
first to demonstrate that the amount of heat created 
in boring cannon was proportional to the amount of 
mechanical work expended. Joule was the first to 
prove experimentally the equivalence of heat and 
mechanical energy. He used an apparatus in which 
falling weights rotated a set of paddle wheels in a 
container of water (Figure 25-3). The mechanical 
energy was computed from the value of the 
weights, and their distance of descent and the heat 
Thermometer 
Weight 
generated was computed from the mass of water 
and its temperature rise. 
We may state the results of Joule's experiment in 
the form of an equation 
W = JQ, 
(25-7) 
where W is the work expended in joules, Q is the 
heat generated in kilocalories, and J is a constant 
called the mechanical equivalent of heat. The best 
experiments to date give 
/=4186 joules/kcal 
= 4.186 joules/cal. 
The conversion of work into heat energy is easily 
accomplished, and it is possible to approach 100% 
efliciency, as in the Joule experiment. However, the 
conversion of heat energy into work is much more 
difficult and, as will be shown in the following chap-
ter, it is impossible to have 100% efficiency. It is 
necessary to have a working substance in order to 
convert heat energy into work. In the steam engine, 
the working substance is water vapor; in the 
gasoline engine, it is a mixture of gasoline vapor 
and air. 
25-4 
THE FIRST L A W O F 
T H E R M O D Y N A M I C S 
The first law of thermodynamics is the conserva-
tion of energy principle, which states that in the 
transformation of thermal energy into another form 
of energy, no energy is created or destroyed. If we 
have a system into which we put heat or on which 
work is done, its energy is increased by the heat put 
into it or the work done on it. Similarly, if we take 
heat out of it, or if we allow it to do work, its energy 
is decreased. This energy is that which we have 
called the internal energy of the system. In other 
words, the change in the internal energy of a system 
is equal to the sum of the heat put into or taken out 
of the system and the work done on or by the sys-
tem. Symbolically, this statement, which is the first 
law of thermodynamics, can be written as follows: 
AL7 = 0 +W. 
(25-8) 
Figure 25-3 Joule's paddle-wheel experiment. 
Δ LT, Q, and W may be either positive or negative. If 
Q units of heat are put into the system, Q is taken 
as positive (+) since it tends to increase the internal 
energy U of the system; if Q units of heat are taken 
out of the system, Q is taken as negative (-) since it 

222 
The First Law of Thermodynamics and Thermodynamic Processes 
tends to decrease the internal energy U of the sys-
tem. The sign of Δ17 will depend on the signs and 
relative magnitudes of Q and W, The first law of 
thermodynamics may be written in differential form 
if we assume that an infinitesimal amount of heat is 
transferred and an infinitesimal amount of work is 
done in the process. The differential form of the 
first law is 
tem at constant pressure is 
W = pAV, 
(25-11) 
dU = dQ-^dW. 
(25-9) 
Example 2. What is the change in the internal 
energy of a system when 100 cal of heat are sup-
plied to it at the same time that it does 200 joules of 
work? 
SOLUTION 
AU=Q 
+ W 
= (100 cal)(4.186 joules/cal) - (200 joules) 
= 418.6 joules - 200 joules 
= 218.6 joules. 
25-5 
T H E R M O D Y N A M I C 
P R O C E S S E S 
We shall consider four thermodynamic processes 
here—namely, adiabatic, isobaric, isovolumic, and 
isothermal processes. It should be pointed out that 
these four processes are not the only processes that 
can occur in thermodynamics, but they are among 
the more common ones. 
An adiabatic process is one that occurs without 
transfer of heat to or from the system. Applying the 
first law of thermodynamics to an adiabatic pro-
cess, we have 
AU=W. 
(25-10) 
If a gas expands adiabatically, it does work on the 
surroundings and therefore loses internal energy. 
Conversely, if a gas is compressed adiabatically, its 
internal energy increases since work is done on the 
gas. 
Example 3. During an adiabatic compression, 50 
joules of work is done on the gas. What is the 
change in internal energy? 
SOLUTION 
Δ Ι / = 0 + W= 0 joules + 50 joules 
= 50 joules. 
An isobaric process is one that takes place at 
constant pressure. The work done on or by the sys-
where Δ V is the change in volume. The first law of 
thermodynamics for a constant pressure process 
takes the form 
Δ17 = (3+ρΔ ν . 
(25-12) 
Many chemical processes take place at constant 
pressure. 
Example 4. If 100 cal of heat are added to 
10"' m' of CO2 in a cylinder with a movable piston, 
and the gas expands against an external pressure of 
1 atmosphere (101 x 10' nt/m') to a volume of 1.5 x 
10"' m', what is the change in the internal energy of 
the gas? 
SOLUTION 
AU=Q-W 
= 
Q'pAV 
= (100 cal)(4.186 joules/cal)-(1.01 x 10'nt/m') 
x(1.5x 10' m ' - l . O x 10'm') 
= (418.6 joules - 1.01 x 0.5 x 10' joules) 
= 418.6 joules-50.5 joules 
= 368.1 joules. 
An isovolumic process is one that takes place at 
constant volume. It is sometimes called an iso-
choric process. Since the volume does not change, 
no work is done on or by the system, and the first 
law takes the form 
AU = Q. 
(25-13) 
AU the heat that is added or removed from the sys-
tem goes into a change in the internal energy. 
Example 5. If 10 cal of heat are added to a gas in 
a rigid container so that its volume cannot change, 
what is the change in the internal energy of the gas? 
SOLUTION 
AL7=Q + W = Q + p A V = Q + 0 
= (10 cal)(4.186 joules/cal) 
= 41.9 joules. 
An isothermal process is one that takes place at 
constant temperature. The performance of an 
isothermal process requires that the system be in 
mechanical and thermal equilibrium with a heat 
reservoir. No real process is ever perfectly isother-
mal just as no real gas is ever a perfect or ideal gas; 

Thermodynamic Processes 
223 
Porous 
plug 
P2 
Initial state 
V x 
Porous 
plug 
\ί I I I I I I 
I I I I I 
Final state 
Figure 25-4 Throttling process. 
but many processes are quite close to being isother-
mal. A simple and important isothermal process is 
the isothermal expansion or compression of an 
ideal gas. The internal energy of an ideal gas de-
pends only on the temperature; its change is, there-
fore, zero in an isothermal process. Hence, for an 
isothermal process involving an ideal gas, Δ Í/ = 0 
in the first law and 
0 = 
(25-14) 
where both Q and W have the same units and can 
be + or - according to the sign rule adopted for the 
first law. 
Example 6. An ideal gas does 100 joules of work 
in an isothermal expansion. How many calories of 
heat must be added to the gas during the process? 
SOLUTION 
Since the gas is ideal, ALT = 0 in the expansion. 
Therefore, 
0 = W + Q 
and 
0 = -100 joules+Q 
Two other thermodynamic processes deserve 
brief mention. One of these is the jree expansion, 
which is an internal energy conservation process, in 
which a gas is permitted to expand freely into a 
vacuum without any heat flow into or out of the gas. 
In a free expansion. 
= 
Uf, 
(25-15) 
where Ui is the initial internal energy and Uf is the 
final internal energy. The free expansion process 
has some theoretical importance but no practical 
importance. 
The other thermodynamic process is known as a 
throttling process. 
A throttling process is per-
formed by forcing a gas through a porous plug to a 
region of lower pressure in an insulated cylinder, as 
illustrated in Figure 25-4. A continuous throttling 
process can be performed by a pump that maintains 
a constant high pressure on one side of a porous 
wall and a constant lower pressure on the other 
side, as illustrated in Figure 25-5. It can be shown 
that in a throttling process 
17, + P.V,= 
172 + P 2 V 2 , 
(25-16) 
where Ui, pu and V i are the internal energy, pres-
sure, and volume of the gas on one side of the 
porous wall and U2, Pi, and Vi are the internal 
energy, pressure, and volume on the opposite side 
of the wall. The quantity LT 4-ρV is called the en-
thalpy. The throttling process and Eq. (25-16) are 
very important in steam engines and refrigerators. 
Figure 25-5 Continuous throttling process. 

224 
The First Law of Thermodynamics and Thermodynamic Processes 
25-6 
S P E C I F I C HEAT OF A N IDEAL G A S 
The specific heat of a gas, unlike a liquid or solid, 
depends on whether it is measured at constant pres-
sure or constant volume. At constant pressure, the 
volume of the gas increases when heated, and ex-
ternal work equal to pdV is done by the gas. At 
constant volume, the pressure increases but no ex-
ternal work is done and all of the heat goes into 
increasing the internal energy of the gas. Therefore, 
the specific heat at constant pressure is greater than 
the specific heat at constant volume. 
By the use of the first law of thermodynamics, we 
can obtain an expression that gives the difference 
between these specific heats for an ideal gas. Sup-
pose Q units of heat are added to η  moles of an 
ideal gas at constant pressure, and the gas under-
goes a change in temperature dT. Then, 
dQ = nQdT, 
(25-17) 
where Cp is the molar specific heat at constant pres-
sure. The first law of thermodynamics, applied to 
the case when heat is added to the gas and the gas 
expands at constant pressure or does work, is 
dU = 
dQ-dW, 
(25-18) 
Since dU = nCvdT, where C„ is the molar specific 
heat at constant volume, and dW = pdV, we have 
nadT=nCpdT-pdV 
or 
n(Cp - a)dT 
= pdV. 
(25-19) 
Differentiating the general gas law, pV = nRT gives 
pdV-^Vdp 
= nRdT. 
At constant pressure dp = 0 and 
pdV = nRdT. 
(25-20) 
Substituting Eq. (25-20) into Eq. (25-19), gives 
n(Cp - a)dT 
= nRdT 
or 
Cp 
Cv — R' 
(25-21) 
Equation (25-21) shows that the molar specific heat 
at constant pressure exceeds the molar specific heat 
at constant volume for an ideal gas by the universal 
gas constant R, The relation agrees well with ex-
perimental results for simple molecules but not so 
well for complex ones. The ratio Cp/C is important 
since it is easier to measure experimentally than 
either Cp or C„, independently. It is denoted by the 
Greek letter y. The value of y decreases with the 
complexity of the gas. For a monatomic gas, y = 
1.67; for a diatomic gas, y = 1.40. Approximate val-
ues of Cp, Cv, Cp - C„, and CplCv are given in Table 
25-1 for a number of real gases. 
25-7 
A D I A B A T I C P R O C E S S FOR 
A N I D E A L G A S 
In an adiabatic process, the pressure changes 
more rapidly with volume than it does in an isother-
mal process. In addition to the volume change, 
there is also a temperature change in an adiabatic 
process. This leads to a greater change in pressure 
than is the case for an isothermal process. To see 
this, consider an ideal gas contained in a perfectly 
non-conducting cylinder fitted with a movable pis-
ton. Let the gas undergo a small adiabatic expan-
Table 25-1 
Approximate Molar Specific Heats of Gases. 
Type of gas 
Gas 
c( 
) 
^"Vmole'^Cy 
C Í 
"Vmole° C/ 
C p - a 
Cp 
Monatomic 
A 
5.0 
3.0 
2.0 
1.67 
He 
5.0 
3.0 
2.0 
1.67 
Diatomic 
CI. 
7.2 
6.0 
2.1 
1.35 
CO 
7.0 
5.0 
2.0 
1.40 
H . 
6.9 
4.9 
2.0 
1.41 
N . 
7.0 
5.0 
2.0 
1.40 
o. 
7.0 
5.0 
2.0 
1.40 
Polyatomic 
CO. 
8.8 
6.8 
2.0 
1.30 
H^S 
8.3 
6.2 
2.1 
1.34 
SO. 
9.7 
7.5 
2.2 
1.30 
Ethane ( C Ä ) 
12.4 
10.3 
2.1 
1.20 
Ether [(C2H5)20] 
33.3 
30.8 
2.5 
1.08 

Problems 
225 
sion so that there is a change in volume dV. The 
work done by the gas will equal pdV. Since dQ =0 
in an adiabatic process, and dU = nQdT for an 
ideal gas, we have from the first law 
nadT 
= -pdV, 
(25-22) 
In Section 25-6, we saw that 
pdV-^Vdp = nRdT. 
Eliminating dT between these equations, we obtain 
naipdV + Vdp) + nRpdV = 0. 
Since R = 
Q-C, 
aVdp 
+ QpdV = 0. 
Dividing by QV 
ρ 
av 
" 
or 
where 
y = α* 
Integration of this equation gives 
In ρ -I- γ In V = constant 
or 
p V = constant. 
Equation (25-23) may also be written 
(25-23) 
(25-24) 
for any two points of the process. By applying the 
general gas law to Eq. (25-24), relations between Γ 
and V and Τ  and ρ may be obtained. These rela-
tions are 
T,vr'= 
(25-25) 
and 
„  a\/y)-\)>T, _ „ ((Ι/γ)-!)^· 
Pi 
I1—P2 
Í2. 
Figure 25-6 is a p - V diagram for isothermal and 
adiabatic processes. It can be observed that the 
adiabatic curves have steeper slopes than the 
isothermal curves at any one point, and that an 
adiabatic curve cuts a set of isothermal curves. 
Example 7. A volume of CO2 gas at a temperature 
of 27°C is compressed adiabatically to 1^ of its 
original volume. Find the final temperature. 
SOLUTION 
(
TZ \ 1.30-1 
loj 
T2 = (300°Κ )(10'') = 597°K = 324°C. 
Adiabatic curves 
Isothermal 
curves 
Figure 25-6 p-V diagram for adiabatic and isothermal 
processes. 
(25-26) 
P R O B L E M S 
1. A sample of CO2 in a cylinder with a movable piston is heated by a burner. If 60 cal of heat are added to 
the gas and the gas expands against an external pressure of 1 atmosphere (1.01 x 10^ nt/m') from an 
initial volume of 1000 cm^ to a final volume of 1600 cm^ how much work does the gas do on the piston? 
What is the change in internal energy of the gas? 

226 
The First Law of Thermodyriamics and Thermodynamic 
Processes 
2. The temperature of 1 mole of nitrogen gas is raised from 10°C to HOT. Find the heat required to 
accomplish this at constant volume. If the process is carried out at constant pressure, find: 
(a) the heat added, 
(b) the increase in internal energy, and 
(c) the external work done by the gas. 
3. Derive the following formulae for an adiabatic process of an ideal gas, assuming γ to be a constant. 
TV 
constant 
and 
constant. 
4. In a Wilson cloud chamber at a temperature of 20° C, particle tracks are made visible by causing conden-
sation on ions by an adiabatic expansion of the gas to 1.40 of its initial volume. If γ = 1.40 for the gas, 
what is the gas temperature after expansion? 
5 . Find the work done by a gas initially at a pressure p©  = 10^ nt/m\ temperature ίο  = 27° C, and volume 
Vo = 1 m^ when it: 
(a) expands to a volume of 2 m^ at constant pressure, 
(b) expands to a volume of 2 m^ at constant temperature, and 
(c) has its pressure increased to 2 x 10^ nt/m' at constant volume. 
6. Niagara Falls is about 50 m high. What approximate temperature increase would you expect for the 
water at the bottom of the Falls? 
7. Starting with the definition of work W done by an expanding gas, show that for an adiabatic expansion 
of a gas from a state (pi, Vi) to a state (pi, V2) the work done is given by 
7 - 1 
8. When a system is taken from state a to state b along the path acb shown in Figure 25-7, 20 cal of heat 
flow into the system and 30 joules of work are done by the system. 
(a) How much heat flows into the system along 
the path adb if the work done by the sys-
tem is 10 joules. 
(b) When the system is returned to a along the 
curved path, the work done by the system is 
20 joules. Does the system absorb or liber-
ate heat, and how much? 
Figure 25-7 
9. How much heat must be added to helium gas initially at a pressure of 10' nt/m', temperature 27'*C, and of 
volume 1 m^ when the gas: 
(a) expands to a volume of 2 m^ at constant pressure, 
(b) expands to a volume of 2 m^ at constant temperature, and 
(c) has its pressure increased to 2 x 10' nt/m' at constant volume. 

Problems 
227 
10. When a system is taken from state i to state / along the path iaf in Figure 25-8, it is found that G = 50 
cal and W/¡ = 20 cal. Along the ibf path 
= 36 cal. 
(a) What is Wn along the path 
ibp 
(b) If Wif = -13 cal for the curved return path 
fi, what is Qif for this path? 
(c) Take 17¡ = 10 cal, what is 17/? 
(d) If Ut = 22 cal, what is Qti ? What is 
? 
Figure 25-8 
η  
f 
Η  
f 
i i 
V 
11. Helium at 27°C and 1 atmosphere pressure is compressed adiabatically to a pressure of 5 atmospheres. 
What is the final temperature? 
12. A boy pumps up his bicycle tires on a day when the temperature is 27°C and the atmospheric pressure is 
15 lb/in'. Find the maximum temperature of the air in the bicycle pump if the tire pressures are to be 30 
lb/in' and the air in the pump is assumed to be compressed adiabatically. 
13. A block of copper weighing 0.50 kg initially at 500° A is placed in a tank containing 3.78 kg of water at 
300° A. The tank is made of copper weighing 1 kg. If these items do not exchange heat with the external 
surroundings, calculate: 
(a) the change of internal energy of the copper block, 
(b) the change of internal energy of the water, and 
(c) the change of internal energy if all the components are considered a single system. 
14. The volume of 2 moles of an ideal gas is increased isothermally from 1 to 10 liters at 27°C. How many 
joules of work are done? 
15. The density of air at 0°C and 1 atmosphere pressure is 0.001293 gm/cm\ If 10 gms of air are heated from 
0°C to 100°C at constant pressure: 
(a) How much will the volume change? 
(b) How many joules of work will be done by the gas on expanding? 
(c) The specific heat or air at constant pressure (Cp) and constant volume (c„) is 0.24 and 0.17 
cal/gm°C, respectively. From the work done in expanding the gas, determined the value of γ. 
16. Compute the specific heats Cp and c„ in cal/gm °C for helium, hydrogen, nitrogen, and oxygen. 
17. An atomic bomb explosion causes a ball of fire of about 20 m in diameter at a temperature of 200,000°K. 
Estimate the diameter of the fireball at 2000°C, and state the assumptions you made. 
18. Two insulated containers are connected by a pipe with a valve. One container contains 1 mole of helium 
and the other contains 1 mole of nitrogen. If the valve is opened, what is the final temperature of the 
mixture? 
19. During a process in a closed system, the internal energy of a fluid changes from an initial value of 500 
kcal/kg to a final value of 400 kcal/kg. If 90,000 joules/kg of work are performed by the fluid, compute 
the quantity of heat added to or removed from the fluid during the process. 
20. If 22.4 liters of air at 0°C and 1 atmosphere pressure has its pressure suddenly doubled, what is the new 
temperature and new volume? 

26 
The Second Law of 
Thermodynamics 
and Entropy 
26-1 
The Second Law of 
26-4 
The Thermodynamic or Kelvin 
Thermodynamics 
229 
Temperature Scale 
232 
26-2 The Carnot Cycle 
229 
26-5 
Entropy 
233 
26-3 
Carnot's Theorem 
231 
26-1 
THE S E C O N D L A W OF 
T H E R M O D Y N A M I C S 
The first law of thermodynamics, which is a con-
servation of energy law, places no limitations on 
the possibility of transforming heat into work, or 
vice versa. In the transformation of work into heat, 
a complete conversion is possible. However, there 
are definite limitations in the transformation of heat 
into work. The second law of thermodynamics 
rules out the possibility of converting heat entirely 
into work. 
There are two statements of the second law, one 
postulated by Kelvin and the other by Clausius. 
The Kelvin statement is: 
It is impossible to construct an engine that 
will continuously operate by extracting heat 
from a source at a higher temperature and 
converting it entirely into mechanical work. 
In other words, it is impossible to construct an 
engine that is 100% eñicient, and that some of the 
heat taken in by the engine must be ejected. We 
shall see from the next statement of the second law 
(the Clausius statement) that the ejected heat must 
be into a region of a lower temperature than that of 
the source. The Clausius statement is: 
It is impossible to construct a self-acting 
machine that will continuously operate by 
extracting heat from a region at a lower 
temperature and ejecting it into a region at a 
higher temperature. 
In other words, for a machine to transfer heat 
from a region of lower temperature to a region of 
higher temperature, there must be energy supplied 
to the machine from an outside energy source. An 
everyday example of this is a home electric re-
frigerator. In order for heat to be removed from the 
inside of the refrigerator and ejected into the room, 
electrical energy must be supplied by an outside 
source. The Kelvin statement and the Clausius 
statement of the second law can be shown to be 
equivalent. The first and second laws of ther-
modynamics, like all physical laws, are not subject 
to direct proof. However, all attempts to disprove 
them have failed. 
26-2 
THE C A R N O T 
C Y C L E 
If a substance is taken from an initial state back 
to the same state, the substance is said to have un-
dergone a cycle of operations. Any cycle that can 
be made to occur in the reverse direction by very 
small changes in the conditions is called a rever-
sible cycle. A Carnot cycle is a reversible cycle 
consisting of two isothermal processes and two 
adiabatic processes, the details of which are discus-
sed below. 
229 

230 
The Second Law of Thermodynamics and Entropy 
i y / / / / A 
M
Z
Z
Z
Z
Z 
Figure 26-1 The Carnot engine (schematic). 
According to the Kelvin statement of the second 
law of thermodynamics, in order for an engine to 
convert heat extracted from a source at a higher 
temperature into work, it is necessary to eject some 
of this heat into a source of lower temperature. If 
we have two such sources at temperatures Ti and 
T2, with TI greater than T2, we can transform heat 
into work by the following process, which is called 
a Carnot cycle, or the cycle of a Carnot engine. 
Consider a fluid contained in a cylinder with in-
sulating walls and a conducting bottom fitted with 
an insulating piston. Let Ri and R2 be two reser-
voirs of heat at temperatures Γ ι and T2 ( T i > T2), 
and S—an insulatmg stand (Figure 26-1). We shall 
assume that the heat reservoirs are so large that 
their temperatures remain unchanged by the addi-
tion or subtraction of finite amounts of heat. 
With the cylinder on Ru we raise the piston 
very slowly so that the fluid undergoes a reversible 
isothermal expansion at temperature T i , rep-
resented by AB in Figure 26-2. The fluid absorbs an 
amount of heat Qi from Ri. 
We now place the cylinder on the insulating stand 
S and again raise the piston very slowly. Because 
the system is thermally insulated during the pro-
cess, the fluid undergoes a reversible adiabatic ex-
pansion with a decrease in temperature from T i to 
Γ 2. The process is represented by B C in Figure 
26-2. 
The cylinder is now transferred to JR2, and the 
piston is pushed down very slowly so that the fluid 
undergoes a reversible isothermal compression at 
temperature Γ 2 represented by CD in Figure 26-2. 
The fluid gives up an amount of heat Q2 to R2. 
Finally, the cylinder is again placed on the in-
sulating stand 5, and the piston is pushed down 
very slowly until the temperature of the fluid is 
again T,. The fluid undergoes a reversible adiabatic 
compression and is now in its initial state again. The 
process is represented by DA in Figure 26-2. 
The net amount of heat absorbed by the system 
during the cycle is Qi - Q2. From the first law of 
thermodynamics, the net work done (W) by the 
Ρ  
Ol 
\ w 
α 
V 
Figure 26-2 The Carnot cycle. 

Carnot's Theorem 
231 
system is 
W 
= 
Q,-Q2 
( 2 6 - 1 ) 
The efficiency (η ) of the Caraot cycle is defined as 
the ratio of the net or useful work done by the cycle 
to the heat absorbed from the high temperature 
source. 
( 2 6 - 2 ) 
Since the Carnot cycle is reversible, the above op-
erations can be carried out in the reverse directions. 
When this is done, work is done on the system; it 
absorbs heat Q2 at a temperature T2, and rejects 
heat Q i at a temperature T, greater than T2. This is 
the process performed in the operation of a re-
frigerator. 
Figure 2 6 - 3 is a schematic diagram of a Carnot re-
frigerator. External work W is done on the re-
frigerator, heat Q2 enters the refrigerator at low 
7 
Ol 
\Ν  
^ 
Figure 26-3 Schematic diagram of a refrigerator. 
temperature T2, and heat Q i leaves at a high tem-
perature T i . From the first law of thermodynamics, 
W = Q . - Q2 = 
( 2 6 - 3 ) 
The best refrigerator is the one that removes the 
greatest amount of heat (Q2) from the cold reser-
voir for the least amount of external work (W). We, 
therefore, define the "coefficient of performance" 
(ω ) of a refrigerator as 
Q2 
Q2 
(26-4) 
26-3 
C A R N O T ' S T H E O R E M 
Carnot's theorem is stated as follows: 
No other engine working between the same 
temperatures Ti and T2 can be more efficient 
than a reversible engine. 
We shall now proceed to prove this theorem. Sup-
pose an irreversible engine I has efficiency η / 
greater than the efficiency 
of a reversible Carnot 
engine, 
working between the same temperatures 
T, and T2, and that both engines are adjusted to do 
the same amount of work. Let Q, be the heat 
absorbed and Q2 the heat rejected by the Carnot 
reversible engine (J?), and Q\ and Q2 the heat 
absorbed and rejected by the irreversible engine (7). 
Because both engines are adjusted to do the same 
amount of work and η / is assumed greater than η *. 
and 
W = 
Q i - Q 5 = ( 3 , - Q 2 
( 2 6 - 5 ) 
Since the numerators of the above equation are 
equal, 
Q\<Qx 
and 
Q 5 < Q 2 . 
( 2 6 - 6 ) 
Let engine I drive engine R backwards, as illus-
trated in Figure 26-4. The combined system does 
nothing but transfer O2 - Q2 = Q i - Qi from Γ 2 to 
T i . Since this is a violation of the Clausius state-
ment of the second law of thermodynamics, the as-
Q\\ 
Jo, 
Figure 26-4 Schematic diagram of an irreversible 
engine driving a reversible Carnot engine backwards. 

232 
The Second Law of Thermodynamics and Entropy 
sumption that η ι > η κ is false. The conclusion is 
that 7)1 < r)R, and that Carnot's theorem is true. 
We leave it to the student to prove an important 
corollary to Carnot's theorem, namely: 
All Camot engines operating between the 
same two temperatures have the same effi-
ciency. 
From this statement, it is obvious that the efficiency 
of a Carnot cycle is independent of the working 
substance. 
26-4 
THE T H E R M O D Y N A M I C O R K E L V I N 
T E M P E R A T U R E 
S C A L E 
The important corollary to Carnot's theorem 
stated above shows that the ratio Q2/O1 has the 
same value for all reversible engines that operate 
between the same temperatures ii and t i . There-
fore, Q2/O1 depends only on the temperatures ti 
and Í2. We may, therefore, write 
(26-7) 
where / is an unknown function of ti and Í2. Con-
sider a sequence of engines, each receiving heat re-
jected by the previous one (Figure 26-5). 
The efficiencies of the first two engines are given 
by 
and 
mi 
_ Q . - Q , 
f(t2, h). 
From Eq. (26-8), 
Qi. 
1 - η ΐ 2 
= 
F ( Í„ Í, ) , 
(26-8) 
(26-9) 
(26-10) 
where F is also an unknown function. The first two 
engines working together constitute a third engine 
and 
Therefore, 
^ 
= 
F ( f „ f , ) . 
Qz 
Q3 Q2 
F(h, hY 
(26-11) 
Hence, from Eqs. (26-10) and (26-11), 
PI, 
η  
F ( f „ t 3 ) 
^ ( '" '^ ) - F ( f . . t 3 ) ' 
Figure 26-5 Schematic diagram of a sequence of 
Carnot engines. 
for any Í3. Since Í3 does not appear on the left-hand 
side of the above equation, it may be cancelled, and 
we may write 
Qi^Htd 
Qi 
eitiY 
(26-12) 
where θ is another unknown function. The ratio on 
the right side of Eq. (26-12) is defined as the ratio of 
the two Kelvin temperatures, and is denoted by 
T,/T2. We may, therefore, write Eq. (26-12) 
Q2 
T2 
(26-13) 
It should be pointed out that the temperatures in 
Eqs. (26-7) through (26-12) refer to arbitrary tem-
perature scales. Equation (26-13) defines the Kelvin 
temperature scale. Equation (26-13) states that the 
ratio of two Kelvin temperatures is the ratio of the 
heat absorbed to the heat rejected by a reversible 
engine operating between these two temperatiu-es, 
and is independent of the working substances. 
The definition of the Kelvin scale is completed by 
assigning to T i the value of 273.16°K, the tem-
perature of the triple point of water as in Chapter 1. 
For a Camot engine operating between tempera-

Entropy 
233 
tures Τ  and Τ ι, we have 
It is stated that 
r = 273.16° K^. 
(26-14) 
The smaller the value of Q , the lower the corre-
sponding temperature T. The smallest value of Q is 
zero, and the corresponding Τ  is absolute zero. It 
was pointed out that in a Carnot cycle the heat is 
transferred during the isothermal processes. There-
fore, if a system undergoes a reversible isothermal 
process without heat transfer, the temperature at 
which the process takes place is absolute zero. This 
is a fundamental definition of absolute zero, since it 
is independent of the properties of the substance. 
Making use of Eq. (26-13), we may now write the 
efficiency of a Carnot engine and the coefficient of 
performance of a Carnot refrigerator in terms of 
the Kelvin temperatures as follows: 
and 
T . - T 2 
(26-15) 
(26-16) 
For a Carnot engine, it is only possible to have 
100% efficiency if T2 = 0. Since absolute zero tem-
perature is impossible to attain, a 100% efficient 
heat engine is impossible. In a real engine, the max-
imum efficiency is attained by making the intake 
temperature as high as possible and the exhaust 
temperature as low as possible. For a Carnot re-
frigerator, Eq. (26-3) may be written 
From this equation, it is clear that the greater the 
temperature ratio T1/T2, the more work is required 
to remove a given quantity of heat from the cold 
reservoir. This is true for real refrigerators as well 
as Carnot refrigerators. 
Example 1, Is it possible to have an engine that 
takes in 25 x 10^ cal of heat from its fuel, rejects 
5 X 10^ cal of heat in the exhaust, and delivers 10^ 
joules of mechanical work? 
SOLUTION 
Ol - Q2 = (25 X 10' cal) - (5 χ  10' cal) = 20 x 10' cal 
= (4.186 joules/cal)(20 x 10' cal) 
= 8.372x 10'joules. 
\V= 10'joules. 
This is impossible, since the engine would deliver 
more energy than it received from the fuel. 
26-5 
E N T R O P Y 
If a substance undergoes a reversible process, 
and takes in an amount of heat dQ at a temperature 
T, its entropy (S) is said to increase by an amount 
dQIT. We therefore define the change in entropy of 
a substance by the relation 
(26-18) 
Let us first consider the change in entropy of a 
substance that takes place when it is carried 
through a Carnot cycle. For the adiabatic portions 
of the cycle, no heat enters or leaves the substance, 
so here the change in entropy is zero. During the 
isothermal expansion part of the cycle, the sub-
stance takes in heat Q i at a temperature T,, and its 
entropy increases by an amount < ? i / T i . During the 
isothermal compression part of the cycle, the sub-
stance gives out an amount of heat Q2 at a tempera-
ture T2, and its entropy decreases by an amount 
Q2IT2. The total change in entropy during the whole 
VT. 
T2)' 
From Eq. (26-13), we have 
T. 
Τ 2· 
Therefore, the total change in entropy is zero. 
Any reversible cycle may be replaced by a set of 
isothermal and adiabatic processes or divided into a 
series of Carnot cycles as illustrated in Figure 26-6. 
The total change in entropy for all the Carnot 
cycles will be 
T e = 0, 
(26-19) 
where AQdTc 
is the entropy change for each Carnot 
cycle. By enclosing a large number of Carnot cycles 
in the reversible cycle, we may approximate the 
reversible cycle more closely. In the limit Eq. (26-19) 
transforms to an integral around the reversible cycle 
and 
(26-20) 

234 
The Second Law of Thermodynamics and Entropy 
Figure 26-6 A reversible cycle divided into a series of 
Carnot cycles. 
It follows that the change in entropy is the same 
for any reversible path leading from one state to 
another, otherwise Eq. (26-20) would not be true. 
We may then write that the entropy change of a 
system taken from state 1 to state 2 along a rever-
sible path is given by 
s2-si=¡; 
dQ 
Τ ' 
(26-21) 
So far we have discussed entropy in terms of a 
reversible process, which is only a hypothetical 
process and can never be realized in practice no 
matter how closely actual changes may approxi-
mate it. Some books on thermodynamics discuss 
entropy almost solely on reversible processes and 
only discuss irreversible changes in a meager way. 
This seems to us a serious error as it gives the 
reader a confused notion of the subject. 
It can be shown that any process in which the 
total entropy change of the system and its sur-
roundings is negative violates the second law of 
thermodynamics. Therefore, the entropy change 
must be equal to or greater than zero, and the sec-
ond law of thermodynamics may be written 
mathematically as 
S / - & > 0 
(26-22) 
when Si is the initial entropy of the universe and S/ 
is its final entropy. If only reversible cyclic proces-
ses occur, the entropy change is zero, otherwise it 
is positive. Therefore, since all natural processes 
are hreversible, and hence increase the entropy of 
the universe, we may make the statement that the 
entropy of the universe is continually increasing. 
The term entropy is used as a measure of energy 
unavailable for work. If we have two separate res-
ervoirs of water, one hot and the other cold, a heat 
engine could be devised to perform mechanical 
work by removing heat from the hot reservoir and 
giving heat to the cold reservoir. However, if the 
hot and cold water are mixed, and come to a uni-
form temperature, this energy is no longer avail-
able, since there is no temperature difference. In 
this example, the energy of the system has re-
mained constant, but the entropy has increased. 
The warm water will never unmix itself, and we 
have an irreversible process. Since irreversible pro-
cesses are continually going on in nature, energy is 
continually becoming unavailable for work. This is 
known as the principle of the degradation of energy. 
Entropy is also a measure of the disorder of a 
system. If heat is added to a system, there is a 
greater random motion of the molecules, and the 
system becomes more disordered. Adding a given 
quantity of heat to a system at a low temperature 
increases the disorder more than adding the same 
quantity of heat to the same system at a high tem-
perature. It is reasonable, therefore, to define the 
change in disorder of the system when dQ units of 
heat are added to it at a temperature Τ  by dQIT 
which is equal to the change in entropy of the 
system. 
We showed that the change in entropy of a sys-
tem for any reversible path is given by Eq. (26-21). 
We shall conclude the discussion on entropy by 
saying that the change in entropy is the same for an 
irreversible path as for a reversible one between the 
same two states, since change in entropy is a func-
tion of the initial and final states only and not on the 
path followed between the two states. 
We can, therefore, calculate the change in en-
tropy for an irreversible process by replacing it 
with a reversible one and using Eq. (26-21). The 
change in entropy can also be calculated if S2 and 
SI are known for the two states as a function of, 
say, V and Τ  or from tables. Here the change in 
entropy will be just 
S2-S1. 
Example 2. What is the change in entropy of 
1 mole of an ideal gas that expands to twice its 
original volume into an evacuated vessel? 
S O L U T I O N 
This is an irreversible process, and must be re-
placed by a reversible one. Since there is no tem-
perature change in the free expansion of an ideal 

Problems 
235 
gas, it can be replaced by a reversible isothermal 
_ 
^ 
^ 
—dV 
expansion if we imagine that the gas is in a cylinder 
^ ~ 
V 
* 
with a piston which is allowed to be pushed out 
„ 
„ 
1 
1 
J 11 
J 
· 
XL 
. ' 
From Eq. (26-21), 
slowly by gradually reducmg the exterior pressure 
on it. The internal energy does not change, but heat 
enters the gas and work is done by it. From the 
first 
^ _ 5 = 
É Q= ^ 
dV_ 
2 
law of thermodynamics, 
' 
' 
J v Τ  
Jv 
V 
P R O B L E M S 
1. A Carnot engine, whose high temperature reservoir is at 400°K, takes in 100 cal of heat at this 
temperature in each cycle and gives up 80 cal of heat to the low temperature reservoir. 
(a) What is the temperature of the low temperature reservoir? 
(b) What is the efficiency of the engine? 
2. A Carnot engine absorbs 400 cal of heat from a source and rejects 300 cal to a reservoir. 
(a) Find the eflftciency. 
(b) Find the ratio of the temperature of the source to the temperature of the reservoir. 
(c) Reverse the Carnot engine given above so that it becomes a refrigerator. Find the coefficient of 
performance. 
3. A Carnot engine during each cycle accepts 14 cal from the hot reservoir and rejects 6 cal to the cold 
reservoir. 
(a) How much mechanical work is done per cycle? 
(b) What is the thermal efficiency? 
(c) What is the ratio of the Kelvin temperature of the hot reservoir to the Kelvin temperature of the 
cold reservoir? 
4. The motor in a refrigerator has a power output of 200 watts. If the freezing compartment is at 270°K and 
the outside air is at 300°K, assuming ideal efficiency, what is the maximum amount of heat that can be 
extracted from the freezing compartment in 10 minutes? 
5. The entropy of saturated water at 100°C is 0.31cal/gm°C and that of saturated steam at the same 
temperature is 1.76cal/gm°C. What is the heat of vaporization at this temperature? 
6. When 1 kg of water at 0°C is mixed with 0.5 kg of water at 50°C, what is the change in entropy? 
7. A refrigerator has a coefficient of performance equal to one-quarter that of a Carnot refrigerator. The 
refrigerator is operated between two reservoirs at temperatures of 200°K and 300°K. It absorbs 400 
joules from the low temperature reservoir. How much heat is rejected to the high temperature reser-
voir? 
8. What is the thermal efficiency of an engine which operates by taking η  moles of an ideal gas through the 
following cycle? 
(a) Start at po, Vo, To; 
(b) change to 3po, Vo; 
(c) change to 3po, 3Vo; 
(d) change to po, 3Vo; 
(e) change to po, Vo. 
9. A Carnot refrigerator takes heat from water at 0°C and rejects 1000 kcal of heat to the room at 27°C. 
(a) How much water is converted to ice? 
(b) How much energy must be supplied to the refrigerator? 
10. How much work must be done by a refrigerator to remove 1 kcal of heat from a reservoir at - 73°C and 
reject it to a reservoir at 27'*C if the refrigerator has a coefficient of performance equal to 80% of a 
Carnot refrigerator? 

236 
The Second Law of Thermodynamics and Entropy 
11. An engine of 20% thermal efficiency is used to drive a refrigerator having a coefficient of performance of 
4. What is the ratio of the heat input to the engine to the heat removed from the cold reservoir by the 
refrigerator? 
12. Calculate the change in entropy when a 2 kg block of aluminum at 727°C is brought in contact with 1 kg 
of water at 27°C. 
13. Calculate the change in entropy of 2 kg of water when it is raised from the freezing point to the boiling 
point temperature. 
14. Show that, for any cycle, 
15. Show, by applying the result of Problem 14, that the entropy of an irreversible process is 
dS>f, 
and in the limiting case of the process in an isolated system, 
<iS>0. 

27 
Kinetic Tlieory 
of Matter 
27-1 
Matter 
237 
27-2 
Kinetic Theory of Gases 
237 
27-3 
Kinetic Theory Interpretation of 
Temperature and Specific Heat 
239 
27-4 
Avogadro's Law and Dalton's Law 
of Partial Pressures 
240 
27-5 
Real Gases in Contrast to Ideal 
Gases 
240 
27-6 
Liquids 
241 
21-1 
Solids 
241 
In this chapter we will discuss pressure, tempera-
ture, and specific heat in terms of molecular mo-
tion. 
27-1 
MATTER 
The material of which all bodies are made is called 
matter. There are three different states of matter— 
soHd, liquid, and gas. So far we have discussed the 
macroscopic properties of matter. However, all 
matter is made up of very small particles called 
molecules. There are spaces between the molecules, 
which are called intermolecular spaces; all macro-
scopic properties of matter are really based on the 
properties of these molecules. Forces called inter-
molecular forces exist between the molecules. Al-
though the intermolecular forces are very strong, the 
distance through which these forces act is very 
small. The molecules of matter are in more or less 
rapid motion, and the temperature of a given sub-
stance is determined by the average velocity of its 
molecules. Differences exist in the kind of motion 
possessed by the molecules in three states. In gases, 
the intermolecular forces are very small and there is 
complete random motion. In solids, where the 
intermolecular forces are relatively large, the 
molecules oscillate about certain fixed points, being 
held by the attractive forces of their neighbors. In 
liquids, the molecules have no fixed positions and 
slip about with ease over one another. 
27-2 
KINETIC T H E O R Y O F G A S E S 
We have previously discussed the macroscopic 
behavior of gases and the equation of state (the 
general gas law) for an ideal gas. We are now in 
a position to develop a microscopic theory of 
gases and to derive an equation of state for an 
ideal gas from a molecular model. We shall first 
make a few assumptions: 
1. All molecules are in motion in all directions. 
2. Molecules move in straight lines between 
collisions. 
3. All collisions are perfectly elastic. 
4. Diameters of molecules are neglected in 
comparison to the distance traveled between 
collisions. 
5. Intermolecular forces are very small. 
6. Time spent during collision is much less 
than time spent between collisions. 
Now let us consider η  ideal gas molecules. 
237 

238 
Kinetic Ttieory of Matter 
m 
v,^ j 
A 
^^u: 
y 
y 
Β  
X 
Figure 27-1 Molecule with a velocity v, in a cubical 
container. 
each of mass m, contained in a cubical box of unit 
edge length as illustrated in Figure 27-1. Con-
sider a molecule moving from face A to face Β 
with a velocity Vu Let the component of the vel-
ocity in the X-direction be Vxi. The change in 
momentum on impact with face Β will equal 
2mt;xi. Since the time for the molecule to travel 
across the cube and back will be 
llvxu 
the 
number of impacts per second with face Β will 
equal VxJl. From the above information, we now 
know that the rate of change of momentum pro-
duced at face Β in the χ -direction, which equals 
the average force F^i, is 
(27-1) 
Since pressure is force divided by area and the 
area of each face of the cube is unity, the aver-
age pressure due to this molecule is 
Pi = mvli. 
(27-2) 
Let us now consider other molecules with x-
components of velocities Vx2, Vx3, etc. Since pres-
sures 
add 
the 
total 
pressure, 
p, 
with 
all 
molecules of the same mass, is given by 
ρ = P i + P 2 + - = m(rL + Vx2 + - ) . 
(27-3) 
Let η  be the total number of molecules in the 
unit cube. The average value of the square of 
the X-component of velocities of all the molecules is 
Vx =• 
and Eq. (27-3) can be written 
ρ = nmv^. 
(27-4) 
(27-5) 
Similar expressions can be obtained for the y-
and ζ -directions, namely. 
Since 
ρ = nmv' 
and 
ρ = nmij?. 
v' = 
Vx'+v'+v' 
(27-6) 
and because the molecules do not move in any 
preferred direction. 
Ϊ 7 = ΐ 7 = ϊ? . 
Hence, Eq. (27-5) becomes 
1 
ρ =^nmi; . 
If the gas is contained in a volume V , 
1 Ν  
—  
or 
(27-7) 
(27-8) 
(27-9) 
(27-10) 
pV 
= jNmv\ 
(27-11) 
where Ν  is the total number of molecules in the 
volume V . 
Since density equals mass per unit volume, we 
can write Eq. (27-10) as 
1 
- I 
ρ =3Pt)' 
or 
(27-12) 
Therefore, the value of ν  can be calculated if we 
know the pressure and density of the gas at a given 
temperature. We call v' the mean-square 
velocity 
of the gas molecules and 
= Vrms—the root-
mean-square 
velocity. Equation (27-12) states that 
the root-mean-square 
velocity of the molecules in 
a gas varies inversely with the square root of the 
density of the gas at constant pressure and temper-
ature. 
Example 1. Calculate the root-mean-square vel-
ocity of the molecules in air at 0°C and 1 atmos-
phere pressure. 
SOLUTION 
For air at 0°C and 1 atmosphere pressure, 
p = 1.296kg/m' 

Kinetic Theory Interpretation of Temperature and Specific Heat 
239 
p = latm = 1.013 X 10'nt/m' 
/3xl.013xl0^ntM? 
ρ 
V 
1.2%kg/m' 
= 485 m/sec. 
27-3 
KINETIC T H E O R Y I N T E R P R E T A T I O N 
OF T E M P E R A T U R E A N D S P E C I F I C 
HEAT 
For one mole of an ideal gas, Eq. (27-11) may be 
written 
(27-13) 
where No is the number of molecules in a mole of 
the gas (Avogadro's number) and Μ  is the molecu-
lar mass of the gas. The quantity \Mp is the total 
translational kmetic energy of the gas molecules; 
for an ideal gas, it equals the internal energy (U) of 
a mole of the gas. 
The equation of state for one mole of an ideal gas 
IS 
pV = RT. 
Combining Eqs. (27-13) and (27-14), 
(27-14) 
(27-15) 
The important deduction from this equation is that 
the internal energy of an ideal gas varies directly 
with the absolute temperature, and depends only on 
the temperature. It defines temperature on the basis 
of the kinetic theory, and gives us an understanding 
of the temperature of a gas in terms of the motion 
of its molecules. 
Since R and N o are both universal constants, 
their ratio is also a universal constant, called the 
Boltzmann constant k. 
J. 
R 
8-31 joules/mole-°K 
No 
6.02 X 10" molecules/mole 
= 1.38X lO^'joules/molecule-^K. 
Equation (27-15) may now be written as 
-mv 
=2'^T, 
(27-16) 
which gives the average kinetic energy 
per 
molecule of a gas in terms of its temperature. 
Since the translational kinetic energy of a 
molecule depends on the values of the three rectan-
gular components of its velocity, we say that the 
molecule has three translational degrees of free-
dom. On the average, it is to be expected that the 
energy associated with each degree of freedom is 
the same. Since the total average kinetic energy per 
molecule is IkT, each molecule possesses an aver-
age kinetic energy of feT/2 per degree of freedom. 
This is a special example of the Principle of 
equipartition of energy. 
For a monatomic gas, the molecular energy is all 
kinetic energy, and its internal energy per mole can 
be represented by Eq. (27-15)—namely. 
U = JRT. 
(27-17) 
If the temperature of the gas is increased by dT, the 
kinetic energy increases by 
dU = ^RdT. 
Since the specific heat at constant volume ( C ) is 
equal to dl7/dT, 
a=^R. 
(27-18) 
Also, since Cp=Cv-^R, 
the ratio C p / C „ = y is 
equal to 
jR+R 
5 
. 
γ = - ^ 
= 3 = 1.67. 
The above results for C „ and y agree well with ex-
perimental results. 
Boltzmann pointed out that the heat required to 
raise the temperature of a gas depends on the 
number of degrees of freedom—that is, the number 
of independent coordinates necessary to describe 
its motion. For a diatomic gas, there are 5 degrees 
of freedom (3 translational and 2 rotational), and 
for a polyatomic gas—6 degrees of 
freedom 
(3 translational and 3 rotational). Therefore, for a 
diatomic gas. 
and 
jR+R 
7 
. 
γ =
-
^ 
= 3 = 1.40. 
These values are in good agreement with experi-
ment. 
Example 2. What is the average kinetic energy 
of a molecule of a gas at a temperature of 27'*C? 

240 
Kinetic Theory of flatter 
If the gas is CO2, what are the values of C , C p , and 
SOLUTION 
^mv' = | K T = (|)(1.38 X 1 0 " joules/°K)(300°K) 
= 6.20x10"'· joules. 
Carbon dioxide (CO2) is a triatomic gas with 3 
atoms per molecule and 6 degrees of freedom. 
a=^R=3R=6 
cal/mole-°C, 
Cp = Cv^R=4R=S 
cal/mole-°C, 
^Cp^ 
8 cal/mole-°C ^ 
a 
6cal/mole-T 
^•'^^· 
27-4 
A V O G A D R O ' S L A W A N D D A L T O N ' S 
LAW O F PARTIAL 
P R E S S U R E S 
Avogadro's law is stated as follows: 
Equal volumes of different or the same gases 
under the same conditions of pressure and 
temperature contain the same number of 
molecules. 
The law is obtained from the kinetic theory ap-
plied to two different gases at the same pressure 
and temperature. Let us apply Eqs. (27-9) and 
(27-16) to a unit volume of each of the gases. 
1 
1 
p — ^ W i m i ü i , 
p=^n2m2Ü2 
and 
Therefore, 
2^ΐϋι = 2 ^2^2 . 
Ml — n2. 
(27-19) 
where Π γ and n2 are the number of molecules in a 
unit volume of each of the gases. 
Dalton's law of partial pressures is stated as fol-
lows: 
For a mixture of gases in a given volume at a 
given temperature, the pressure exerted by 
each gas is the same as if it alone occupied 
the volume; and the total pressure is equal to 
the sum of the partial pressures exerted by 
each gas. 
This law can also be obtained by applying the ki-
netic theory to two different gases at the same 
temperature. Apply Eq. (27-9) to each of the gases 
confined separately in a unit volume. 
p, = ^n,mii?i, 
p2 = ^n2m2V2. 
(27-20) 
The pressure exerted by a mixture of the two gases 
in a unit volume is 
ρ =^nimit;i +^n2m2Ü2. 
(27-21) 
Therefore, 
ρ = P i + P2, 
which is Dalton's law. 
Example 3. A volume of 100 cm' of hydrogen is 
collected over water at a temperature of 20*^ and a 
pressure of 75 cm of Hg. Find the volume of dry 
hydrogen at a temperature of 0®C and a pressure of 
76 cm of Hg. 
SOLUTION 
Since the hydrogen is collected over water, it is 
saturated with water vapor. The pressure of satu-
rated water vapor at 20°C, from Table 5-1, is 
1.75 cm of Hg. The partial pressure of hydrogen 
(PH), from Dalton's law, is 
PH = 75 - 1.75 = 73.25 cmof Hg. 
Applying the ideal gas law in the form of Eq. (3-17), 
(76 cm Hg)( V) _ (73.25 cm Hg)(100 cm') 
(273°K) 
(293°K) 
V=90cm'. 
27-5 
R E A L G A S E S IN C O N T R A S T T O 
IDEAL G A S E S 
The assumptions made in the kinetic theory deri-
vations of Section 27-2 are never quite realized in 
the case of real gases. Experiment shows that the 
equation of state for the ideal gas only holds for a 
real gas at low pressures. For example. Figure 27-2 
is a plot of pV/nRT 
versus ρ in atm for hydrogen 
and oxygen. The horizontal line indicates the be-
havior expected for an ideal gas. The differences 
exhibited are not difficult to explain. As the pres-
sure is increased, the volume is decreased, and the 
volume of the molecules becomes a more signifi-
cant part of the total volume. The product ρ V is no 
longer constant, but increases with pressure for 
both gases. Also, the cohesive forces play a domi-
nant role when the gas is at an intermediate pressure 

Solids 
241 
nRT 
2.0 
1.0 
Ideal gas 
1 
1 
Ideal gas 
1 
1 
0 
500 
1000 ρ (atm) 
Figure 27-2 
drogen. 
pV/nRT versus ρ for oxygen and hy-
and the molecules are closer together. These forces 
account for the dip in the O2 curve of Figure 27-2. At 
higher pressures, the repulsion of the molecules 
overcomes the cohesive forces. The lack of a dip in 
the hydrogen curve can be attributed to the small 
size of the hydrogen molecule. 
At sufficiently high pressure and low tempera-
ture, the cohesive forces are able to prevent the 
elastic collisions of the molecules to a large extent, 
and we no longer have the kinetic properties of a 
gas. At this point, the gas condenses into a liquid. 
27-6 
LIQUIDS 
In terms of kinetic theory, a liquid may be con-
sidered as a continuation of the gas phase into a 
region of small volumes and high molecular attrac-
tions. The molecular attractions must be high 
enough to keep the molecules confined to a definite 
volume. The molecules of a liquid have less free-
dom of motion than the molecules of a gas. This can 
be observed directly by Brownian motion, discov-
ered by Robert Brown in 1827. Using a microscope, 
he observed small pollen grains floating in water, 
and found that they underwent random motion due 
to the bombardment of the water molecules. The 
same type of observations can be performed with 
small particles suspended in a gas, such as smoke 
particles in air. On comparison of the two observa-
tions, it can be concluded that the gas molecules 
have much greater freedom of motion than liquid 
molecules. 
At temperatures near the critical temperature, it 
is possible to regard a liquid as essentially a gas 
whose molecules have a finite volume and attract 
one another. The equation of state here resembles 
van der Waal's equation [Eq. (5-1)]. The theory of 
liquids is in a much less satisfactory state than the 
theories of gases and solids, but important progress 
is being made in our understanding of the structure 
of liquids. 
27-7 
S O L I D S 
The attractive forces between the atoms of solids 
are stronger than in liquids. In fact, the forces be-
tween the atoms or molecules are so strong that 
they are not free to move about as in a gas or liquid. 
They are held together as if by springs, and oscillate 
back and forth continuously. Thus, a solid is elastic 
because of the elastic behavior of the atoms making 
up the solid. 
As noted in Section 5-1, most solids have a 
periodic arrangement of atoms, and are called cry^-
talline solids. Most metals are crystalline solids. We 
may ask why there is a periodic arrangement of the 
atoms in a crystalline solid. The answer is that the 
potential energy of such a system is a minimum if 
the atoms occupy the smallest possible volume. It is 
observed that the most stable configuration of a 
system is the one with a minimum potential energy. 
A system of balls (or atoms) can occupy the small-
est volume only by the formation of a periodic 
structure. A few solids, however, do not have a 
periodic structure, and thus do not have crystalline 
forms. These solids are called amorphous 
solids, 
two common examples of which are glass and 
rubber. 
Since the atomic forces in a solid are very strong, 
the following question may be asked. What is the 
source of the forces that hold the atoms together in 
a solid? The answer is that the source is elec-
tromagnetic since each of the atoms is made up of a 
positively charged nucleus and negatively charged 
electrons. We may distinguish between three types 
of interatomic forces in solids. 
(a) The Metallic-Bonded Solid 
In this type of solid, the force may be simply ex-
pressed as electrostatic forces between the positive 
nucleus and the electron cloud, which not only sur-
rounds the parent nucleus but also the nucleii of 
neighboring atoms. A material in which the atoms 
are held together by this type of bond has the high 

242 
Kinetic Ttieory of Matter 
electrical and thermal conductivity characteristic of 
metals. 
(b) The Covalent-Bonded Solid 
In this type of solid, the neighboring atoms share 
valence electrons. It is a very strong bond. Coval-
ent crystals have poor electrical and thermal con-
ductivity. Diamond and carborundum are examples 
of such solids. 
(c) The Ionic-Bonded Solid 
The ionic bond is characteristic of an exchange 
of electrons between atoms and the formation of 
ions. It is a strong bond, and the solid has ionic con-
ductivity at high temperatures. Salts such as 
sodium chloride and potassium chloride are exam-
ples of ionic-bonded solids or ionic crystals. 
It should be pointed out that there must be strong 
repulsive forces as well as attractive forces in a 
solid, since large forces are requked to compress a 
solid. Figure 27-3 is a graph of interatomic forces 
versus atomic separation. At a separation ΓΟ , a sta-
ble state exists in which the resultant force is zero. 
The source of the repulsive force that balances the 
attractive force is not so easily explained. It is 
made up of several effects, the most important of 
which occurs when the outer filled electron shells 
of both atoms begin to overlap one another. When 
two completely filled shells overlap each other, the 
number of electrons in the single group formed by 
these shells is too large for them all to be contained 
in states of low energy, and many are forced to 
occupy higher energy levels. The increase of 
energy of these electrons is enough to give strong 
repulsive forces which oppose the attractive forces 
that tend to bring the ions together. 
Repulsive force 
Resultant force 
Attractive Force 
Figure 27-3 Interatomic forces versus atomic separa-
tion. 
P R O B L E M S 
1. Are the following statements true or false? 
(a) The lighter molecules in the air move faster, on the average, than the heavy ones. 
(b) The pressure of air in a closed container might be doubled by heating the air, so as to make the 
molecules go twice as fast. 
2. Consider 1 mole of an ideal gas at 0°C and 1 atmosphere pressure. Imagine each molecule to be, on the 
average, at the center of a small cube. 
(a) What is the length of an edge of the cube? 
(b) If the diameter of an average molecule is about 3 x 10"* cm, what is the ratio of the length of an edge 
of the cube to the average diameter of the molecule? 
3. Using the kinetic theory of matter, explain how the vapor forms in a closed space above a liquid, and 
how this vapor exerts a pressure on the enclosing walls. Why does this pressure have a certain fixed 
value at a given temperature? 
4. (a) If we wish to double the root-mean-square velocity of a molecule, by what factor do we change the 
temperature? 
(b) If, at a given temperature, a molecule A has a mass double that of a molecule B, what is the ratio of 
their root-mean-square velocities? 
5. (a) What is the translational kinetic energy of the molecules in lOgms of oxygen gas at 27°C? 
(b) What is the root-mean-square velocity of a molecule of hydrogen at 273°C and 1 atmosphere pres-
sure? 

Problems 
243 
6. Calculate the mass, root-mean-square velocity, and kinetic energy of helium contained at 1 atmosphere 
and 20°C in a balloon of volume 10 m^ 
7. How much heat must be added to 3 kg of O2 gas to raise its temperature from 0**C to 25°C at constant 
pressure? What is the increase in internal energy? How much external work is done? How much heat 
would need to be added in raising the temperature by the same amount at constant volume? 
8. Two grams of oxygen and 6 gms of carbon dioxide are put into an evacuated 5 Hter flask. Find the 
pressure inside the flask if the temperature is 27°C. 
9. Show that 
^ 
RIM 
, 
^ 
yRiM 
where y = CplCv. 
10. The density of oxygen is 1.43 x 10"^gm/cm^ at 0°C and 1 atmosphere pressure. At what temperature 
would the root-mean-square velocity of oxygen equal the escape velocity from the Earth, which is 
approximately 11.2 km/sec. 
11. The density of hydrogen is 9 x 10"' gm/cm^ at OT and 1 atmosphere pressure. Calculate the root-mean-
square velocity of hydrogen at 1000°C. If temperatures approaching 1000°C are present in the upper 
atmosphere, and the escape velocity from the surface of the Earth is 11.2 km/sec, can there be much 
hydrogen in the Earth's atmosphere? 
12. The velocity of a gas forced through a small hole in a wall of a container is proportional to the velocity 
of the molecules. Compare the time required to force out equal volumes of hydrogen and oxygen under 
the same pressure. 

28 
Sound 
28-1 
Production and Speed of Sound 
245 
28-2 
Intensity of Sound 
246 
28-3 
Psychological Effects of Sound 
and Origins of Musical Sounds 
247 
28-4 
The Doppler Effect 
248 
All sounds have one thing in common. Each is 
caused by the vibration of a material medium and is 
transmitted by waves. The speed of sound depends 
on the density, elasticity, and the temperature of 
the medium. Because of the above properties of 
sound, we have left a discussion of it until after we 
discussed wave motion, properties of matter, and 
temperature. 
28-1 
P R O D U C T I O N A N D S P E E D O F S O U N D 
Sound is a vibratory motion that produces the 
sensation of hearing. Sound waves are longitudinal 
vibrations of material media. If the frequency of 
the vibrations is greater than that which can be 
detected by the ear, it is called an ultrasonic vibra-
tion, if less—an infrasonic vibration. The audible 
range for a human is from 20 to 20,000 cycles per 
second (cps). 
In order for sound waves to be produced, there 
must be a source that initiates a mechanical distur-
bance and a material medium through which the 
disturbance can be transmitted. This material 
medium requirement is not difficult to demonstrate. 
Suppose we suspend an electric bell by means of a 
fine string in a container which can be evacuated by 
means of a pump. We start the bell ringing, and at 
the same time pump the air from the container. The 
sound will become fainter and fainter as the air is 
exhausted until the sound cannot be heard. 
The speed of sound depends upon the elastic 
properties of the medium. In general, the speed of a 
wave in a material medium is given by 
V
elastic f ( 
inerti 
force factor 
inertia factor 
The quantities used for both the elastic force factor 
and the inertia factor depend on the kind of wave 
motion and the medium. For sound waves in solids, 
the elastic force factor is the bulk modulus plus ί 
the shear modulus of the solid (see Chapter 23), 
while for fluids it is only the bulk modulus since the 
shear modulus is zero for a fluid. In both solids and 
liquids, the inertia factor is the density of the 
medium. For sound waves in a solid rod, the elastic 
force factor is the Young's modulus of the medium, 
which is also discussed in Chapter 23, and the iner-
tia factor is the density of the medium. 
In the case of gases, the adiabatic bulk modulus 
is used for the elastic force factor since they are 
highly compressible and the heat of compression 
cannot escape at ordinary frequencies. Therefore, 
the speed of sound in gases is expressed by the 
relation 
(28-1) 
245 

246 
Sound 
However, 
ß
a d = - 
V ( d v ) a d ' 
and the equation of state of an ideal gas for an 
adiabatic process is 
pV 
=K. 
Taking logarithms of both sides of this equation, 
we get 
In ρ + γ In V = In X, 
and taking the differential of this equation gives 
from which we obtain 
Therefore, 
(dp\ 
^__yp 
U V / a d 
V 
yp 
Ρ  ' 
(28-2) 
Since the density of a gas varies inversely with the 
absolute temperature at constant pressure, the 
speed of sound in a gas varies dkectly with the 
absolute temperature. For an ideal gas, 
P^RT 
ρ 
M ' 
where R is the universal gas constant, Μ  is the 
molecular weight, and Τ  is the absolute tempera-
ture. Therefore, 
(28-3) 
which gives good agreement with the measured 
speed of sound in gases. 
Example 1. Find the speed of sound in oxygen at 
300°K. 
SOLUTION 
For oxygen, γ = 1.40 and Μ  = 0.032 kg/mole. 
1? =8.31 joules/mole deg. 
/(1.40)(8.31 joules/mole deg)(300°K) 
^ 
V 
(0.032 kg/mole) 
= 330 m/sec. 
The speed of sound also varies directly with the 
square root of the pressure, but since it also varies 
inversely with the square root of the density and 
since an increase in pressure increases the density, 
the effect of pressure depends on the state of the 
medium. The speed of sound in a gas is independent 
of changes in pressure within wide limits since the 
ratio of pressure to density is a constant. However, 
the effect of pressure is much greater in liquids and 
solids. In the case of liquids, it is an even more im-
portant factor than temperature. 
Since sound is a wave motion, its wave length is 
related to the speed v, and is given by the relation 
(28-4) 
where / is the frequency of the sound. Table 28-1 
gives the speed of sound in air at various tempera-
tures, and Table 28-2 gives the speed of sound in 
thin solid rods and in liquids. 
Table 28-1 Measured Speed of Sound in Air. 
Temperature CQ 
Speed (m/sec) 
0 
331.4 
20 
344.0 
100 
366.0 
500 
553.0 
1000 
700.0 
Table 28-2 Measured Speed of Sound in Solid Rods 
and Liquids. 
Speed 
Speed 
Solid rods 
(m/sec) 
Liquids 
(m/sec) 
Aluminum 
5100 
Alcohol 
1143 
Copper 
3560 
Ether 
1032 
Iron 
1322 
Turpentine 
1326 
Nickel 
4973 
Water 
1461 
Glass 
5550 
Sea water 
1500 
28-2 
INTENSITY O F S O U N D 
A sound wave is a series of compressions and 
rarefactions. We therefore have a pressure am-
plitude as well as a displacement amplitude. The 
pressure amplitude (P) is the maximum amount by 
which pressure differs from atmospheric pressure. 
The intensity (J) of a sound wave is the energy 
transported per unit area per unit time perpendicu-
lar to the direction of propagation of the sound 
wave. A loud sound is one of high intensity, even 
though no simple mathematical relation exists be-
tween them. This is so because the loudness of a 

Psychological Effects of Sound and Origins of l^usical Sounds 
247 
sound depends upon the particular sensitivity of the 
ear, while the intensity is a physical measurement. 
It is frequently useful to compare sound inten-
sities. Two sounds having an energy ratio of 1 to 10 
are said to differ by 1 bel. Since the ear is sensitive 
to a very great range of intensities, a logarithmic 
scale is used which is implicit in the definition of the 
bel. It has been found that the decibel (db) is a more 
convenient unit, and the intensity level (β ) meas-
ured in decibels is defined by the equation 
SOLUTION 
β  = 10 log f, 
io 
(28-5) 
where Jo is the intensity corresponding to the 
threshold of hearing or, in other words, the faintest 
sound that can be heard by the average human. The 
maximum that the human ear can tolerate is about 
120 decibels. Referring to Eq. (28-5), this means 
that the ear has a hearing range from the lowest 
audible intensity to one which is 10'^ times greater. 
Actually, this hearing range depends on the sound 
frequency. Figure 28-1 is an audibility diagram for a 
person with good hearing. Equation (28-5) is ordi-
narily used to compare the intensities of two 
sounds. From Eq. (28-5), it is readily seen that a 
sound 10 times as intense as another is 10 db higher, 
one 100 times as intense is 20 db higher, and one 
1000 times as intense is 30 db higher. 
Example 2. A sonar operator on board a ship 
pings on a submarine. The return signal is one-half 
the intensity of the ping. What is the attenuation of 
the ping by the water and the hull of the submarine 
in decibels? 
CO 
1,111 
1 
1 1,11—1 
|120 
Φ  100 
"Ö  
_r 80 
ill 
gi 60 
t ^° 
w 20 
^ 111I 
> 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
ill 
gi 60 
t ^° 
w 20 
- 1 Γ  
lie 
IWIU UI 1 CK 
711 «y 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
ill 
gi 60 
t ^° 
w 20 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
ill 
gi 60 
t ^° 
w 20 
t 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
ill 
gi 60 
t ^° 
w 20 
% 
ι 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
ill 
gi 60 
t ^° 
w 20 
\ \ 
» 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
ill 
gi 60 
t ^° 
w 20 
% 
• 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
ill 
gi 60 
t ^° 
w 20 
\ 
f 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
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t ^° 
w 20 
1 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
"Ö  
_r 80 
ill 
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t ^° 
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V I 
1 1 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
|120 
Φ  100 
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Th 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
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|120 
Φ  100 
"Ö  
_r 80 
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hearing 
] — 
1 1 π  
1 
10-* o 
10- S 
10-^° 
10- 05 
Ζ 
10-^' ^ 
20 
100 
1000 10,000 100,000 
FREQUENCY, cycles/sec 
β  = 10 log (J watts/m') 
i^=101og0.5 = - 3 d b . 
Figure 28-1 Audibility diagram for a person with good 
hearing. 
(Io watts/mO 
Thus, the return signal was 3 db below the ping in 
intensity. Therefore, the attenuation of the ping by 
the water and the hull was 3 db. 
28-3 
P S Y C H O L O G I C A L E F F E C T S OF 
S O U N D A N D O R I G I N S OF 
M U S I C A L S O U N D S 
A sound wave causes the eardrum (a fibrous 
membrane) to vibrate in synchronization with the 
sound vibrations. These vibrations of the eardrum 
are transmitted to the inner ear. The inner ear is 
filled with a fluid in which are located nerve endings 
that transmit the sound vibrations to the brain. 
The quality of a sound refers to the psychological 
effect it produces for a listener. A musical note usu-
ally consists of a number of vibrations of different 
frequencies, consisting of the lowest frequency, 
called the fundamentaU and other vibrations having 
frequencies of two, three, or more times that of the 
fundamental, called harmonics. 
By contrast, a 
noise is a mixture of many unrelated vibrations. It 
is the number of harmonics present and their re-
spective intensities that determine the quality of a 
sound. Two notes with the same fundamental fre-
quency and intensity will differ in quality when 
sounded on a guitar and a violin because of the 
spectrum of harmonics produced. The pitch of a 
musical note refers to the dominant frequency. 
When several musical notes produce a pleasing sen-
sation to the listener, they are said to be in har-
mony. If the effect produced is disturbing, the notes 
are said to be discordant. The explanation is 
roughly as follows: the listener is aware of not only 
the dominant frequencies of the notes but also the 
difference frequencies or beat tones. Low fre-
quency, unrelated beat tones produce a discordant 
sensation. Higher frequency beat tones, in cases 
where the same beat frequency results from more 
than one pair of dominant frequencies in the set of 
musical notes, result in a sound that has a pleasing 
fullness or richness. Musical instruments, whether 
classified as strings (piano, violin), winds (clarinet, 
flute), or percussion (xylophone, glockenspiel) are 
all designed to minimize discordant effects. 
The means of initiating vibrations is different for 
the various classes of instruments. This is in part 

248 
Sound 
necessary because the medium in which the vibra-
tions originate differs. Thus, transverse standing 
waves are excited in stretched strings in string 
instruments. For wind instruments, the mode of 
vibration is standing longitudinal waves in a colunm 
of air. Percussion instruments involve transverse 
standing waves in solid plates or stretched mem-
branes. In Section 10-3, we discussed standmg 
transverse waves of a stretched string. Equations 
(10-22) and (10-23) provide the relations between 
the wavelength, frequency, and length of the string 
for all harmonics, including the fundamental. 
Another reason for the various methods of in-
itiating vibrations is to provide musical notes of 
higher quality. For example, a violinist enhances 
the quality of notes by using his fingers to create 
points of zero amplitude along the vibrating string. 
This prevents the existence of any harmonics that 
would have non-zero amplitudes at those points. 
The same result is accomplished in a ñute by cover-
ing various holes along the length of the instrument. 
The reader should not infer that the physics of 
musical sounds has been fully treated in this sec-
tion. The discussion here is in fact highly idealized, 
and no consideration has been given to such ques-
tions as musical scales and the human voice. How-
ever, the ideas presented do represent a satisfac-
tory basis for further study.t 
28-4 
THE D O P P L E R EFFECT 
When a hstener and/or source of sound are mov-
ing relative to each other, the frequency of the 
sound as heard by the listener is not the same as the 
true frequency of the source. We shall illustrate this 
by discussing two specific cases. 
(a) Source Stationary and Listener IMoving 
Suppose a stationary source emits a note of fre-
quency /, and the listener approaches the source 
with a velocity Vl, where vl is less than the velocity 
of the sound v. This is illustrated in Figure 28-2, 
where we take the initial distance between the lis-
tener L and the source S equal to the distance the 
sound travels in 1 second or, in other words, numer-
ically equal to the velocity of the sound. 
tThe interested reader is referred to the text. Horns, 
Strings and Harmony, by Arthur H. Benade, Doubleday 
& Co., 1960. 
Figure 28-2 Listener moving toward source. 
If the listener was stationary at L, / waves 
would hit his ear per second. However, during 1 
second he travels a distance ÜL to L'. He therefore 
hears an extra vibration for each wavelength in the 
path length LL'. The number of extra vibrations 
per second he hears is equal to LL'lk 
or UL/A. 
Therefore, the frequency he hears (/') is given by 
or 
(28-6) 
The wavelength of the sound is unchanged, but the 
listener goes to meet the oncoming waves and, con-
sequently, more waves hit his ear per second than 
the source emits in 1 second. 
If the listener had been going away from the 
source, he would have heard ÜL/A fewer vibrations 
per second. Therefore, the frequency he would 
have heard is given by 
^ 
^ 
k 
^ 
V 
or 
λ 
^ 
(b) Listener Stationary and Source iVIoving 
Suppose the source S is producing a note of fre-
quency /, and is approaching the listener L with a 
velocity vs < v. As before, we shall let S be a 
distance ν  from L, as illustrated in Figure 28-3 
where ν  is equivalent to the distance traveled in 1 
second by the sound. The first of the / vibrations is 
given out by the source when it is at S, and the last 
of the / vibrations—when it has advanced to S', a 
distance vs from S. Since vs < v, there will be / 
vibrations between S' and L. Hence, the new 
wavelength λ' in front of the source will be less 

Problems 
249 
Figure 28-3 Source moving toward listener. 
than λ for the source at rest. The new wavelength is 
S'L 
But 
where /' is the frequency the listener hears. There-
fore, 
, _ V _ S'L _ V - Vs 
or 
y - f . - 
¡ 
(28-8) 
If the source had been going away from the listener, 
by the same argument as presented above, the fre-
quency the listener would have heard is given by 
(28-9) 
With the source and listener both in motion, we 
can combine Eqs. (28-6), (28-7), (28-8), and (28-9), 
and obtain 
(28-10) 
The signs adopted are such as to increase the ratio 
/'// when the source and listener are approaching 
one another, and to decrease the ratio /'// when 
they are receding from one another. For example, 
to a listener, the frequency of a locomotive whistle 
will appear to be increased when there is relative 
motion toward the listener, and decreased when 
there is relative motion away from the listener. If 
the medium is moving in the same direction as the 
sound wave, its velocity is added to the velocity of 
the sound, and subtracted if moving in the opposite 
direction. This alteration of frequency due to the 
motion of the listener and/or source is known as the 
Doppler effect. 
We have derived the above equations for the 
case where the source and listener are moving 
along a line joining them. In general, it can be 
shown that the equations hold when Vs and VL are 
the components of the velocities along the line join-
ing the source and the listener. This is also true for 
the velocity of the medium. 
When the velocity of the source is greater than 
the velocity of the sound wave, no regular wave 
train results. Instead, a bow and stem wave is set 
up similar to that produced by a boat moving over 
water. Another example is the shock wave as-
sociated with a supersonic boom. 
Example 3. A bat flies straight toward a wall at 
30 m/sec on a day when the speed of sound is 330 
m/sec, while emitting a note of 15,000 cps. What 
frequency does it hear? 
SOLUTION 
The bat is both the source of the sound and the 
listener. 
= 18,000 cps. 
330+ 30) m/sec 
(330-30) m/sec J 
Example 4. A locomotive goes past a station at 
20 m/sec, blowing its whistle at a frequency of 1000 
cps on a day when the velocity of sound is 330 
m/sec. What change in frequency is heard by a boy 
standing on the station platform? 
SOLUTION 
The frequency heard by the boy as the train ap-
proaches is 
The frequency heard by the boy as the train recedes 
Change in frequency = (1065 - 943) cps 
= 122 cps. 
P R O B L E M S 
1. The speed of sound in steel is 5000 m/sec and the density of steel is 7.8 x 10^ kg/m\ What is the elastic 
force factor of steel? 

250 
Sound 
2. What is the adiabatic bulk modulus of a gas if the speed of sound in the gas is 300 m/sec and the density 
is 2 kg/m'? 
3. A sound of frequency 1000 cycles/sec is radiated uniformly in all directions with an intensity of 16 
watts/m' at a radius of 5 m on a day when the speed of sound is 300 m/sec. 
(a) What is the total acoustic power radiated by the source? 
(b) What is the apparent frequency of the sound to an observer approaching the source at a rate of 20 
m/sec? 
4. A cowardly man, running away from a dentist at a speed of 10 m/sec, emits a shriek of terror having a 
frequency of 5000 cycles/sec. What is the frequency heard by the dentist who is pursuing the man at 4 
m/sec, if the temperature is 20°C? 
5. In Problem 3, if the intensity of the sound 50 m away is 16 x 10"' watts/m', what is the intensity level in 
decibels at a radius of 5 m compared to the intensity at a radius of 50 m? 
6. A locomotive, traveling with a speed of 25 m/sec, sounds its whistle as it approaches an observer at a 
grade crossing when the temperature is 20''C. The frequency of the whistle is 250 cycles/sec. What is the 
wavelength of the sound waves in the region between the locomotive and the observer? 
7. If the sound source in Problem 3 is mounted on a vehicle and is carried at a uniform speed of 50 m/sec 
past a stationary observer, what is the difference between the apparent frequency as it approaches and 
the apparent frequency as it goes away? 
8. An automobile moving at 30 m/sec is approaching a factory whistle having a frequency of 500 
cycles/sec. If the speed of sound in ah* is 340 m/sec, what is the apparent frequency of the whistle as 
heard by the driver? 
9. A ship, dead in the water, emits a sound pulse from its sonar gear, and 2 seconds later receives an echo 
identifiable as a submarine. A short while later, the hydrophone picks up the sound of a torpedo being 
fired, and the alarm is immediately given. Assuming the torpedo travels with a constant speed of 50 
m/sec, how much time does the crew have to abandon ship from the time the alarm was sounded to the 
time the torpedo strikes? Take the speed of sound in sea water as 1500 m/sec and neglect the motion of 
the submarine. 
10. A man runs at 5 m/sec on a calm day when the temperature is 20°C between two fire stations. The fire 
• 
whistles sound simultaneously in the two stations at a frequency of 600 cycles/sec. Find the apparent 
frequency of each whistle. 
11. What sound intensity is 3 db louder than a sound of 0.10 watts/m'? 
12. A locomotive travels 30 m/sec in still air, blowing its whistle. The frequency of the whistle is 600 
cycles/sec and the speed of sound in the air is 340 m/sec. 
(a) What frequency would be heard by a passenger on a second train traveling at 20 m/sec away from 
the first? 
(b) What frequency would be heard by the passenger if a wind of velocity 10 m/sec is blowing in the 
direction that the locomotives are traveling? ' 
13. A train passes by a station, blowing a whistle, on a day when the speed of sound is 330 m/sec. A 
stationary observer on the platform notices that the frequency of the note he hears changes in the ratio 
9/8 as the engine passes him. Find the speed of the train. 
14. How rapidly must an automobile be moving away from an observer so that the frequency of its horn will 
drop from 1100 to 900 cps on a day when the temperature is 20°C? 

29 
Charge, Field, and 
Potential 
29-1 
Introduction 
251 
29-2 
Electrical Forces 
252 
29-3 
Units of Charge 
252 
29-4 
Fields 
253 
29-5 
Electric Potential 
254 
29-6 
Calculation of Ε from V 
255 
29-1 
INTRODUCTION 
The earliest observations of electrical phe-
nomena were made by the Greeks, but no careful 
experiments were carried out until the seventeenth 
century. The original observations indicated that 
certain substances, when rubbed with fur or cloth, 
attracted small objects in the neighborhood. Such 
substances are said to be electrically charged. 
Following the observations of the force exerted 
by a charged body on all uncharged bodies, the next 
step was the study of the forces exerted by differ-
ent charged bodies on each other. It was found, for 
example, that if two pieces of glass were each rub-
bed with a separate piece of silk so that all four 
acquired a charge, the two pieces of glass repelled 
one another, but each piece of glass attracted each 
piece of silk. Similarly, glass that had been rubbed 
with silk attracted hard rubber that had been rub-
bed with fur, but repelled the fur. This led to the 
realization that there must be two kinds of electric 
charge. 
To distinguish between the two kinds of charge, 
that produced on glass by rubbing with silk was 
called positive (+), and that produced on hard rub-
ber by rubbing with fur was called negative 
(-). 
Then, by application of the rule that unlike charges 
attract and like charges repel, it was possible to find 
the sign of the charge on any body. 
It was established early in this century that all 
electrical charges are integer multiples of a funda-
mental or smallest unit of charge. The electron is a 
negatively charged particle, and has a charge mag-
nitude that corresponds to this smallest unit of 
charge. Matter is made up of atoms consisting of a 
central core or nucleus surrounded by lighter par-
ticles called electrons. The nucleus contains two 
kinds of particles: protons and neutrons, which are 
roughly 1840 times as heavy as the electrons. The 
protons are positively charged, while the neutrons 
are uncharged. The number of electrons surround-
ing the nucleus is equal to the number of protons, 
and each electron has a negative charge equal in 
magnitude to the positive charge of the proton, so 
that the atom as a whole is electrically neutral. The 
number of neutrons is roughly equal to the number 
of protons in light elements, but is greater in the 
heaviest elements. A given element may have sev-
eral stable forms of different nuclear mass, that is, 
it may have a different number of neutrons but the 
same number of protons. This means that the ele-
ment can exist in forms with different mass but with 
the same charge. Such forms of an element are 
called isotopes. 
For the purpose of electrostatic theory, sub-
stances can be divided into two classes: conductors 
and insulators. In a conductor, the negative electri-
cal charge (electrons) can flow easily throughout 
251 

252 
Charge, Field, and Potential 
the substance under the influence of an electric 
field, while in an insulator, a large electric field is 
required to cause the electrons to flow. 
29-2 
E L E C T R I C A L 
F O R C E S 
Electrical forces, like gravitational forces, vary 
inversely as the square of the distance, but they are 
about 10'^ times stronger. If you were standing at 
arm's length from someone, and each of you had 
about 1% more electrons than protons, or vice 
versa, the repulsive force would be enough to lift a 
weight equal to that of the Earth. Since you feel no 
force at all at this distance from someone else, this 
must mean there is essentially perfect balance be-
tween the protons and electrons of each person. 
We have said that the electrical force varies in-
versely as the square of the distance. This means 
that if we have two point electrical charges, the 
force of attraction or repulsion of one charge on the 
other is inversely proportional to the square of the 
distance between them. The force is also directly 
proportional to the size or magnitude of these 
charges. 
This law was discovered by Coulomb in 1785. In 
his apparatus, the charges were carried on pith balls 
and the force between them was measured with a 
torsion balance as shown in Figure 29-1. Two small 
spheres A and Β are attached to the ends of a rod. 
which is suspended by a torsion fiber attached to 
the center of mass of the rod. If sphere A is 
charged and a sphere C charged with a like charge 
is placed next to it, they will repel one another and 
the fiber will be twisted through a measurable angle. 
The force of repulsion is determined by the amount 
of twist put in the fiber when the charged spheres 
are at a given distance. Coulomb's law states that: 
The force between two point charges is 
directly proportional to the product of the 
charges and inversely proportional to the 
square of the distance between them. 
ρ = κ ψ , 
where X is a constant of proportionality. 
The vector equation for the force is 
F = X ^ r a , 
(29-1) 
(29-2) 
where fo is a unit vector pointing along the line join-
ing the charges. 
If other charges are in the vicinity of q, each will 
exert a force on q, and the total electrostatic force 
on q will be 
F = XíjÉ Sro.. 
(29-3) 
í = i Γί 
The roi's all have unit magnitude, but they may 
point in different directions. The mutual interaction 
of a pair of charges is unaffected by other charges. 
The algebraic sum of the charges within a closed 
system is constant. 
(I qi = constant j . 
However, Coulomb's law is not perfectly true 
when the charges are moving. The electrical forces 
also depend on the motion of the charges. 
29-3 
UNITS O F C H A R G E 
In the MKS units, the force is expressed in new-
tons, the distance in meters, and q in coulombs. 
The constant Κ  is not defined directly but through 
another constant co by the equation 
1 
Figure 
29-1 Torsion 
balance 
to 
demonstrate 
Coulomb's law. 
47760* 
The new constant, called the electric permittivity 

Fields 
253 
of free space, is defined to be 
10' coul' 
^^"47rc^nt-m'' 
where c is the velocity of light in free space; c = 
3 X 10* m/sec. Now we see that in free space 
and 
eo = 8 . 8 5 X 1 0 - ^ ' - ^ ^ 
nt — m 
cour 
In MKS units. Coulomb's law is 
1 
47Γ € ο  Γ 
= 9 x l O ' ^ f o , 
(29-4) 
where 
F = force (nt) 
q = charge (coul) 
qi = charge 1 (coul) 
r= distance between point charges (m) 
fo = unit vector pointing in direction of line joining 
the charges. 
The total force exerted on a given charge qby η  
other charges is given by Eq. (29-3). In a normal 
laboratory situation, the number of 
electrons 
and/or protons distributed in and around the 
laboratory that can affect a given test charge q is 
usually very large. As a result, the calculation re-
quired in the summation in Eq. (29-3) is prohibi-
tively difficult. This difficulty is easily by-passed, 
however. One simply "maps" the electrical influ-
ences in the laboratory by carrying the charge q 
throughout the volume of the laboratory and 
measuring the magnitude and direction of the total 
force experienced by q at each point. If q is re-
placed by a charge q', the force it would experience 
is simply (q'lq) times the force measured for q, and 
a second measurement is unnecessary. 
Any physical quantity that takes on different val-
ues at different points in space is said to define a 
field. If the values change in magnitude only, the 
field is a scalar field. If the values change in mag-
nitude and direction, the field is a vector field. 
29-4 
F I E L D S 
A field is any physical quantity that takes on dif-
ferent values at different points in space. There are 
two kinds of fields: namely, scalar fields and vector 
fields. Temperature is an example of a scalar field. 
while the velocity of a flowing fluid is an example of 
a vector field. Also electric and magnetic fields are 
vector fields. 
A vector field may be represented by drawing a 
set of arrows whose magnitudes and directions in-
dicate the values of the vector field at the points 
from which the arrows are drawn (Figure 29-2). 
Figure 29-2 
arrows. 
Representation of an electric field by 
We can go further and draw lines that are tangent 
to the vectors everywhere, with the density of the 
lines proportional to the magnitude of the vector 
field (Figure 29-3). 
The lines are sometimes called lines of force. In 
electrostatics, they originate on positive charges 
and end on negative charges. Lines of force do not 
intersect one another since the direction of the field 
cannot have two values at one point. 
We adopt the convention that the number of lines 
per unit area over a surface perpendicular to the 
direction of the lines is proportional to the field 
strength. 
Figure 29-3 
lines. 
Representation of an electric field by 

254 
Charge, Field, arid Potení/a/ 
An electric field is a vector point function, that 
is, a vector function of position in space in 
terms of which the force on a charge at rest at 
that point in space may be determined. An 
electric field exists at a point in space if a 
force is exerted on an electric charge placed 
at that point. 
The electric field intensity or the electric field 
strength at a point is a vector having the direction 
of the force that would be exerted on a positive 
point test charge placed at the point, and a mag-
nitude equal to the magnitude of this force divided 
by the magnitude of the test charge. 
(29-5) 
If q is positive, Ε is in the direction of F and equal 
in magnitude and direction to Flq. If q is negative, 
Ε and Flq are opposite in direction to F. The unit of 
Ε in MKS units is newtons/coulomb. 
If the charges causing the field are free to move 
(for example, metallic bodies), the presence of the 
test charge q may change their positions, and hence 
change the field we wish to measure. The change 
will be small if q is small, so an accurate definition 
of Ε in the form of an equation is 
F 
dF 
Ε = lim — or 
Ε = - τ - . 
fl-*o q 
dq 
(29-6) 
Example 1. Find the field strength in magnitude 
and direction at a point on the bisector and 6 m 
from the line joining the charges of 5 x 10"' 
coulombs each for both unlike and like charges 
separated by a distance of 16 m. 
SOLUTION 
Put a test charge q at point Ρ  in Figure 29-4. 
r5xlO;'coul)gl 
(10'm')^ 
= 0.45 nt/coul 
|6 m 
\ 
5x10« coul. 
16m 
5x10« coul 
Ε = (0.45 nt/coul)^ + (0.45 nt/coul)^ 
= 0.72 nt/coul. 
Therefore, Ε = 0.72 nt/coul parallel to line of 
charges. 
Put a test charge q at point Ρ  in Figure 29-5. 
E. = E2=(< 
= 0.45 nt/coul 
coul 
\Γ (5χ  10"'coul)q 
A 
(I0'm')q 
Ε = (0.45 nt/coul)^ + (0.45 nt/coul)^ 
= 0.54nt/couL 
Therefore, Ε = 0.54 nt/coul along bisector. 
1 
^-i-
5x10« coul 
16m 
5x10« coul 
Figure 29-4 
Figure 29-5 
29-5 
E L E C T R I C 
POTENTIAL 
When a charge moves in an electric field, work is 
done on the charge by the electric forces. There is a 
change in the electric potential energy of the 
charge. Just as it was convenient to define Ε as the 
force per unit charge, it is similarly convenient to 
define the electrical potential V as the electrical po-
tential energy per unit charge. 
Electric potential is best understood by consider-
ing first the difference of potential between two 
points in an electric field. 
The difference of potential between two 
points in an electric field Is defined as the 
work per unit charge necessary to carry a test 
charge from one point to another. 
The work done by applying an external force. FA, 
in moving a charge q an infinitesimal distance 
is 
¥A'ds 
= dW = 
-¥'ds, 
where F is the force of the field. 
The negative sign is inserted because the compo-
nent of the motion anti-parallel to Ε is opposed by 

Calculation of Ε from V 
255 
the field, or work is done on the charge by the 
external force. Since 
F = ^ E 
or 
or 
Hence 
or 
dW = 
-qE'ds 
= - E · ds. 
dV = - E d s 
d V = - Ε cos 0 ds. 
(29-7) 
where dV is the infinitesimal increase in potential, 
and θ is the angle between the positive directions of 
the vector Ε and the vector ds. 
The difference of potential between a point A 
and a point Β is given by the equation 
E d s , 
(29-8) 
and is independent of the path because the potential 
energy difference is independent of the path. 
The unit of potential in MKS units is the volt that 
is equal to a joule per coulomb. The electric field 
intensity can now be expressed in volts per meter, 
which is equivalent to newtons per coulomb. 
We shall now consider the potential at a point. 
Potential is usually defined as being zero at a point 
infinitely distant from the charges producing the 
field. Therefore, the potential at a point s is the 
work that must be expended in carrying a test 
charge from infinity to the point s. 
W 
or 
= - | 
F d s = - ^ | 
E d s 
V(s) = - J ' E d s . 
(29-9) 
As an example, let us consider the potential in the 
vicinity of a point charge. Since the field from the 
charge is radial, we can write. 
E d s = E d r = E d r , 
and 
V{r) = -
Edr. 
if we take the charge as the origin of the coordinate 
system. But 
q 
47Γ€ ο  Γ 
Therefore, 
V(r) = 
47Γ€ ο  r 
(29-10) 
The potential due to several charges is given by 
V(r) = ¿ ^ ¿ f , 
(29-11) 
where r, is the position vector of the point r with 
respect to the ith charge. If the potential is due to a 
continuous distribution of charge. 
V(r) _ 
1 
fdq 
47Γ€ ο  J 
Γ * 
(29-12) 
29-6 
C A L C U L A T I O N OF Ε  F R O M V 
The electric field Ε can be obtained easily from 
the potential V. We have established that 
^=dV 
= 
-Eds. 
Q 
In rectangular coordinates, 
E = i E , + j E y + k E , 
and 
Therefore, 
ds = idx+jdy+kd2. 
dV =-(ExdxEydy-^ 
E^dz) 
also 
dV, 
. dV ^ 
. dV ^ 
Therefore 
ρ _ 
dV 
dV 
dV 
Hence 
where 
V = i — + i — + k - ^ 
and is called the gradient operator (see Appendix 
1-8, p. 332). 
Thus, if V is known for all points in space, that 
is, if the function V(x, y,z) is known, the compo-
nents of Ε and hence Ε itself can be obtained. 
Example 2, The potential V of a particular elec-
tric field Ε is given by V = a(2x^ - 3y^ - z \ where 
α is a constant. Find the components of E. 

256 
Charge, Field, and Potential 
SOLUTION 
Therefore, 
Ex =- 4ax, 
Ey = 6ay 
and 
Ez = 2az. 
P R O B L E M S 
1. Two point charges are placed as A = - 3 0 x 10"coul at ( - 9 m , 0 ) and B = + 40x 10"''coul at 
(4- 16m,0). 
(a) What is the magnitude and direction of the force which charge A exerts on charge Β 7 
(b) What is the magnitude of the electric field intensity at the point P(0,12 m)? 
(c) What is the electric potential at the point Ρ  7 
2. Two point charges are placed as A = + 30 x 10"'^ coul at (- 9 m, 0) and Β  = - 40 coul at (+ 16 m, 0). 
(a) What is the electric field intensity at the point P(0,12 m)? Calculate the magnitude, and show the 
direction in a diagram roughly to scale. 
(b) What is the electric potential at the point P ? 
3. A negative point charge of 20microcoul is located at the point A(x =0, y =30 cm), and an equal 
positive charge is located at the origin. 
(a) What is the electric potential at point B(x = 40 cm, y = 0)? 
(b) What is the electric field intensity at point ΒΊ 
(c) What is the electric potential energy of a positive point charge of 5 microcoul placed at the point Β ? 
4. An electron (φ  = 1.60x 10"''coul, m = 9.11 x 10"^'kg) is in the electric field between two charged 
plates. 
(a) What field is required so that the electron has no vertical motion? 
(b) If the field is doubled, determine the time it takes for the electron to move 5 cm vertically upward. 
(c) If the direction of the field is changed by 180° in part (b) determine the time for the electron to move 
5 cm vertically downward. 
5. An electron (mass = 9.11 x 10"^' kg; charge = 1.60 x 10"'' coul) is projected horizontally with a velocity 
of 10^ m/sec between two parallel plates at right angles to a uniform electric field of intensity 
2 X 10^ nt/coul, upward, as in Figure 29-6. 
(a) Indicate in the diagram, by pluses or 
minuses, the charge distribution on the 
plates. 
(b) Calculate the voltage between the plates re-
quired to produce this field. 
(c) What is the net force on the electron in the 
field region? 
(d) What is the acceleration of the electron? 
1 
1 
I. I 1 
I Ί I 
• 
 I
ι 
I
I 
Ι 
i 1^1 
I. I 1 
I Ί I 
• 
 I
ι 
1^ i 
I 
I , 
I 
I 
-Z.=4 cm-
Figure 29-6 
6. Two small 1 gm pith balls are attached to fine threads 1 m long hung from a conunon point. When the 
balls are given equal quantities of positive charge, they repel one another so that the threads are at an 
angle of 30° with one another. 
(a) Draw a diagram showing all the forces on each ball. 
(b) Find the charge on each ball. 

Problems 
257 
7. An electron (mass = 9.11 x 10"^' kg; charge = 1.60 χ  10"'^ coul) is projected horizontally with a velocity 
of 10^ m/sec between two parallel plates at right angles to a uniform electric field of intensity 
5 X 10^ nt/coul, upward, as in Figure 29-7. 
(a) Indicate on the diagram, by pluses or 
minuses, the charge distribution on the 
plates. 
(b) Calculate the voltage between the plates re-
quired to produce this field. 
(c) What is the net force on the electron in the 
field region? 
(d) What is the acceleration of the electron? 
Figure 29-7 
} 
I 
I 
I 
I 
I 
I 
I 
I 
I 
I 
IE 
I 
i: 
~i 
1 
1—τ  
I
I
I 
I 
I 
id=2cm 
h 
-L=10cm-
8. (a) What is the electric field intensity halfway between two isolated positive point charges of 
4 X 10"*° coul, 60 cm apart? 
(b) What is the electric potential (relative to a point at infinity) at the point halfway between the two 
isolated charges of part (a)? 
9. An electron (mass = 9.11 x 10"^' kg, charge = 1.60 x 10'^ coul) released from rest in an electric field of 
3 X 10^ volts/m travels 0.2 cm. 
(a) What is its final speed? 
(b) How long a time elapses before it reaches its final speed? 
(c) What is its final energy? 
10. Charges A, B, C, and D are located at the corners of a square 20 cm on a side as illustrated in Figure 
29-8. Charge A = + 15 x 10"' coul, Β  = + 15 x 10"' coul, C = - 15 x 10"' coul, and D = - 15 x 10"' coul. 
At the intersection of the diagonals (E) find: 
(a) The magnitude and direction of the electric 
field intensity and 
(b) the potential. 
Figure 29-8 
Β  
\ 
^ Β  
\ 
/ 
\ 
/ 
\ 
/ 
\ 
/ 
\ 
/ 
/X 
20 cm 
/ 
\ 
D / 
\ 
D 
c 
11. Two electric charges of opposite sign, but of the same magnitude, 4microcoul are separated by a 
distance of 6 m in a vacuum. What is the electric field intensity and potential at a point midway between 
them on a straight line joining them? 
12. An electron (mass = 9.11 x 10"^* kg, charge = 1.60 x 10"" coul) revolves about a proton which has the 
same charge as the electron in an atom. The radius of the circular orbit is lO''^ m. 
(a) What is the radial acceleration of the electron? 
(b) What is the angular velocity of the electron? 
13. If the electric potential along the x-axis is given by V = (8 + 4x - 2x^) volts, where χ  is in meters, what 
is the electric field along the x-axis at χ  = 0 and at χ  = 2 m? 
14. A vacuum tube consists of a cylindrical cathode of 0.10 cm in radius inside a hollow cylindrical anode of 
inside radius 0.60 cm. The potential of the anode is 400 volts above that of the cathode. An electron 
leaves the surface of the cathode with zero initial velocity. Find its velocity when it strikes the anode. 

258 
Charge, Field, and Potential 
15. In a certain electric field, the potential is found to be given by the formula V = cd, where c is a positive 
constant and d is a coordinate. 
(a) Find the value of Ε in the direction of increasing d, 
(b) What formula would give the work done in moving a charge from -I- d i to -I- d z ( d i greater than d z ) ? 
(c) Give the units of the constant c. 
16. A positively charged ring of radius R lies in the y-z plane. Show that the formula for the electric field 
intensity at a point on the axis of the ring (x-axis) a distance r from the center of the ring is 
where q is the total charge on the ring. 

30 
Gauss's Law, 
Capacitance, 
and Dielectrics 
30-1 
Gauss's Law 
259 
30-2 
Applications of Gauss's Law 
260 
30-3 
Capacitance 
262 
30-4 
The Parallel-Plate Capacitor 
262 
30-5 
The Energy of a Charged 
Capacitor 
262 
30-6 
Capacitors in Series and Parallel 
30-7 
A Dielectric and Polarization of a 
Dielectric 
264 
30-8 
Susceptibility, Dielectric Constant, 
and Displacement 
265 
263 
30-1 
G A U S S ' S LAW 
In Chapter 29, we replaced calculations of forces 
on and potential energy of charges with calcula-
tions of field intensity and potential. This made it 
possible to avoid the serious complications related 
to the large numbers of charges involved in a typi-
cal situation. However, it will not have escaped the 
perceptive reader that the simplifications are more 
apparent than real for large numbers of charges. 
For example, it is no easier to calculate the total 
field intensity due to many charges than it is to 
calculate the total force exerted on a test charge 
due to the same charges. What is needed is some 
further concept that will make possible calculations 
such as the determination of the field intensity at all 
points in space due to an array of charges uniformly 
distributed on the surface of a hollow sphere. 
Such a distribution has a degree of symmetry that 
affords the needed additional simplification. That is, 
because of the spherical symmetry of the charge 
distribution, the field intensity should be the same 
at all points that are the same distance radially from 
the surface of the sphere. This assertion is easily 
verified experimentally. We therefore turn to a 
presentation of Gauss's law, the method by which 
the field intensity is calculated for symmetric 
charge distributions. 
To derive this law, let us take an arbitrary surface 
A, called a Gaussian surface, which encloses a 
positive point charge q, as illustrated in Figure 30-1. 
Let ^ be a distance r from an element of area dA 
on the surface A, and η — a unit vector perpendicu-
lar to dA in an outward direction. The magnitude of 
the field intensity Ε at dA has the value 
E = 
1 
47Τ € ο  r 5. 
Figure 30-1 Gaussian surface surrounding a point 
charge. 
259 

260 
Gauss's Law, Capacitance and Dielectrics 
The product of Ε normal to the surface, Ε · η , and 
the area dA is 
E'ndA=EcosedA=j^i^^^, 
47Γ€ ο  
r 
where θ is the angle between Ε and n. 
The solid angle dω  in steradians subtended by dA 
ai q, is, by definition. 
. 
cos θ dA 
dω  = 
2 — . 
Therefore, 
E n d A 
47Γ€( qdω , 
We now integrate both sides of the equation over 
the entire closed surface. 
EndA 
^ dω . 
Since a closed surface subtends a total solid angle 
of 47Γ steradians, at any point within the volume en-
closed by the surface 
EndA 
A 
eo 
(30-1) 
Equation (30-1) is Gauss's law for the special case 
where the enclosed charge is a single point charge. 
Suppose the enclosed charge is some arbitrary dis-
tribution of charge. Since any distribution of charge 
can be considered the sum of a number of point 
charges, we can v^oite Eq. (30-1) as follows: 
) E-ndA =-
A 
€ o 
(30-2) 
where the summation is taken only over the charges 
lying within the closed surface. The result is inde-
pendent of the way in which the charge is dis-
tributed. It also does not depend on whether the 
charge is positive or negative. Why? 
Equation (30-2) is a mathematical expression of 
Gauss's law, namely: 
The surface integral of the normal component 
of Ε  over any closed surface in an electrosta-
tic field equals 2q/€ o, where the Σ η  is the net 
charge inside the surface. 
In dealing with a continuous distribution of charge, 
we can write Gauss's law as 
^ E n d A = ¿ ^ ^ pdi;, 
(30-3) 
where ρ is the volume charge density inside the 
volume ν  enclosed by the Gaussian surface A. In 
the general case, ρ is a function of position. 
30-2 
A P P L I C A T I O N S O F G A U S S ' S L A W 
Gauss's law can be used to solve a number of 
electrostatic field problems involving a special sym-
metry. We shall now apply it to a few such prob-
lems. 
(a) Field of a Charged Spherical Conductor 
Let us find the magnitude of Ε at a point Ρ  which 
is a distance r from the center of a spherical con-
ductor of radius a bearing a total charge +a uni-
formly distributed over the surface. We first choose 
a spherical surface (Gaussian surface) A of radius 
r > α around the charge distribution and concentric 
with it, as shown in Figure 30-2. We now apply 
Gauss's law. Ε has the same magnitude at all 
points on A since Ε can only be a function of r. 
Therefore, 
i 
= Q 
E n d A = Ε φ  
dA=4Trr'E=-
A 
€ o 
Thus, the electric field has the magnitude 
(30-4) 
which is the same as if all the charge were concen-
trated at the center of the spherical shell. If the 
Gaussian surface had a radius r<a, the charge en-
closed would be zero (why?) so that Ε = 0 inside the 
sphere. The above results are the same for a hollow 
charged spherical conductor, since for a solid con-
ductor aU of the charges are on the outside. 
Figure 30-2 Spherical Gaussian surface surrounding 
a charged sphere. 

Applications 
of Gauss's Law 
261 
(b) Field of a Charged Conducting Piate 
Suppose that a charged plate of finite thickness is 
infinite in extent and that the charge per unit area is 
σ . Considerations of symmetry lead us to believe 
that the field direction is everywhere perpendicular 
to the plate and must be the same on each side. This 
time we choose for our Gaussian surface a cylinder 
with end areas A, one within and one outside the 
plate, and side walls perpendicular to the plate, as 
shown in Figure 30-3. 
No lines of force cut the side walls of the cylin-
der, since they are parallel to the side walls, so that 
the component of Ε perpendicular to these walls is 
zero. Since Ε equals zero everywhere inside the 
plate, the surface integral of Ε calculated over the 
entire surface of the cylinder is EA, where A is the 
area of the end of the cylinder outside the plate. 
The total charge enclosed by the cylinder is σ Α. 
Hence, from Gauss's law. 
from which 
EA = - σ Α , 
€ o 
(30-5) 
Example 1. A soap bubble 10 cm in radius with a 
wall thickness of 3.33 x 10"^ cm is charged to a po-
tential of 100 volts. Show that if it breaks and falls 
as a spherical drop the potential of the drop is 
10,000 vohs. 
Figure 30-3 Cylindrical Gaussian surface in a plane 
conducting plate. 
SOLUTION 
Volume = 4 7 γγ, ' Δ γ, = (47γ)((0.1)' m')(3.33 x 10"' m) 
ΙττΓζ' = (47γ)(0.01 m')(3.33 x 10"' m) 
Γ2' = (3χ 0.01)(3.33χ 10"') m' 
m . 
Therefore, 
Γ2 = 1 X 10"^ m 
v." 
47Γ€ ο Γ2 
4'7Γ€ ο Γι 
0.1m 
100 volts 
0.001m 
V2 = 10 m-volts 
0.001m = 10,000 volts. 
Example 2. A hollow metallic spherical shell 
0.50 m radius is charged uniformly with a negative 
charge of 2 microcoulombs. 
(a) What is the electric field intensity at the 
center of the sphere? 
(b) What is the electric potential at the center of 
the sphere? 
(c) What is the magnitude of the electric field in-
tensity at a point 1 m from the center of the sphere? 
(d) What is the electric potential at a point 1 m 
from the center of the sphere? 
SOLUTION 
(a) Ε = 0 at the center of the sphere, since there 
is no charge in the space inside the sphere. 
(b) The electric potential at the center of the 
sphere is the same as at its surface. It is constant in-
side the sphere. 
V = E r = E r = _L0 
47Γ € ο  r 
=("<>«·ϊ^)( 
= 36x 10' volts, 
(c) E = j ^ ^ = ( 9 x l 0 ^ 
2x lO'^coulX 
0.5 m 
) 
or 
\ 
coul 
= 18 X 10' volts/m. 
nt-m^ 
2x 10"'coul\ 
2x 10"^coul\ 
i m 
; 
= 18x10' vohs. 

262 
Gauss's Law, Capacitance and Dielectrics 
Example 3. What is the magnitude of the electric 
field intensity just outside the surface of a plate 2 m 
long and 1 m wide, which carries a charge of 
4 microcoulombs? 
SOLUTION 
E = ^ = 2xlO-'coul/m' 
8.85x10 TTcö ü F* 
^ 
, 
nt-m 
= 2.26 X 10' volts/m. 
30-3 
C A P A C I T A N C E 
The capacitance of a conductor is defined as the 
ratio of the charge carried by the conductor to the 
potential of the conductor relative to zero. That is. 
C = 
(30-6) 
For the case of two conductors, the charge that 
must be transferred from one to the other per unit 
potential difference is defined as the capacitance 
between the two conductors. 
(30-7) 
In MKS units, Q is in coulombs and V is in volts. 
Therefore, from Eq. (30-6), C is in coul/volt and is 
given the name farad. 
1 farad = 1 coul 
1 volt* 
The farad is too large a unit of capacitance for prac-
tical work, so smaller units are usually chosen, such 
as the microfarad (10'^ farads) and the pico- or 
micromicrofarad (10'^ farads). 
The combination of two nearby conductors hav-
ing equal and opposite charge is called a capacitor. 
30-4 
THE P A R A L L E L - P L A T E 
C A P A C I T O R 
Let us calculate as an example the capacitance of 
the parallel plate capacitor of Figure 30-4. The 
electric field will be uniform and perpendicular to 
the plates everywhere except near the edges, a 
region that we neglect here. If σ  is the surface 
Figure 30-4 Parallel plate capacitor. 
charge density on one plate, Eq. (30-5) tells us that 
(30-8) 
Since the field is uniform, the charge density on the 
other plate must be the same. By definition. 
=5 
A' 
(30-9) 
where Q is the total charge on a plate. Therefore, 
the electric field is equal to 
€ ο Α  
(30-10) 
The potential difference between the plates is 
(30-11) 
Hence, the capacitance is 
A 
^ 
A 
(30-12) 
AV = 
Ed=-d, 
€ o 
^ 
A V 
σ  . 
d ' 
—d 
30-5 
THE E N E R G Y OF A C H A R G E D 
C A P A C I T O R 
In charging a capacitor, it is necessary to do work 
to carry the charge from one plate to the other. The 
amount of work required increases as more charge 
is transferred. Suppose at the beginning of the 
charging both plates are uncharged, and at the end 

Capacitors in Series and Parallel 
263 
they have a charge Q. The element of work at any 
time to take an element of charge dQ from one 
plate and deposit on the other through a potential 
difference AV is 
dW = 
AVdQ=^dQ, 
The total work to increase the charge from zero to 
which may be written 
W 
1 
(30-13) 
The energy supplied to a capacitor in charging it is 
stored by the capacitor and released when it dis-
charges. It is reasonable to assume that the energy 
stored by a capacitor is stored in the electric field, 
since the electric field increases as Q or Δ V in-
crease. 
Let us choose a parallel plate capacitor of area A 
and plate separation d. Its capacitance, we have 
seen, is C = β ο Α  Id. Also, Q = σ Α. Substitution of 
these two expressions for C and Q into Eq. (30-13) 
gives 
a'Ad 
2eo 
· 
Since the volume ν  containing the field is just Ad 
and Ε  = afeo, we can write 
υ 
2 
(30-14) 
where WIv is the energy per unit volume stored in 
the field. We will later show this to be a more gener-
ally valid statement concerning energy storage in an 
electric field. 
30-6 
C A P A C I T O R S IN S E R I E S A N D 
P A R A L L E L 
Figure 30-5 shows two capacitors connected in 
series. What is the equivalent capacitance of this 
combination if a potential difference AV is main-
tained across it? The total work done per unit 
charge taken from one plate to the other on both 
capacitors is 
AV = AV, + A V 2 , 
Figure 30-5 Two capacitors in series, 
and, since Q = CA V, we have 
c 
C, 
C2 
or 
c 
c / C 2 -
(30-15) 
Thus, the reciprocal of the equivalent capacitance 
of a set of capacitors connected in series equals the 
sum of the reciprocals of the individual capaci-
tances. 
Figure 30-6 shows two capacitors connected in 
parallel. What is the equivalent capacitance of this 
combination if a potential difference AV is main-
tained across it? If Q i is the charge put on one 
capacitor and Q2 the charge put on the other, the 
total charge supplied by the source is 
Each capacitor, however, has the same potential 
difference so that 
Since 
or 
AV = AV, = A V 2 . 
Q = CAV, 
CAV 
= C,AV 
+ C 2 A V 
C = C. + C2, 
(30-16) 
Figure 30-6 Two capacitors in parallel. 

264 
Gauss's Law, Capacitance and Dielectrics 
o o o o o 
o o o o o 
o 
o 
o 
o 
o 
o 
o 
o 
o 
o 
(a) No field 
(b) Field 
Figure 30-7 Array of non-polar molecules. 
Thus, the equivalent capacitance of a set of 
capacitors connected in parallel is equal to the sum 
of the individual capacitances. 
30-7 
A D IEL EC T RIC A N D POLARIZATION 
O F A D I E L E C T R I C 
A dielectric, or insulator, is a material containing 
no free charge. However, a charge appears on the 
surface of a dielectric when it is placed between the 
plates of a charged capacitor due to the rotation of 
or production of dipoles. A dipole consists of two 
separated charges of equal magnitude and opposite 
sign. There are two kinds of dielectrics. They are 
called polar and non-polar dielectrics. A polar 
dielectric consists of permanent dipoles, while a 
non-polar dielectric only possesses dipoles when it 
is in an electric field. These dipoles are called in-
duced dipoles. 
We shall first discuss non-polar dielectrics. In the 
absence of an external electric field, the electrons 
of a given atom of a non-polar dielectric are distrib-
uted symmetrically around the nucleus, as illus-
trated in Figure 30-7(a). When a field is applied, the 
electrons are displaced in the direction opposite to 
that of the field [Figure 30-7(b)]. The center of grav-
ity remains fixed since there is no translational 
force on the atom as a whole. Each atom thus ac-
quires an electric dipole moment (p), which is paral-
lel and in the same direction as the applied field. 
The dipole moment ρ of a pair of charges is defined 
as 
p = qá, 
(30-17) 
where q is the magnitude of one of the charges and 
d is the charge separation. The action of the elec-
tric field in giving each atom of the dielectric an 
induced dipole moment is called polarization. 
The polarization (P) of a substance is defined as 
the electric dipole moment per unit volume, and it is 
proportional in magnitude to the applied electric 
field. 
In a polar dielectric with permanent dipoles, the 
dipoles are randomly oriented when no external 
field is present, as shown in Figure 30-8(a). When an 
external field is applied, an orientation of these di-
poles in the direction of the field takes place, as 
illustrated in Figure 30-8(b). The degree of orienta-
tion is increased as the electric field strength is 
increased. 
Consider a dielectric slab with a field Ε applied 
perpendicular to it as illustrated in Figure 30-9. If 
the slab is a non-polar dielectric, the field polarizes 
it; that is, it induces dipole moments throughout the 
slab. If the slab is a polar dielectric, the field orients 
the dipoles in the direction of the field. We can 
readily see that the net effect of the field is the same 
for both non-polar and polar dielectrics. The in-
terior of the dielectric remains neutral since the 
0 
0 
0 
0 ^ 
€ ^ 
Q 
0 
0 
^ 
o 
o 
o 
σ  
o 
Ε  
o 
0 
o 
(a) No field 
(b) Field 
Figure 30-8 Array of polar molecules. 

Susceptibility, 
Dielectric Constant, and Displacenient 
265 
o o o o o 
o o o o o 
o o o o o 
Figure 30-9 Dielectric slab in an electric field. 
positive and negative charges of adjacent dipoles 
cancel out each other's effects. The net result is the 
production of a layer of bound negative charges on 
one surface of the slab and a layer of bound posi-
tive charges on the other, as depicted in Figure 
30-9. From the definition of polarization and dipole 
moment, it can be seen that Ρ  = σ ί,, where ab is the 
bound charge per unit area. 
30-8 
SUSCEPTIBILITY, 
D I E L E C T R I C 
C O N S T A N T , A N D D I S P L A C E M E N T 
It is possible to discuss the consequences of 
polarization without a consideration of the atomic 
processes involved. The polarization Ρ  depends on 
the nature of the substance and, for most dielec-
trics, is proportional and parallel to the electric field 
in the dielectric ED. We define a property of the 
dielectric, called its susceptibility Xe, by the equa-
tion 
^ 
" 
--XeeoEo. 
(30-18) 
Xe = 
or 
P = 
Since only a material substance can be polarized, 
the susceptibility of a vacuum is zero. One may 
question why Xe is not just defined as Xe = P / E D 
since Ρ  is proportional to ED. The answer to this 
question is that with Xe defined by Eq. (30-18) we 
may write Xe as a pure dimensionless number, 
which is the same regardless of the system of units 
used for measuring Ρ  and ED-
The surface density of bound charge at any point 
on the surface of a dielectric is equal to the normal 
component of Ρ  at the surface. For the special case 
of a surface perpendicular to E , 
ab = P = Xe€ oED 
(30-19) 
Consider a dielectric between the plates of a 
capacitor, as illustrated in Figure 30-10. In the reg-
ion between the plates and the dielectric, we en-
counter two different kinds of charge. On the sur-
face of the dielectric, there is a concentration of 
bound charge per unit area ab equal to P. On the 
surface of the conductor, there is a concentration 
of free charge per unit area σ / from an external 
source equal to € oE/. Since these charge concentra-
tions are opposite in sign, we can write 
eoED = € o(E/ -Eb) 
= af- 
ab. 
(30-20) 
Since ab = Ρ , we see that 
or 
Since 
af = (af-ab) 
+ ab = € OED + Ρ  
€ OE/ = € O B D + P . 
(30-21) 
Ρ  = XeCoEo, 
€ oEf = € O ( H - ; ^ . ) E D . 
D = € OED+P 
P=0 
D = € oE, 
(30-22) 
Figure 30-10 Showing E,, 
plate capacitor. 
Eb, Eo, and Ρ  in a parallel 

266 
Gauss's Law, Capacitance and Dielectrics 
The quantity € o(l 
is called the permittivity of 
the dielectric e. Therefore, 
(30-23) 
Since the permittivity of a dielectric is always 
greater than the permittivity of vacuum, it is conve-
nient to have the ratio of the permittivity of the 
dielectric to that of vacuum. This ratio is called the 
relative permittivity or the dielectric constant ( X ) 
of the dielectric. Thus, 
(30-24) 
The dielectric constant is a pure dimensionless 
number. It is equal to unity for vacuum and is 
greater than unity for a material substance. Its num-
erical value can be obtained for a particular dielec-
tric by measuring the capacitance (C) of a capacitor 
with the dielectric between the plates and the 
capacitance (Co) with free space between them. 
The ratio C/Co then gives the dielectric constant of 
the dielectric directly. 
It is convenient to set the quantity (COED 
+ P ) of 
Eq. (30-21) equal to D. D is given the name electric 
displacement 
and, in general, is a vector quantity 
defined by the vector equation 
D = € OED + Ρ  = € OE/ = σ /. 
(30-25) 
Equation (30-22) can now be written 
O = € o{l-^Xe )ED = €  ED = σ /. 
(30-26) 
The electric displacement D has some similarity to 
ED, but depends only on the free charge, while E D 
depends on both the free and bound charges. 
From Eq. (30-25), as illustrated in Figure 30-10, 
we see that while the magnitude of E D in the 
dielectric differs from that in the gap between the 
dielectric and the plate, the magnitude of D does 
not differ from one location to the other. If we 
know σ / on the plates of a capacitor and e for the 
dielectric, we may find E D from Eq. (30-26). 
We have seen that in free space the energy per 
unit volume stored in the field equals koE^ For a 
dielectric material, we replace co by € , which leads 
us to the expression for the energy stored per unit 
volume in a dielectric—namely. 
W 
1 
^2_DE_D' 
(30-27) 
that for free space. This is due to the extra work 
that must be done to polarize the dielectric. 
Example 4. A parallel plate capacitor in air with 
plates separated by 0.4 mm when charged with 1 
microcoulomb is found to have a potential differ-
ence of 40 volts between the plates. What is the 
area of each plate? 
SOLUTION 
6ο Α  
and 
C = 
C 
€ oA 
or 
We can readily see that the stored energy per unit 
volume with a material dielectric is greater than 
^ ^ 
Qd ^ 
(10'coul)(4X 10'm) 
' ^ ^ ^ 
\ ( 8 . 8 5 X 1 0 - Ä 4 0 " ^ 
nt-m / 
10-'* coul-m 
8.85 X 10-*-"^^^^ 
nt-m^ 
= 1.13 m'. 
Example 5. A 2 microfarad 
capacitor (1), 
charged to a potential difference of 100 volts, 
is suddenly connected to and shares its charge 
with an uncharged 1 microfarad capacitor (2). 
What is the loss of energy in this process? 
SOLUTION 
In order for capacitor (1) to share its charge with 
capacitor (2), they must be connected in parallel. 
Q = C,A V, = (2 X 1 0 ' farads) (100 volts) 
= 2 X 10"' coul 
QI + Q2 = 2 X 10"' coul = CAV + CzA V 
= (2 X 10"' farads + 10"' farads) Δ V 
2 X10"' coul 
^ ^ = 3 x l O - ' f a r a d s = ^'^^^^^^ 
Wbefore = | CjA V = | ( 2 X 10"' farads) 
X [(100)'volts'] = 10"'joules 
Wafcer = I CA ν  = I [(2 -K 1) X 10"' farads] 
X [(66.7)' volts'] = 0.67 x 10"' joules 
A W = (10"' joules) - (0.67 x 1 0 ' joules) 
= 3.3 X 10"' joules. 

Problems 
267 
Example 6. Two parallel conducting plates have 
equal and opposite charges. The space between the 
plates contains a dielectric with a dielectric con-
stant of 4. The resultant electric field intensity in 
the dielectric is 10^ volts/m. Compute: 
(a) the free charge density (σ /) on the plates, 
(b) the bound charge density (σ *,) on the surface 
of the dielectric, 
(c) the susceptibility (xe) of the dielectric, 
(d) the permittivity (c) of the dielectric, 
(e) the polarization (P) in the dielectric, 
(f) the displacement (D) in the dielectric, 
(g) the energy density in the dielectric. 
SOLUTION 
(a) σ / = eEo = 
€ OKED 
= (8.85 X 1 0 - ' ' ^ ) ( 4 ) ( 1 0 ^ volts/m) 
= 354xl0-'coul/m'. 
(b) ab = 
af-€ ü ED 
= (354xlO-'coul/m') 
- 
(8.85X 10'^^?^VlO'volts/m) 
Λ  
nt-m / 
= 265.5 X 10 'coul/ml 
(c) ;^, = - ^ - l = X - l = 4 - l = 3 . 
(d) € =Ke„ = ( 4 ) ( 8 . 8 5 x l O - ' ^ ^ ) 
= 3.54x 10"" coul'/nt-m'. 
(e) Ρ  = σ ,- 
COJSD = at = 265.5 x 10"" coul/ml 
(f) D = COED = σ / = 354 x 10"' coul/m'. 
( 3 5 4 X 1 0 " ' ) ^ ^ " 
, . „ c o u l \ 
= 1770joules/m'. 
P R O B L E M S 
1. A long solid cylinder of radius a has a uniform volume charge density p. Find expressions for the field 
and for the potential at all points outside the cylinder if the zero potential is defined to be at r = ΓΟ . 
2. Find expressions for the field and for the potential for all points inside the cylinder of Problem 1. 
3. Prove that the potential at a point Ρ  surrounded by charges qi, ^2, q 3 , . . . , at distances ΓΙ, Γ2, Γ3, . . . , from 
the point Ρ  is given by 
\r, 
Γ2 
Γ3 
/ 
where the zero potential is taken at infinity. 
4. Figure 30-11 represents two long coaxial con-
ducting cylinders. The inner conductor is 
grounded and the outer cylinder carries a posi-
tive charge per unit length of λ. Sketch Ε and V 
as a function of r for all values of r from 0 to oc. 
Figure 30-11 
5. A solid sphere of radius R has uniform charge density p. Find expressions for the field and for the 
potential at all points inside the sphere if the zero of potential is defined to be at r = ro. 
6. Find expressions for the field and potential for all points outside the sphere of Problem 5. 

268 
Gauss's Law, Capacitance and Dielectrics 
7. Two concentric conducting spherical shells of radii R\ and Ri iRi<R2) 
bear charges 
Q i and Qi, 
respectively. Find expressions for the electric field at: 
(a) 
r<Ri, 
(b) Ri<r<R2, 
and 
(c) 
r>R2. 
8. Find expressions for the potential, relative to zero potential at infinity, in Problem 7 at 
(a) 
r<Ri, 
(b) Ri<r<R2, 
and 
(c) 
r>R2. 
9. A capacitor has a capacitance of 2 microfarads when its plates are separated by a layer of air. It is 
charged to 400 volts by means of a battery. 
(a) Find the charge on the plates. 
(b) Find the energy stored in the capacitor. 
10. Compute the energy stored in a 3 microfarad capacitor with air between the plates 
(a) when charged to a potential of 100 volts and 
(b) when the charge on each plate is 2 x 10"' coulombs. 
11. The charged capacitor in Problem 9 is first disconnected from the battery and then immersed in oil 
having a dielectric constant of 3. 
(a) What is the difference of potential between the plates? 
(b) What is the energy of the capacitor? 
(c) What is the source of energy change? 
12. The capacitor in Problem 10, after being charged to a potential of 100 volts, is immersed in a liquid of 
dielectric constant 2. 
(a) What is the potential difference between the plates? 
(b) What is the energy of the capacitor? 
13. Show that the energy of a dipole is given by the relation 
W = - p . E 
and the torque on the dipole is given by 
Τ  = ρ X E, 
where ρ is the dipole moment. 
14. A 2 microfarad capacitor charged to 100 volts 
and a microfarad capacitor charged to 200 volts 
are connected as in Figure 30-12. Find the 
difference of potential and charge on each 
capacitor and the loss of energy that has taken 
place. 
Figure 30-12 
15. Two capacitors, of capacitances C i = 3 micromicrofarads and C2 = 6 micromicrofarads, are connected 
in series, and the resulting combination is connected across 1000 volts. Compute: 
(a) the equivalent capacitance of the combination, 
(b) the total charge on the combination and the charge on each capacitor, 
(c) the potential difference across each capacitor, and 
(d) the energy W stored in the capacitors. 
16. (a) Calculate the capacitance of a capacitor consisting of 2 parallel plates 100 cm' in area separated by a 
layer of paraffin 1 mm thick and dielectric constant 2. 
(b) If the capacitor is connected to a 200 volt source, calculate the charge on the capacitor and the 
energy stored in the capacitor. 

Problems 
269 
17. Two oppositely charged conducting plates, having numerically equal quantities of charge per unit area, 
are separated by a dielectric 5 mm thick of dielectric constant 1.5. The value of Ε in the dielectric is 
lOSolts/m. Compute: 
(a) the free charge per unit area on the conducting plates, 
(b) the bound charge per unit area on the surface of the dielectric, 
(c) the polarization Ρ  in the dielectric, 
(d) the displacement D in the dielectric, and 
(e) the energy density in the dielectric. 
18. Calculate the capacitance of a spherical capacitor, formed from two thin metal spheres of radii R i and 
R i , the space between which is filled with a dielectric of dielectric constant K . 
19. For the capacitor of Problem 18, find Ε, P, and D for a point in the dielectric medium when the 
capacitor bears a charge Q. 
20. In Figure 30-13, find the capacity between the 
points A and B. A 40 volt battery is connected 
between A and B, What is the energy stored in 
the 4 μ f capacitor? 
Figure 30-13 

31 
Direct Electric 
Currents 
31-1 
Current and Current Density 
271 
31-5 
Electric Circuits and Kirchhoff's 
31-2 
Ohm's Law, Resistance, Resistivity, 
Rules 
275 
and Conductivity 
272 
31-6 
The Wheatstone Bridge and the 
31-3 
Resistances in Series and Parallel 
273 
Potentiometer 
276 
31-4 
Energy and Power in Electric 
31-7 
Series Circuit with Resistance and 
Circuits, Electromotive Force 
274 
Capacitance 
277 
In this chapter we make the transition from static 
electricity to current electricity. We will discuss the 
flow of a charge in conductors and introduce the 
concepts of current resistance and electromotive 
force. Following this, we will give a detailed discus-
sion of direct current electric circuits and some 
measuring instruments. 
31-1 
C U R R E N T A N D C U R R E N T 
D E N S I T Y 
Electric current is deñned as the time rate of flow 
of charge across any cross section of a conductor. 
1 = dt' 
(31-1) 
The MKS unit of current is the ampere. One am-
pere is equal to one coulomb per second. Direct or 
steady currents are constant in time; that is, equal 
quantities of charge pass a given cross section of a 
conductor in equal intervals of time. The direction 
of the current is taken conventionally as being that 
in which a positive charge would move in an elec-
tric field. In a majority of cases, it is actually a 
negative charge flow in the opposite direction that 
produces the current. 
If we are dealing with a current distributed uni-
formly over a cross-sectional area Λ  of a conductor 
perpendicular to the direction of the current, we 
can express the current in terms of the current 
density j , which is the current per unit area by the 
equation 
I = jA. 
(31-2) 
The general relationship for the total current 
flowing through a surface area at any orientation, 
with the current direction as shown in Figure 31-1, 
is 
ndA, 
(31-3) 
where η  is the unit vector normal to the surface and 
dA is an element of the area A. The integral is 
taken over the surface through which we are cal-
culating the current, and is independent of the 
shape of the surface. 
Figure 31-1 Relationship between current and cur-
rent density. 
271 

272 
Direct Eiectric Currents 
Since current is due to charged particles of 
charge e (electron charge), with a mean drift veloc-
ity ν  and density Ν  per unit volume, the current 
density is 
} = Ney. 
(31-4) 
Thus, j is a vector whose direction is that of the 
velocity ν  of the particles. 
31-2 
O H M ' S LAW, R E S I S T A N C E , 
RESISTIVITY, A N D CONDUCTIVITY 
In many conductors (for example, metals), the 
current I is proportional to the difference of poten-
tial AV, 
or 
laAV 
AV 
= 
IR, 
(31-5) 
where 1? is a constant called the resistance. It de-
pends upon the dimensions, material, and tempera-
ture of the conductor. The above equation is the 
usual expression of Ohm's law. The unit of R is the 
ohm (Ü) if A V is in volts and I is in amperes. 
Another form of Ohm's law is obtained from the 
fact that the current through a conductor is propor-
tional to the electric field. This fact may be expres-
sed mathematically in terms of the current density 
as 
jaE 
or 
E = pj, 
(31-6) 
where ρ is a constant called the resistivity of the 
material. This relation contains the same informa-
tion as the familiar form of Ohm's law [Eq. (31-5)], 
but it has a more general application. It is a vector 
equation that applies to any point in the material. It 
can be applied to a conductor of any shape in which 
neither the field nor the current density are uniform 
and integrated to give Eq. (31-5). 
The reciprocal of the resistivity 1/p is called the 
conductivity of the material σ . Equation (31-6), 
written in terms of the conductivity, is 
] = σ Ε . 
(31-7) 
It should be pointed out that, in many materials, ρ 
and consequently σ  are not constant at a particular 
temperature but vary with the current in the mater-
ial. These materials are said to be non-ohmic or 
non-linear. 
For a uniform isotropic conductor of length / and 
cross-sectional area A, with a potential difference 
A V between its ends, the electric field and the cur-
rent density will be constant throughout the con-
ductor and will have the values 
E=— 
and 
j = ^ . 
The resistivity will be given by 
Ε 
^V|l 
AV 
Since 
= - j - , we can write 
(31-8) 
which shows that the resistance of a uniform con-
ductor is directly proportional to its length and in-
versely proportional to its cross-sectional area. We 
also see from the previous equations that the units 
of resistivity and conductivity are ohm-m and 
(ohm-m)-', respectively. 
Example 1. A wire l(X)m long and 2 mm in 
diameter has a potential difference of 1.5 volts be-
tween the ends of the wire, and carries a current of 
1 ampere. 
(a) What is its resistance? 
(b) What is the resistivity of the material from 
which the wire is made? 
SOLUTION 
(a) 
R 
(b) ρ 
AV^ 1.5 volts 
/ 
1 amp = 1.5 ohms 
R A ^ (1.5 ohms)(7r)[(0.001)' m'] 
/ 
(100 m) 
= 4.7 X 10'^ ohm-m. 
The resistivities of various materials at room 
temperature are given in Table 31-1. 
Most metals increase in resistivity with a rise in 
temperature. The increase is linear over a fairly 
large temperature interval. The size of the tempera-
ture interval and its position on the temperature 
scale varies from metal to metal. As an example, 
the interval for copper is from -200°C to -f-300^C 
and the interval for platinum is from 0°C to 400°C. 

Resistances in Series and Parallel 
273 
Table 31-1 Resistivity and Temperature Coefficient 
of Various Metals. 
Temperature 
Resistivity 
coefficient of 
Metal 
at 20°C (ohm-m) 
resistivity per C° 
Aluminum 
2.8 X 10-« 
4.0 X 10-' 
Copper 
1.7X10-' 
4.0X10"' 
Gold 
2.4X10-' 
3.4X10"' 
Iron 
1.0 X 10-' 
5.0X10"' 
Nickel 
7.8 X 10-' 
6.0X10"' 
Platinum 
9.8 X 10-' 
3.6 X 10"' 
Silver 
1 . 5 X 1 0 ' 
3.8 X 10"' 
Tungsten 
5.5 X 10"' 
4.5 X 10"' 
The temperature coefficient of resistivity may be 
defined as follows: 
The temperature coefficient of resistivity is the 
fractional change in resistivity per degree 
change in temperature. 
The definition is written mathematically as fol-
lows: 
or 
Ap 
ρ = p2o(l + α2ο Δ ί), 
(31-9) 
(31-10) 
where Δ ί is the temperature interval. 
The index «20 means the temperature 20°C has 
been chosen as the reference level. Table 31-1 gives 
the temperature coefficient of resistance for various 
metals. 
Because the changes in physical dimensions are 
much smaller than the change in resistivity with 
temperature, we can neglect dimensional changes 
in considering the change in resistance with temper-
ature of an extended material, and write Eq. (31-9) 
as 
Ai? 
«20 = 1?2θΔ ί' 
(31-11) 
since ρ = R A H and the quantity A H appears both 
in numerator and the denominator when we substi-
tute R A H for ρ in Eq. (31-9). 
It should be pointed out that while the resistance 
of most conductors increases with an increase in 
temperature, there are a group of materials called 
intrinsic semiconductors, where the resistance de-
creases with increase in temperature. Carbon is an 
example. Also, the resistance of some conductors 
disappears at very low temperatures. These materi-
als are called superconductors. There are also al-
loys with a resistance independent of temperature 
over a wide temperature range. Examples are con-
stantan and manganin, which are used in making 
standard resistance coils. 
31-3 
R E S I S T A N C E S IN S E R I E S A N D 
P A R A L L E L 
We will now begin a discussion of direct current 
circuitry. First we will discuss resistance combina-
tions. Consider three resistances connected in 
series as illustrated in Figure 31-2. The current 
must be the same in each resistance since there are 
no other paths through which some of the current 
might go. Hence, the potential drops across the re-
sistances R i , R2, and R i will be 
AV, = I1?„ 
AV2 = IR2, 
AV3 = J1?3. 
The total potential drop (A V) across the three re-
sistances will be the sum of the individual potential 
drops, and will equal the current times the total re-
sistance ( R ) , that is. 
AV = IR = IRi + IR2 
+IR3 
and 
R = Ri^R2 
+ R3. 
(31-12) 
Therefore, the total resistance of a number of resis-
tances in series is the sum of the individual resis-
tances. 
Figure 31-2 Resistances in series. 
Next, consider three resistances connected in 
parallel, as illustrated in Figure 31-3. The total cur-
rent through the unbranched part of the circuit 
splits up when it reaches the branch point, with a 
fraction of the total current passing through each 
branch. Therefore, the total current I must be equal 
to the sum of branch currents I i , I2, and I3, that is, 
/ = J, + J2 + l3. 
The potential drop in each branch must be the 
same since all the branches rejoin at a common 

274 
Direct Electric Currents 
¡2 
a 
-m 
W W W V -
VAAAAAR-
Figure 31-3 Resistances in parallel. 
junction. Thus, the currents in each branch are 
^'-:r7' ^'-:r7' ^'-:r7' 
and the total current / = Δ VIR, where R is the total 
resistance. Combining these equations gives 
Δ ν ^Δ ν _^Δ ν _^Δ ν  
R 
R\ 
Ri 
R3 
or 
R 
R\ 
R2 
R2 
(31-13) 
Therefore, the reciprocal of the total resistance of a 
number of resistances in parallel is the sum of the 
reciprocals of the individual resistances. 
The total resistance of a number of resistances in 
parallel is less than that of the smallest individual 
resistance. Why? 
31-4 
E N E R G Y A N D P O W E R IN E L E C T R I C 
C I R C U I T S , E L E C T R O M O T I V E 
F O R C E 
Now we will consider energy and power in elec-
tric circuits. When a constant electric current I 
passes through an ohmic resistance, the energy dis-
sipated appears in the form of heat. When the po-
tential drop across a resistance RisAV, 
the energy 
W dissipated in the resistance per unit charge 
equals AV, that is. 
Since 
Q=IAt, 
where Δ ί is the time interval, we have 
W = AVIAt. 
The rate at which energy is dissipated, or the 
power P , is 
P = - ^ = AVI = I ' R 
(31-14) 
When a charge is taken around a circuit, work is 
done on it. To provide this work, a source of energy 
is required. The source of energy may be chemical 
as in a battery, mechanical as in a generator, or 
thermal as in a thermocouple, although the latter 
situation is of less practical importance. When con-
sidering sources of electrical energy in a circuit, a 
term called the electromotive force (^) of the source 
is used and is defined as follows: 
The electromotive force (emf) in a circuit is 
equal to the energy supplied per unit positive 
charge in carrying the charges around the 
circuit. 
It should be pointed out that the work done on 
the charge or the energy dissipated in the circuit 
may occur within the source of the emf as well as 
outside it. In a steady state, the rate at which elec-
tric energy is put into an electric circuit is equal to 
the rate of dissipation of the electric energy in the 
circuit. Consider a simple circuit containing a bat-
tery Β  of emf (?) and an external resistance R , as 
illustrated in Figure 31-4. In the battery, the elec-
trolyte and the plates have a certain resistance, re-
ferred to as the internal resistance Ri. The power 
supplied must equal the power dissipated and, 
therefore, 
I^ = 
fR-^l'Ri. 
The potential difference across the external circuit. 
Figure 31-4 A circuit with a battery of emf ^, terminal 
voltage V, and Internal resistance f?i, in series with an 
external resistance R. 

Electric Circuits and Kirchhoff's Rules 
275 
which is the potential difference across the termi-
nals of the battery, is given by Δ V = IR. Using this 
in the above relation and dividing by J, we have 
(31-15) 
When current is being drawn from the battery, the 
potential difference between the terminals is less 
than the emf by an amount IR,. The potential differ-
ence equals the emf when the current has been re-
duced to zero. 
31-5 
E L E C T R I C C I R C U I T S A N D 
K I R C H H O F F ' S R U L E S 
Many circuits are more complex than the simple 
circuit illustrated in Figure 31-4. There are some-
times several conducting paths each with resistors, 
batteries, or generators. Many complex circuits can 
be broken down into somewhat more simple series 
and parallel combinations allowing a somewhat 
simple solution for the current in the circuit and the 
difference of potential across each element. How-
ever, in many complicated circuits this is not 
possible. 
A complicated network can be solved with the 
help of two general principles, known as Kirch-
hoff's rules. They are: 
1. The total current flowing into a junction of 
two or more conducting paths in a circuit is 
equal to the total current flowing out of the 
junction. 
2. Around any closed loop in a circuit, the alge-
braic sum of the potential differences is equal 
to the albegraic sum of the emf's in the loop. 
In the application of Kirchhoff's rules, we first 
assign a current direction in each loop. This choice 
may be correct or incorrect. If it is incorrect, the 
current comes out negative, which means that the 
current flows in a direction opposite to the chosen 
direction. It is also necessary to choose the signs of 
the emf's and potential differences as we go around 
the loop or, in other words, we must adopt a set of 
sign conventions. The sign conventions are the fol-
lowing: 
1. An emf is taken as positive if we proceed from 
the positive to the negative terminal of the 
source in going around the loop, and negative 
if we proceed from the negative to the positive 
terminal. 
2. A potential difference is taken as positive if 
we proceed in the same direction as the as-
signed loop current and negative if we proceed 
in an opposite direction. 
The method of applying Kirchhoff's rules is best 
illustrated by an example. 
Example 2. Find the current through the 50 ohm 
resistor in Figure 31-5. 
10 ohms 
lO^^ohms 
30 ohms 
— \ W — 
25 ohms 
Λ Λ Λ /^— 
25 volts 
Figure 31-5 Circuit with a divided current path. 
SOLUTION 
Apply Kirchhoff's first rule to the junction A: 
I. + Í2 = L 
Applying Kirchhoff's second rule to the three loops 
gives 
± (n)(amp) ± (n)(amp) ± (n)(amp) = ± (volts) 
lOi, + 50(i, + 1*2) + 30(i, - ¿3) = 0, 
IOÍ2 + 50(1,4-1%) + 25(Í2 +13) = 0, 
30(Í3-íi) + 25(Í2 + Í3) = 25. 
These three equations may be written 
± (n)(amp) ± (n)(amp) ± (n)(amp) = ± (volts) 
90i, + 50í2-30Í3 = 0, 
50i, + 85Í2 + 25Í3 = 0, 
-30i, + 2 5 Í 2 + 55Í3 = 25. 
Solving for ιΊ and 1*2 from the above three equa-
tions, gives ii = -?fÍ2 and 1*2 = -1.244 amp. Sub-
stituting i I and I2 into the equation from Kirchhoff's 
first law, gives 
/ = 0.017 amp downward. 

276 
Direct Electric Currents 
31-6 
THE W H E A T S T O N E B R I D G E A N D 
THE 
P O T E N T I O M E T E R 
We shall now discuss two important measuring 
instruments—namely, the Wheatstone bridge and 
the potentiometer. The Wheatstone bridge is widely 
used for measuring resistance. It consists of a cir-
cuit as illustrated in Figure 31-6. JRi, 1?2, and Ri are 
adjustable calibrated resistors, and Rx is an un-
known resistance. The resistors are adjusted so that 
no current flows through the galvanometer G, 
which is a current detecting instrument. Since no 
current flows in fed, the potential difference be-
tween b and d is zero. Therefore, 
Vat = Vaä 
and 
Vtc = Vdc, 
where Vat is the potential difference between 
points a and b, etc. 
Using Ohm's law, these equations can be re-
written as 
LRi 
= hRi 
and 
J,R = 
hRi. 
When the second equation is divided by the ñrst, 
we find 
Kc^Ry 
R^ 
R2 
(31-16) 
If we know Ri, R2, and R2 we can calculate Λ . 
Actually, it is only necessary to know one of the 
resistances and the ratio of the other two. 
Several types of the Wheatstone bridge are made 
commercially. A simple form of the Wheatstone 
Figure 31-7 Slide wire bridge. 
bridge used in elementary work has two of the re-
sistors replaced by a bare wire of uniform diameter. 
The wire is divided into two sections by a sliding 
knife edge, as illustrated in Figure 31-7. Since the 
wire is of uniform diameter the resistance of either 
section is proportional to its length. Therefore, 
when the bridge is balanced, 
Rx^Rs^Kh 
Ri 
R2 
KU 
or 
(31-17) 
where l?i is a resistor of known resistance. 
A potentiometer is an instrument that can be 
used to compare the emf's of two sources, or to 
measure the emf of one source if the other is known 
since it does not draw any current from the sources. 
A simple form of the potentiometer is shown in 
Figure 31-8. XY is a high resistance bare wire, Ζ is 
a sliding point contact, and Β  is a battery that pro-
vides a steady current through XY. It must have a 
greater terminal voltage than the emf of either the 
unknown battery X or the standard battery S. The 
point contact is positioned so that no current flows 
Figure 31-6 Wheatstone bridge circuit. 
Figure 31-8 Simple potentiometer. 

Series Circuit with Resistance and Capacitance 
277 
through the galvanometer when S is in the circuit 
and then repositioned so that no current flows in the 
galvanometer when X is in the circuit. Let Ζ and 
Z' be these positions. This indicates that the poten-
tial drop across the wire section t and L is equal to 
the emf of the batteries S and X, respectively. 
Therefore, 
or 
= IRx=IKh 
and 
% = IRs = IKls 
(31-18) 
% 
Rs 
Is' 
31-7 
S E R I E S CIRCUIT WITH R E S I S T A N C E 
A N D 
C A P A C I T A N C E 
In the application of Kirchhoff's rules to D C cir-
cuits, only batteries, generators, motors, and resis-
tors are important. When a capacitor is in a D C 
circuit in series with a resistor, there will be a 
transient current while the capacitor charges or dis-
charges. When a steady state is reached, the current 
through the capacitor will be zero. 
Consider a capacitor of capacitance C in series 
with a resistor R connected to a battery of voltaget 
V, then 
or 
V = 
^-^RI 
(31-19) 
(31-20) 
To solve the differential Eq. (31-20), multiply by C 
and rearrange the terms 
dQ 
^ 
dt 
CV-Q 
RC 
Integration of Eq. (31-21), gives 
(31-21) 
- In (CV -Q) = J^-^ constant. 
(31-22) 
The constant is determined by setting Q = 0 at ί = 
0, and Eq. (31-22) becomes 
tFor simplicity in notation we use the term voltage and 
the symbol V rather than the term potential difference 
and the symbol AV. Both notations, however, refer to the 
same concept. 
- l n ( C V - Q ) = - ^ - l n ( C V ) 
or 
If we now take the antilogarithm of both sides of 
this equation, we obtain 
^-..c^CV-Q^ 
Q 
CV 
' 
CV 
or 
Q = CV(l-e-'^'''^). 
(31-24) 
This equation shows that the charge on the 
capacitor approaches exponentially to a value equal 
to CV. At a time t = RC, the charge has increased 
to within l/e of its final value. The product RC is 
called the time constant (τ ) of the circuit. The cur-
rent at any time is given by 
This shows that the current decreases exponen-
tially, and has the value He at ί = τ . 
If we have an isolated charged capacitor, initially 
at a voltage Vo, and resistance R is then connected 
across it at time t = 0, the differential equation for 
the charge on the capacitor at any time as it dis-
charges is given by Eq. (31-20) with V = 0, which is 
A solution of this equation gives 
Q = CVoe""''' 
= CVoe-"\ 
(31-26) 
which shows that the charge decays exponentially 
to zero. The discharge current is 
I = 
dQ^_Vo^.,Rc^_Vo^ 
dt 
(31-27) 
and is the same as the expression for the charging 
current except that it has a minus sign indicating 
that it is oppositely directed. 
Example 3. In the circuit of Figure 31-9, V = 10 
volts, jR = 10*^ ohms, and C = 2 microfarads. After 
the capacitor is fully charged, switch Si is opened 
and Si is closed. 
(a) Find the charge on the plates of C at times 
ί = 0, 2 seconds, and ^. 
(b) What is the initial value of / ? 

278 
Direct Electric Currents 
Figure 31-9 
SOLUTION 
(a) Apply Eq. (31-26); namely, Q = CVe'^'^^. At 
time t = 0, 
(? = ( 2 x 10-'farads)(10volts)[e-^'*"^^^^'^"^^'^'°'''"*^'"] 
= (2xl0-'farads)(10volts) 
= 2x10'coul. 
At time t = 2 seconds, 
0 = (2 X 10"' farads)(10 volts)[e 
= (2 X 1 0 ' farads)(10 e"' volts) 
= 7.32xlO-'couL 
At time t = 00 
-(2 sec )/l(106 Ω )(2χ  10-« farads)]-! 
-(00 sec)/[(106 Ω )(2χ  I0-* farads)]-
Q = (2 X 1 0 ' farads)(10 volts)[e 
= (2xl0-'farads)(0volts) 
= 0 coul. 
(b) Apply Eq. (31-25); namely, I = 
^e"'''', 
J ^ 10 volts -(0sec)/l(106nK2xl0-^farads)] _ 10 VOltS 
10'Ω  ^ 
1 0 Ώ  
= 10' amps. 
P R O B L E M S 
1. A wire of 5 ohms resistance is stretched to increase its length by 10%, the volume and conductivity of 
the wire remaining constant. What is the new resistance? 
2. A rod of material A (p = 2 x 10"^ ohm-m) of uniform cross section 0.5 cm' and 2 m long is connected in 
series with a rod of material Β (ρ = 1.5x 10"^ ohm-m) of cross section 0.3 cm' and 1.8 m long. If a 
current is passed through the combination, find the ratio of the potential drop across A to the potential 
drop across B. 
3. At 0° C, the resistance of a certain wire is 200 ohms; at 40''C, it is 220 ohms. What is the temperature 
coefficient of resistance of this wire? 
4. A current of 1 ampere flows in the 15 ohm resis-
tance of the circuit shown in Figure 31-10. What 
is the potential difference between X and Y ? 
Figure 31-10 
5. Three 4-ohm resistors are connected as in Fig-
ure 31-11. What is the resistance between the 
points A and Β ? 
Figure 31-11 
10 ohms 
5 ohms 
10 ohms 
I — \ W — 
15 ohms 
4 ohms 
4 ohms 
Α Λ Λ τ -
4 ohms 
-β  

Problems 
279 
6. Two cells, one of emf 1.5 volts and internal resistance 0.2 ohms, the other of emf 3 volts and internal 
resistance 0.3 ohms, are connected in parallel. The combination is connected in series with an external 
resistance of 5 ohms. What current flows through the external resistance? 
7. The potential difference across the terminals of a battery is 10.8 volts when there is a current of 4 
amperes in the battery from the negative to the positive terminal. When the current is 3 amperes in the 
reverse direction, the potential difference becomes 12.9 volts. 
(a) What is the internal resistance of the battery? 
(b) What is the emf of the battery? 
8. When the switch S is open in the circuit shown 
in Figure 31-12, the voltmeter V, connected 
across the terminals of the dry cell, reads 1.52 
volts. When the switch is closed, the voltmeter 
reading drops to 1.37 volts and the ammeter 
reads 1.5 amperes. Find the emf and the internal 
resistance of the cell, assuming that the volt-
meter reading with the switch open is the emf of 
the dry cell. 
Figure 31-12 
9. A series circuit consists of a 6 volt battery with an internal resistance of 0.4 ohms, a 3 volt battery with 
an internal resistance of 0.6 ohms connected so as to aid the 6 volt battery, and a resistor of 5 ohms. 
What will be the reading of a voltmeter requiring negligible power to operate connected across the 3 volt 
battery? 
10. A 600 ohm resistor and a 400 ohm resistor are connected in series across a 90 volt line. A voltmeter 
across the 600 ohm resistor reads 45 volts. 
(a) Find the voltmeter resistance. 
(b) Find the reading of the same voltmeter if connected across the 400 ohm resistor. 
11. In a resistor of 100 ohms, there is a constant current of 0.50 amperes for 3 minutes. 
(a) What is the power expended in the resistor? 
(b) How much energy is expended in the resistor? 
12. How would you change, with the help of a resistor, a 10 volt voltmeter of internal resistance 1000 ohms 
to an ammeter which has a full scale reading of 1 ampere. Show the circuit diagram and compute the 
magnitude of the resistor. 
13. The uniform wire of a slide wire Wheatstone bridge is 100 cm long, and a balance is obtained when the 
sliding contact is 25 cm from one end. If the known resistor of 200 ohms connects to the slide wire at the 
end nearest the sliding contact, what is the resistance of the unknown resistor? 

280 
Direct Electric Currents 
14. In the DC circuit shown in Figure 3 1 - 1 3 , 
J2= 1.5 amps 
h = 1 amp 
? , = 4 volts 
? 2 = 12 volts 
Γ2 = 2 ohms 
Ri = 2 ohms 
R2 = 2 ohms 
1?3 = 3 ohms 
1?4 = 3 ohms 
R5 = 1 ohm 
Figure 31-13 
Find: 
(a) the current J i , 
(b) the power expended in the resistance R2, 
(c) the potential difference Vat, 
(d) the internal resistance ri of the 4 volt battery, and 
(e) the terminal voltage across the battery in part (d). 
15. In the circuit shown in Figure 3 1 - 1 4 , 
Ji = 3 amps 
V,b = 2 1 volts 
Ri = 3 ohms 
J?2 = 6 ohms 
JR3 = 8 ohms 
1?4 = 4 ohms 
r = 2 ohms 
Figure 31-14 
Find: 
(a) the emf of the battery, 
(b) the power expended in the resistance Ru 
(c) the effective resistance of R2, Ri, and R4, 
(d) the potential difference Vcd between c and d, and 
(e) the current h through the resistance -R2. 
16. Draw the circuit diagram of a potentiometer, label emf's, currents, resistances, and use Kirchhoff's laws 
to write down equations from them from which solutions could be made for the currents in the circuit. 
17. A capacitor of 2 microfarads charged so that the potential difference across its terminals is 1 2 0 volts, is 
suddenly connected to a resistor of 2 5 0 , 0 0 0 ohms. What is the time constant of this circuit? 
18. A 5 m length of a potentiometer wire is required to balance the emf of a battery. When a 10 ohm resistor 
is connected across the battery, the length required for balance is 4 m. What is the internal resistance of 
the battery? 
19. The differential equation for the instantaneous charge Q of a capacitor after being disconnected from a 
source and connected to a resistance R is given by the equation 
Show that 
Q = CVe-
and 

Problems 
281 
20. In the circuit shown in Figure 31-15, V =40volts, R=4x 
10^ohms, and C = 10microfarads. 
(a) Find the initial value of the current when S i 
is closed, the maximum charge on the 
capacitor plates, and the time constant of 
the circuit. 
(b) After the capacitor is fully charged, 5 i is 
opened and Si is closed. Find the charge on 
the plates of the capacitor C 4 seconds 
after closing Si. 
Figure 31-15 

32 
Magnetic 
Fields 
32-1 
Magnetism 
283 
32-5 
The Cyclotron 
287 
32-2 
Magnetic Force on a Current 
283 
32-6 
Torque on a Loop, Dipole 
32-3 
The Biot-Savart Law 
285 
Moment 
288 
32-4 
The Magnetic Field of and the 
32-7 
Electric Meters 
288 
Force on a Moving Charge 
286 
32-1 
M A G N E T I S M 
We turn now to another facet of electromagnetic 
interactions—namely, magnetism and magnetic 
fields. Some knowledge of magnetism existed at the 
time of the ancient Greeks, about 600 B.C. Experi-
mental studies on magnetism began about the six-
teenth century. In 1600, William Gilbert published a 
collection of experimental facts about magnetism. 
We shall outline here the experimental facts known 
by Gilbert. 
It was observed that certain natural stones called 
lodestones possess the property of attracting pieces 
of iron. Also a piece of iron can be made into a 
magnet by stroking it with a lodestone. It was also 
found that the attractive force of a magnet is con-
centrated at regions called poles. When a magnet-
ized needle is freely suspended on a vertical axis, 
one end of the needle points north and the other 
end south. The end of the needle that points north is 
called the north-seeking pole or north pole. The 
other end is called the south-seeking pole or south 
pole. Like poles repel and unlike poles attract each 
other. The magnetic properties of a magnet can be 
destroyed in several ways, one of which is by heat-
ing it. 
The space outside a magnet in which its influence 
can be detected contains a magnetic field. The 
shape of the magnetic field can be mapped by scat-
tering iron filings in the vicinity of the magnet, as 
illustrated in Figure 32-1. The direction of the 
magnetic field at any point is chosen as the direc-
tion that a north pole of a freely suspended magnet, 
or compass needle, points. 
^ 
LJ. 
^ 
- 1 . 
Figure 32-1 Magnetic field in the vicinity of an iso-
lated bar magnet. 
32-2 
M A G N E T I C F O R C E O N A C U R R E N T 
The first experiments on the magnetic effects of 
currents were performed by Ampere in 1820. The 
283 

284 
Magnetic Fields 
/ 
/ 
/ 
\ 
/ 
\ 
Ν  
Fields 
add 
- / ·— 
Ν  
Fields 
oppose 
(a) Attraction 
\ 
\ 
Χ -, 
Μ  
\ 
Fields 
Fields 
add 
oppose 
Fields 
add 
N / 
Fields 
oppose 
(b) Repulsion 
Figure 32-2 Magnetic fields around two wires carrying current. 
work was continued by Oersted and others. They 
found that two long wires carrying currents in the 
same direction attract one another, whereas when 
carrying currents in the opposite direction they 
repel one another. 
The magnetic field around a wire carrying a cur-
rent forms closed circular loops around the wire. 
The direction is that in which a north pole of a freely 
suspended magnet or compass needle points, and 
can easily be remembered by means of the righi-
hand rule. With the thumb pointing in the direction 
of the conventional current, the fingers of the right 
hand encircling the wire point in the direction of the 
magnetic field. 
The reason why wires carrying currents attract or 
repel one another is best illustrated with the aid of 
Figure 32-2. In Figure 32-2(a), the currents are flow-
ing in the same direction (out of the paper, as illus-
trated in the figure by a dot · representing the head 
of an arrow). The fields oppose one another or are 
in the opposite duection in the space between the 
wires and reinforce one another or are in the same 
direction outside the wires. Therefore, the field is 
stronger outside the wires. If the field lines are con-
sidered to represent magnetic flux lines, it may be 
said that the magnetic flux density is greater outside 
the wires. It is experimentally observed that the 
wires move toward one another. In Figure 32-2(b), 
the currents are flowing in the opposite direction, as 
is illustrated in the figure by a dot representing the 
head of an arrow and a cross ®  representing the 
tail of an arrow. The fields here are in the same 
direction between the wires and in the opposite 
direction outside the wires. In this case, it is experi-
mentally observed that the wires move apart. 
Let us now consider a quantitative relation for 
the force on a current element in a magnetic field. 
Let an element of length dl which carries a current 
J, be placed in a magnetic field of flux density Β  at 
an angle θ to the direction of the field, as illustrated 
Figure 32-3 The force on a current element in a 
magnetic field. 

in Figure 32-3. The force on this element as inferred 
from experiments on a closed current loop is writ-
ten in terms of the vector-product or cross-product 
as 
dF = Id\xB, 
(32-1) 
where dF is the force on the element d\ in newtons, 
I is the current in amperes, Β  is the magnetic flux 
density or magnetic induction in newtons/ampere-
meter, and dl is the element of length in meters. 
The dh-ection of dF is perpendicular to the plane 
containing dl and B, and is directed into the paper. 
32-3 
T H E B I O T - S A V A R T L A W 
Experiments of Ampere and others showed that 
the basic relation for magnetic flux density or 
magnetic induction (B) at a point Ρ  produced by a 
current carrying element, as illustrated in Figure 
32-4, is given by 
d B = c i ^ . 
(32-2) 
where C is a constant of proportionality, r is the 
distance from the current carrying element to P, 
and fo is a unit vector in the direction of r. Equa-
tion (32-2) is known as the Biot-Savart law. 
The constant of proportionality C is given as 
47γ' 
where μ  is called the permeability constant. The 
permeability 
of free 
space 
is 
μ ο  = 47r x 
10"^nt/amp\ Now, Eq. (32-2) may be written as 
dB = ^ ^ ^ ^ . 
47Γ 
For the entire circuit, 
β ^μ ο /Xdlxro 
47Γ J r^ ' 
(32-3) 
(32-4) 
From Eqs. (32-1) and (32-3) the force on an ele-
ment of current Jidli, due to the magnetic induction 
from another current element /2dl2, is 
dF, = /,dl, 
x d B 2 , 
where 
Then 
d B 2 = i^/: 
47r 
d l z X r o 
The Biot-Savart Law 285 
(inward) 
1 
r 
θ 
y 
d\ 
Wo 
Figure 32-4 Magnetic flux density dB at a point Pdue 
to a current element dl. 
or 
d F , = ^ ¿ : f [dl,x(dl2Xro)]. (32-5) 
Example 1. What is the magnitude of the force 
per unit length on a long wire carrying a current of 
1 ampere placed parallel to and 1 m away from 
another long wire carrying the same current? 
SOLUTION 
We shall first find an expression for |B| at a 
perpendicular distance R meters from a long 
straight wire. 
From Eq. (32-3), the magnetic flux density at Ρ  due 
to the current element d l has the magnitude 
I^BI ^ μ ο  IJdl X fol 
The magnitude of the total magnetic flux density at 
Ρ  is then 
47Γ 
I I _ f" 
μ ο / 8 ί η Μ  

286 
Magnetic Fields 
Here, 
and 
Therefore, 
dl sin θ = rde 
R 
^ 
sino' 
Jo 4 
4 ^ Ä t - c o s e ] „ ' 
Now we shall consider the two parallel wires as 
shown 
The magnetic flux density a distance R from wire 
(2) has the magnitude 
The force due to B2 on dl of wire (1) is 
dF = J,dlxB2 
= n l , l . ^ d l , 
where η  is a unit vector along R pointing toward 
wire (2). [If the currents were in the opposite direc-
tion, π  would be negative or pointing away from 
wire (2).] Therefore, 
and 
/ 
" 27Γ1? 
With / i = J2 = 1 ampere and R = 1 m, 
μ ο υ 2 ^ /47Γ  X 10"^nt/amp'\/(l amp)(l amp)\ 
I 
"^ITTR 
\ 
27Γ 
A 
(Im) 
/ 
= (2xl0-' nt/m)n. 
32-4 
THE M A G N E T I C FIELD O F A N D T H E 
F O R C E S O N A M O V I N G C H A R G E 
If a charge q is displaced an amount dl with a 
velocity v, we have 
/ d l = ( ^ ) v d í = díjv. 
The Biot-Savart law, namely. 
μ ο  Jdl X ro 
becomes 
_ 
f μ o J d l X ro _ 
f μ odq(\xro) 
For a point charge, ν  and r are constants in the 
integration over dq, and we have 
ρ _ μ ο ς ( ν Χ
Γ ο ) 
47Γ Γ ' 
· 
(32-7) 
The force on a current element in an external 
magnetic field is given by Eq. (32-1), namely, 
dF = J d l x B . 
For a moving charge, Id\= dqy and 
F = I d p v x B 
or 
F=<?(vxB). 
(32-8) 
If an electric field is also present, the total elec-
tromagnetic force on q is 
F = q(E + vxB), 
(32-9) 
which is often called the Lorentz force law. Con-
sider the motion of a charge in a uniform magnetic 
field of flux density B. If the charge is initially mov-
ing in a plane normal to B, then the force on it 
(assuming Ε = 0) is also in this plane and perpen-
dicular to its direction of motion. The charge, there-
fore, will move in a circle in this plane. 
By Newton's second law, F = ma. Since a = v'lr 
and F = qvB, we obtain 
mv = qvB, 
(32-10) 
where m is the mass of the charged particle and r is 
the radius of the circle. The radius of the circle r 
and the angular velocity ω  of the particle will be 
given by 
mv 
r = 
(32-11) 

The Cylotron 
287 
and 
The equation shows that the angular velocity 
and, hence, the time taken to make one revolution 
is independent of the velocity of the particle so long 
as the relativistic change of mass with velocity is 
neglected. 
32-5 
T H E C Y C L O T R O N 
The cyclotron, an ion accelerator developed by 
E. O. Lawrence in 1931, illustrates Eqs. (32-10) to 
(32-12). A cyclotron consists of an evacuated 
chamber placed between the poles of a large elec-
tromagnet. The chamber contains two semicircular 
hollow electrodes called "dees" as illustrated in 
Figure 32-5. A high frequency alternating potential 
difference is applied to the dees, which produces an 
alternating electric field between them. Ions enter 
the chamber at the ion source F placed between the 
dees. 
The ions are accelerated across the gap by the 
electric field between the dees, but are shielded in-
side from the action of the field. The magnetic field 
causes each moving ion to follow a semicircular 
path, the radius of which depends on the velocity, 
mass, and charge of the ion, and the magnetic field 
strength according to Eq. (32-11). The velocity of 
the ion increases each time it crosses a gap. As the 
velocity increases, the radius of the ion path in-
creases. After many trips around, the high energy 
particles are ejected out of the window W. From 
Eq. (32-11), the velocity of the particles are given 
by 
m 
(32-13) 
and the final energy of the ions will be given by 
(32-14) 
One might think that by increasing the radius that 
an unlimited energy could be obtained. This is not 
so, since when the velocity of the ion approaches 
the speed of light, the relativistic increase in mass 
causes the ions to cross the space between the dees 
out of step with the changing electric field. As a 
result, the gain in velocity and, thus, in energy does 
not occur. 
Ν  
F ΑΑΑΑΛ  
Evacuated 
chamber 
Side view 
Filament 
\ 
/ (ion source) 
Top view 
High voltage, 
high frequency 
generator 
Figure 32-5 Schematic diagram of a cyclotron. 

288 
Magnetic Fields 
Figure 32-6 Rectangular loop in a magnetic field. 
32-6 
T O R Q U E O N A LOOP, DIPOLE 
M O M E N T 
We discuss now the torque on a current carrying 
wire loop in a magnetic field, which is the principle 
underlying the operation of electric meters. Con-
sider a rectangular loop, carrying a current /, with 
sides of length / and d situated in a magnetic field 
of uniform flux density B, as illustrated in Figure 
32-6. 
The force on any element of the loop is 
dF = JdlxB. 
The component of the force perpendicular to the 
loop F„  is 
F„ = F cos Ö = IB cos θ 
IBl cos Θ , 
where θ is the angle between the plane of the loop 
and the direction of B. The magnitude of the total 
torque τ  on the loop is given by 
τ  = F„d = IBld cos Θ . 
Since Id equals the area of the loop A , we can write 
τ  = IBA cos θ. 
The product lA is defined as the magnetic dipole 
moment (m) of the loop. Since m = JA, the torque 
can be expressed as 
τ  = mB cos θ = mB sin φ . 
where φ  is the angle between the normal to the 
plane of the loop and the direction of B. 
If the magnetic moment is regarded as a vector m 
with direction n, where π  is a unit vector peφ en-
dicular to the plane of the loop, the torque relation 
can be expressed in the form 
where 
τ  = m X B, 
m = n m 
= 
n l A 
(32-15) 
(32-16) 
is the magnetic dipole moment of the loop. The di-
rection of m is determined by the right-hand rule 
for a loop or coil. Grasp the loop with the fingers of 
the right hand following the loop and pointing in the 
direction of the current flow, then the extended 
thumb points in the direction of the magnetic field 
due to current in the coil or in the direction of the 
magnetic dipole moment m. The effect of the torque 
is to rotate the loop toward its equilibrium position 
with its magnetic moment m in the same direction 
as the field B. 
32-7 
E L E C T R I C M E T E R S 
Meters make use of the torque on a current 
carrying coil in a magnetic field. Figure 32-7 is a top 
view of a meter movement. 
The pole pieces are shaped so that the field is 
parallel to the plane of the coil for any position of 
the coil. When the ciurent passes through the coil. 

Electric Meters 
289 
Figure 32-7 Schematic diagram of a meter move 
ment. 
the magnitude of the torque on the coil is given by 
the relation 
τ  = NBIA, 
(32-17) 
where Ν  is the number of turns on the coil. The coil 
will turn until the torsional restoring torque due to 
the spring balances the magnetic torque τ . The re-
storing torque is proportional to the angle of deflec-
tion φ  and can be put equal to Κ φ , where Κ  is the 
torsion constant of the spring. Therefore, the coil 
comes to rest when 
Κ φ  = NBIA 
or 
Φ  = NBIA 
Κ  
' 
(32-18) 
Since N, B, A, and Κ  are all constants for a 
particular meter, 
φ  = € Ι 
(32-19) 
or the deflection of the pointer is proportional to the 
current. Galvanometers, which are instruments 
used to measure currents, work on the above prin-
ciple. Therefore, Figure 32-7 also shows the rudi-
ments of a galvanometer. 
A voltmeter used to measure potential differ-
ences in a circuit is a galvanometer with a large 
number of turns on the moving coil so that there is a 
large deflection of the coil for small currents. It 
must be designed to draw as little current as pos-
sible so as not to affect signiñcantly the current in 
the rest of the circuit. A high resistance is put in 
series with the coil as illustrated in Figure 32-8. 
The current through the coil is given by 
A V 
R + r' 
Figure 32-8 Internal connections of a voltmeter. 
series resistance. Since the deflection is propor-
tional to I, it is also proportional to AV. 
The range of the meter can be modified by vary-
ing the resistance JR. If we wish the meter to have a 
full scale reading A V/ with a current /„, through the 
meter movement, we must have a series resistance 
R so that 
r + R 
or the series resistance R required is 
(32-20) 
(32-21) 
An ammeter, which is used to measure currents 
in a circuit, is a galvanometer with a very low resis-
tance, so as to cause a low potential drop across the 
instrument. This is achieved by putting a low resis-
tance (R) in parallel or shunt with the coil of the 
meter as illustrated in Figure 32-9. Only a small 
fraction of the current goes through the coil so a 
coil with a large number of turns is necessary. 
where r is the resistance of the coil and R is the 
Figure 32-9 Internal connections of an ammeter. 
As in the case of the voltmeter, the range of the 
meter can be modified by varying the resistance R. 
The shunt resistance R required to give a full-scale 
reading If with a current I through the shunt resis-
tance and a current L through the meter movement 
is determined as follows: Since the same potential 
difference exists across the meter movement and 

290 
Magnetic Fields 
the shunt resistance, 
Lr 
= 
LR, 
where r is the resistance of the meter movement or 
coil. Also, 
Solving these two equations for R, gives 
^ = 7 - ^ . 
(32-22) 
The total resistance of the meter is the sum of R 
and r in parallel and is less than R, which is much 
less than the resistance r of the meter movement. 
Example 2, The moving coil of a meter move-
ment of height 3 cm and width 2 cm has 100 turns 
and is pivoted in a field of 5(X)0 nt/amp-m. 
(a) Calculate the torque when 2 milliamperes 
flow in the circuit. 
(b) If the movement is controlled by springs that 
exert a restoring torque of 0.02 nt-m/deg, what is 
the steady deflection produced by a current of 2 
milliamperes? 
SOLUTION 
(a) Assume that the field is parallel to the plane 
of the coil for any position of the coil. Apply Eq. 
(32-17), 
T = NBIA 
= (100)(5000 nt/amp-m)(2 x 10' amp) 
X (0.03 m)(0.02 m) 
= 0.6 nt-m. 
(b) Deflection produced by the current of 2 mil-
liamperes is obtained by applying Eq. (32-19), 
Φ
= ^ = 
0.6 nt-m 
C 
0.02 nt-m/deg = 13 deg. 
Example 3. A meter movement has a resistance 
of 100 ohms and takes 3 milliamperes for full scale 
deflection. 
(a) Calculate the series resistance necessary if 
the meter is to act as a voltmeter having a full scale 
deflection of 6 volts. 
(b) Calculate the shunt resistance necessary if 
the meter is to be used as an anuneter with a full 
scale deflection of 2 amperes. 
SOLUTION 
(a) Applying Eq. (32-21), 
i m 
6 volts 
3 X 10"' amp -100 Ω  =1900 ohms. 
(b) Applying Eq. (32-22), 
Lr 
3x 10'ampX 100Ω  
, 
^ = V ^ = 
(2-3xlO-')amp 
=0.15ohms. 
P R O B L E M S 
1. What is the force on a conductor of length 15 m carrying a current of 5 amperes in a field of 4 nt/amp-m 
at right angles to the conductor? 
2. Calculate the force on the conductor of Problem 1 if the conductor is 
(a) parallel to the field and 
(b) 30° to the direction of the field. 
3. With 10 closely spaced turns carrying a current of 5 amperes, a square coil 40 cm x 40 cm is suspended 
vertically in an east-west plane in a region, where there is a uniform horizontal magnetic flux density B, 
of 1.5 nt/amp-m directed from west to east. What is the torque acting on this coil? 
4. A uniform magnetic field of 2 nt/amp-m is directed vertically upward. A proton (charge = + 1.60 x 
10'^ coul) is directed into this field at a speed of 3 x 10^ m/sec. Describe quantitatively how it will move 
if it is projected 
(a) eastward in a horizontal plane, 
(b) northward in a horizontal plane, 
(c) vertically downward, and 
(d) at an angle of 30° with the upward direction of B. 

Problems 
291 
5. A single flat loop of area 25 cm^ lies in the x-y plane, carrying a current of 5 amperes, clockwise as 
viewed from above. A uniform magnetic field of density 4nt/amp-m is in the positive ζ -direction. 
(a) What is the torque exerted by the magnetic field on the loop? 
(b) What is the magnetic moment of the loop? 
6. A long, straight wire carries a current of 1.5 amperes. An electron (charge = - 1.60 x 10"'^ coul) travels 
with a velocity of 5 x 10^ cm/sec parallel to the wire, and in the same direction as the conventional cur-
rent. What is the force (magnitude and direction) exerted on the electron by the magnetic field? 
7. A circular loop of 0.2 m^ area in the x-y plane carries a clockwise current of 4 amperes. 
(a) What is the magnitude and direction of its magnetic moment? 
(b) What is the torque on the loop due to a uniform magnetic field of 0.25 nt/amp-m parallel to the x-z 
plane and making an angle of 30° with the positive χ -direction? 
8. (a) What is the maximum torque on a rectangular galvanometer coil 5 cm x 12 cm of 600 turns 
when carrying a current of 10"^ amperes in a field of Β  = 0.10 nt/amp-m? 
(b) What is the torque on the coil when the normal to the plane of the coil makes an angle of 60° with the 
direction of B? 
9. (a) What current must an infinitely long straight wh-e carry in order that Β  = 2.5 x lO"'* nt/amp-m at a 
distance of 2 cm from it? 
(b) What current must a circular coil of 100 turns and 10 cm radius carry in order that Β  = 0.001 nt/amp-
m at its center? 
10. A long, straight wire carries a current of 100 amperes. 
(a) What is the magnetic flux density at a point 5 cm from the wire? 
(b) What is the force per unit length on another wire parallel to and 5 cm from it carrying a current of 
10 amperes in the opposite direction? 
11. A straight wire 25 cm long is at right angles to a uniform magnetic field of 0.7 nt/amp-m. What current 
must flow in the wire such that the force on it be 0.4 nt? 
12. For the meter of Problem 13, calculate the shunt resistance necessary if the meter is to be used as an 
ammeter with a full scale deflection of 1 ampere. 
13. A meter movement has a resistance of 50 ohms and takes 1.2 milliamperes for full scale deflection. Cal-
culate the series resistance necessary if the meter is to act as a voltmeter having a full scale deflection of 
3 volts. 
14. The magnetic field density in a cyclotron, which is accelerating protons (charge = 
1.60x 10"'^ coul, 
mass = 1.67 X 10'^^ kg), is 2 nt/amp-m. 
(a) Calculate the number of revolutions per second of the protons. 
(b) If the maximum radius of the cyclotron is 0.30 m, what is the maximum kinetic energy of the 
proton? 
(c) Through what potential difference would the proton have to be accelerated in order for it to have the 
same kinetic energy obtained as with the cyclotron? 
15. What shunt resistance is required in a milliammeter with a movement resistance of 100 ohms giving full 
scale deflection for 1 milliampere in order for it to give full scale deflection for 60 milliamperes? 
16. What is the velocity of a beam of electrons (charge on electron = - 1.60 x 10"'^ coul) when the simul-
taneous influence of an electric field of intensity 34 χ  10"* volts/m and a magnetic field of flux density 
2 X 10"^ nt/amp-m, both fields being normal to the beam and to each other, produces no deflection of the 
electrons? 
17. The resistance of a galvanometer is 30 ohms and the current for full scale deflection is 10 milliamperes. 
(a) What shunt resistance is required to make it into an ammeter of 1 ampere full scale deflection? 
(b) What series resistance is required to make it into a voltmeter of 150 volts full scale deflection? 
18. How would you change a 10 volt voltmeter with an internal resistance of 1000 ohms to an ammeter of 
current carrying capacity 1 ampere with the help of a resistor? Compute the magnitude of the resistor. 

292 
Magnetic Fields 
19. In Figure 32-10, Ρ  and Q are cross sections of 
two long, parallel, straight wires perpendicular 
to the plane of the paper with Ρ  carrying a 
current of 20 amperes into the plane of the 
paper and Q carrying an equal current out of 
the plane of the paper. 
Compute the magnetic flux density (B) at 
point R, 
Figure 32-10 
20. The earth's magnetic field at a certain location has a northerly component of BN = 0.1 x 10"* nt/amp-m 
and an upward component of Bv = 0.4 x 10"* nt/amp-m. Calculate the magnitude and dúection of the 
force a straight 1 m long wire would experience, if it were oriented parallel to the south-north direction 
and was carrying a current of 10 amperes toward the north. 

33 
Magnetic Field 
of a Current 
33-1 
Magnetic Flux 
293 
33-5 
Energy and the Magnetic Field 
296 
33-2 
Ampere's Law 
293 
33-6 
Generators and Motors 
297 
33-3 
Electromagnetic Induction; 
33-7 
Series Circuit with Resistance 
Faraday's Law and Lenz's Law 
294 
and Inductance 
299 
33-4 
Mutual Inductance and Self-
Inductance 
296 
In this chapter, we shall discuss magnetic fields and 
energy stored in a magnetic field along with other 
topics. In addition, we shall point out the strong 
analogy with electric fields. 
33-1 
M A G N E T I C FLUX 
In Chapter 32, we mentioned briefly the terms 
magnetic flux and magnetic flux density. However, 
more should be said about both these terms. The 
total number of magnetic field lines that cut any 
area A as illustrated in Figiu-e 33-1 is the magnetic 
flux (Φ ) through the area, that is. 
φ =^ 
B n d A , 
(33-1) 
where η  is a unit vector perpendicular to A. 
The unit of φ  in the MKS system of units is the 
weber, which is equal to a nt-m/amp, and the unit of 
the flux density Β  is webers/m'. 
Experimentally, when we plot the magnetic field 
about a wire as illustrated in Figure 32-2, we see 
that the magnetic flux lines are closed lines. This 
implies that if a closed surface A is taken in the 
field, every line entering the surface must leave it. 
This means that if we calculate the net flux through 
the surface by integrating over the surface, the 
Figure 33-1 Magnetic flux through an area A 
value of the integral must be zero. 
B n d A =0. 
33-2 
A M P E R E ' S L A W 
In this section, we will discuss the relationship 
between the line integral of Β  around any closed 
293 

294 
Magnetic Field of a Current 
path and the current I enclosed by the path. In the 
example given in Chapter 32, we saw that the mag-
nitude of Β  at a point a distance R from an infinitely 
long straight wire was given by 
IITR' 
and that the direction of Β  was peφ endicular to the 
wire. Now let us calculate the integral of Β  along a 
circular arc of radius R about a ciurent carrying 
wue as illustrated in Figure 33-2. 
B / 
A 
\ 
^^Λ ι 
— 
¿>^^Λ \ 
—  
Figure 33-2 A circular path for integrating Β around a 
wire. The current links the path of integration. 
Referring to Figure 33-2, 
dl = Rde 
2 7 Γ 
' 
which is independent of J? or 
B - d l = M „ / ^ . 
Integrating completely around the circle gives 
> B d l = M„I. 
(33-2) 
This equation states that the integral of Β  · dl 
around any closed path is equal to μ ο  times the 
current passing through any area bounded by the 
path of integration. This equation is called Am-
pere's law. If there is no enclosed ciurent, 
^ B d l = 0. 
(33-3) 
It should be pointed out that the above derivation 
is not a general proof of Ampere's law. A general 
proof is beyond the scope of this book. The student 
will have to accept here that the law is true in 
general for any magnetic field shape, for any as-
sembly of currents, and for any path of integration. 
33-3 
E L E C T R O M A G N E T I C 
I N D U C T I O N ; 
F A R A D A Y ' S L A W A N D L E N Z ' S L A W 
Faraday showed experimentally that when the 
magnetic flux through a closed circuit is changing, 
there is an emf induced in the circuit that is directly 
proportional to the time rate of change of the flux. 
In analytical form, Faraday's law states that 
' dt' 
(33-4) 
where ^ is in volts, φ  is in webers, and t is in 
seconds. The minus sign is an indication of the di-
rection of the induced emf. The direction of the 
induced emf is a consequence of the principle of 
conservation of energy. The induced current al-
ways flows in such a direction that its magnetic field 
opposes the change in conditions giving rise to the 
induced current. This statement is known as Lenz's 
law. The induced current is set up by the induced 
emf. 
We shall now consider some experimental verifi-
cations of both Faraday's law and Lenz's law. First, 
we shall consider a current in a coil of wire induced 
by a moving magnet, as illustrated in Figure 33-3. If 
the north pole of the magnet is pushed into the coil, 
the induced current will flow in such a dúection so 
as to set up a magnetic field that opposes the inser-
tion of the magnet, as illustrated in Figure 33-3(a). 
If the magnet is pulled out of the coil, the induced 
current flows in the opposite direction, or in such a 
direction that its magnetic field opposes the with-
drawal of the magnet, as illustrated in Figure 
33-3(b). The magnitude of the induced current will 
be proportional to the speed with which the magnet 
is pushed into or pulled out of the coil. 
Let us consider for the moment the reason for 
the above experimental facts based on energy con-
servation. The coil has a resistance and the induced 
current produces heat, which requires a source of 
energy. The only other source of energy is the 
mechanical work done on the magnet in pushing it 
into the coil. In order for work to be done on the 
magnet, it must experience an opposing force as it 
enters the coil. This means that the field produced 
by the current in the coil must be in such a direction 

Electromagnetic Induction: Faraday's Law and Lenz's Law 
295 
Figure 33-3 Currents induced by moving magnets. 
Figure 33-4 Currents induced by currents. 
Figure 33-5 Currents induced by closing and breaking the circuit. 
as to repel the magnet. If the magnet were attracted 
by the field in the coil, mechanical energy and heat 
energy would be produced by the system without 
any work being done on it, which is a contradiction 
of the principle of conservation of energy. A similar 
argument can be presented for the dh-ection of the 
induced current when the magnet is withdrawn 
from the coil. 
If we insert or withdraw a current carrying coil 
instead of a magnet into another coil, an induced 
current will flow in the other coil. The magnitude 
and direction of the induced current will be gov-
erned by Faraday's and Lenz's laws and the effects 
are the same as with the magnet. This is illustrated 
in Figure 33-4. 
Let us consider what happens if, instead of in-
serting and withdrawing the current carrying coil 
into the larger coil, we leave it inside the larger coil 
and open and close the circuit leading to the inside 
coil as illustrated in Figure 33-5. It is then observed 

296 
Magnetic Field of a Current 
that the induced current flows in the opposite direc-
tion to the current producing it when the circuit is 
closed, and in the same direction when the circuit is 
broken. Why? 
33-4 
MUTUAL I N D U C T A N C E A N D 
S E L F - I N D U C T A N C E 
Let us next consider two coils arranged so that 
part of the flux produced by a current in the ñrst 
(primary) coil threads the other (secondary) coil. 
What emf will be induced in the secondary coil by 
this flux linkage with the primary coil? (If the cur-
rent in the primary coil is varied, an induced emf is 
set up in the secondary coil.) In general, the emf 
induced in the secondary coil is directly propor-
tional to the time rate of change of the current in 
the primary coil. Symbolically, 
We can write this as an equation by introducing a 
proportionality constant 
dt' 
(33-5) 
where Μ  is the coefficient of mutual inductance of 
the two coils. The minus sign indicates that the emf 
in the secondary coil is opposite in direction to the 
emf in the primary coil. 
Two coils have a mutual inductance of 1 henry if 
a change in current of 1 ampere per second in the 
primary coil induces an emf of 1 volt in the secon-
dary coil. 
The value of % is also given by Eq. (33-4). 
Equating Eqs. (33-4) and (33-5) gives 
dt 
dt 
or 
dip 
(33-6) 
where dφ s is the total flux linkage between the two 
coils produced by the current change dip in the 
primary coil. 
An induced emf is also set up in single coil by a 
current change in the coil. The induced emf op-
poses the change in current and is sometimes called 
a "back e m f (^b). It is proportional to the time rate 
of change of the current, which permits us to write 
the following equation: 
(33-7) 
where L is the coefficient of self-inductance of the 
coil and is expressed in henrys. 
A coil has a coefficient of self-inductance of 1 
henry if a change in the current of 1 ampere per 
second induces a back emf of 1 volt in the coil. 
The value of ^j, is also given by Eq. (33-4). 
Equating Eqs. (33-4) and (33-7) gives 
^~ 
dV 
(33-8) 
where dφ  is the total change in flux and dl is the 
total change in current. 
33-5 
E N E R G Y A N D THE M A G N E T I C FIELD 
When we build up a current in an inductor, an 
external source of emf must do work against the 
induced emf. This suggests the possibility of energy 
storage in the inductor. Neglecting resistance, the 
rate at which the external source does work is 
P = - f J = L 7 f . 
The total work done in building the current up to 
a value of / amperes in time t is 
or 
W = \ L I \ 
Idl 
(33-9) 
As an example, consider a solenoid as illustrated 
in Figure 33-6. At a point inside the solenoid and 
sufficiently far from the ends, the magnetic field is 
parallel to the axis and uniform. At a point outside 
the field is very weak. Let us calculate the line in-
tegral of the induction around the rectangular path 
shown in Figure 33-6. Over the two radial sides, the 
path is perpendicular to the field, so their contribu-
tion is zero. The same will be true for that part of 
the field that is parallel to the axis, outside, since 
the field there is very weak. The only contribution 
then comes from that part of the path of length d 
parallel to the axis and inside the solenoid, where 
Φ Β' d\ = Bd = ^ondl 
or 
Β = μ ο η ΐ, 
(33-10) 

Generators and Motors 
297 
energy per unit volume stored in the solenoid is 
Figure 33-6 Magnetic field in a solenoid. 
where η  is the number of tiwns of wire per unit 
length on the solenoid. The induced emf in the sol-
enoid is 
where I is the length of the solenoid. The magnetic 
flux through the coil from Eq. ( 3 3 - 1 0 ) is 
φ  = ^ Β ' η ά Α = Β Α  
= μ ο η ΙΑ , 
( 3 3 - 1 2 ) 
where A is the cross-sectional area of the solenoid. 
From Eq. ( 3 3 - 1 1 ) , the induced emf is 
21A 
dl 
= - μ ο η  
lA-^, 
( 3 3 - 1 3 ) 
Comparing Eqs. ( 3 3 - 1 1 ) and ( 3 3 - 1 3 ) , we obtain a 
relation for the self-inductance of the solenoid 
dl 
dl 
—  = - μ , η  
l A ^ 
or 
L=μ onHA. 
dt 
( 3 3 - 1 4 ) 
If we substitute Eq. ( 3 3 - 1 4 ) into Eq. ( 3 3 - 9 ) , we get 
the energy stored in the solenoid to be 
Ψ  = 
^μ ο η Η Α Ι\ 
( 3 3 - 1 5 ) 
From this, we can conclude that this energy is 
stored in the magnetic field. It should be pointed out 
that this is not a proof. However, all experimental 
work supports it. Since the quantity lA is the vol-
ume of the solenoid, the energy density or the 
W 
1 
2T2 
Since 
from Eq. ( 3 3 - 1 0 ) , 
Β = 
μ ο π ΐ, 
W^IB^ 
V 
2 μ ο * 
( 3 3 - 1 6 ) 
We may, therefore, state that the energy density 
stored in a magnetic field is given by Eq. ( 3 3 - 1 6 ) . 
This is the analogue of the expression for the 
energy density in an electric field of a capacitor 
l e o E ^ with free space as a dielectric. 
33-6 
G E N E R A T O R S A N D M O T O R S 
Generators and motors both make use of a coil 
called the armatiue, rotating in a magnetic field. 
The magnetic field of a generator is usually created 
by electromagnets, which consist of soft iron 
wound with many turns of wire called the field coils 
through which pass an electric current. A generator 
is a machine which, when driven mechanically, gen-
erates an electromotive force, while a motor is a 
machine that does mechanical work at the ex-
pense of electrical energy. 
We shall first discuss the operation of two prin-
ciple kinds of generators. These two kinds are: al-
ternating current ( A C ) and direct current ( D C ) . If a 
single coil is placed between the north and south 
poles of two magnets and rotated, an induced emf is 
set up in the coil as illustrated in Figure 3 3 - 7 . The 
emf and, consequently, the current changes direc-
tion every half revolution, and an alternating cur-
rent is generated. Figure 3 3 - 8 is a graph of the 
induced emf versus position of the coil as deter-
mined in Section 3 2 - 6 . If an alternating current is 
desired in the external circuit, the current is re-
moved from the coil by collecting rings and 
brushes, as illustrated in Figure 3 3 - 9 , where ß i and 
0 2 are the brushes and C i and C 2 are the collecting 
rings. The armature of a practical alternating ciu-
rent generator consists of a large number of coils. If 
a direct current is desired in the external circuit, a 
commutator and brushes are required to remove 
the current, as illustrated in Figure 3 3 - 1 0 . The com-
mutator C is a divided ring. The segments are insu-
lated from one another, and the brushes Bi and B2 
are so arranged that they slip from one segment to 

298 
Magnetic Field of a Current 
Figure 33-7 Rotating coil in a magnetic field. 
_Position 
1° of coil 
Figure 33-8 External emf versus position of coil for a 
coil rotating in a magnetic field. 
Figure 33-9 Simple alternating current generator. 
the other at the instant the current in the coil of the 
armature changes direction. Figure 33-11 is a graph 
of the emf in the external circuit versus position of 
the coil when such a commutator is used. 
To obtain a practically steady direct current from 
a dh-ect current generator, a large number of coils 
are placed on the armature, and the commutator is 
divided into a correspondingly large number of seg-
Figure 33-10 Simple direct current generator. 
Ε 
φ  
lö  
ι 
Χ
/ ^ 
. 
Λ . 
Position 
Ε 
Φ  0° 
90** 
180*» 
270*» 
seo*» 
of coil 
ώ  
Figure 33-11 External emf versus position of coil for 
a single coil rotating in a magnetic field when a 
commutator is used. 
ments. Figure 33-12 is a graph of the emf versus 
position of the armature for a direct current gen-
erator with a number of coils on the armature. 
We shall now briefly discuss electric motors. An 
electric motor is the reverse of an electric gen-
erator. Any direct current generator can serve as a 
direct current motor because if a current is passed 
through the armature from an external source there 

Series Circuit witti Resistance and Inductance 
299 
Resultant emf 
Ε  
g V.—>^ 
0) 
ννν\ 
, 
Position 
C 
θο^'Χ ^ιβο^ 27oy 7360° 
of coil 
Φ  
Χ  
UJ 
Figure 33-12 External emf for a four coil armature 
with a commutator. 
will be a force on each armature coil and the coil 
will start to turn. Some motors, called universal 
motors, will run either on alternating or direct cur-
rent. In order to have a motor run on alternating 
current, the armature and field currents must be in 
phase with each other so that they reverse at the 
same instant. 
In a motor, in addition to the applied emf, we 
have an induced or back emf when the armature 
rotates, since the coils are cutting across lines of 
magnetic flux. This back emf opposes the im-
pressed or applied emf. If V is the external applied 
voltage, and ^j, the back emf, then the resultant 
voltage available to cause a current in the armature 
is V - 
Therefore, 
ν -Έ ,=ΙΚ  
or 
V = ^t,+/I?, 
(33-17) 
where I is the current in the armature, and R is the 
resistance of the armature coils, or the sum of the 
resistance of the armature and field coils, depend-
ing upon the kind of motor. 
33-7 
S E R I E S CIRCUIT WITH R E S I S T A N C E 
A N D I N D U C T A N C E 
We must now discuss the effects that the inser-
tion of an inductor in a circuit entails. Consider an 
inductor of inductance L in series with a resistor of 
resistance R connected to a battery of voltage V. 
V = 
IR+L^ 
at 
or 
Ldl 
= dt. 
V-IR 
Integrating both sides, we obtain 
(33-18) 
(33-19) 
-^\n(V-IR) 
= j - - ^ constant. 
(33-20) 
Let us assume that there is no current flowing at 
ί = 0. We then obtain the constant of integration, 
and Eq. (33-20) becomes 
Rt 
l n ( V - J I ? ) - l n V = - - ^ , 
or 
In V-IR^ 
Rt 
V 
L' 
(33-21) 
If we now take the antilogarithm of both sides of 
Eq. (33-21), we obtain 
' 
V 
^ 
or 
(33-22) 
Equation (33-22) shows that the current increases in 
the same way as the charge on a capacitor in the 
R-C 
circuit. The quantity L/R 
is called the time 
constant (τ ) of the inductor, and Eq. (33-22) may be 
written 
/ = | ( i - ^ 
η  
(33-23) 
The larger the value of τ , the slower is the build up 
of current in the inductor. 
If, after a steady ciurent is reached in the induc-
tor, the battery or inductor is shorted out, then V = 
0, and Eq. (33-19) becomes 
Ldl 
IR = dt. 
(33-24) 
which can be solved in a similar manner to that 
above to give 
(33-25) 
Equation (33-25) shows that the current decreases 
exponentially to zero, and that this decay is gov-
erned by the same time constant τ = LIR. 
Example 1. 
(a) Find the flux through a coil of 10 turns of 
average diameter 2 cm when placed in a field of 
lOOOwebers/m'. 
(b) If the field strength is reduced to 200 
webers/m^ in 1 second, what is the induced emf? 

300 
Magnetic Field of a Current 
SOLUTION 
(a) The total flux is 
φ  = ΒΑ = (1000 webers/m'XirXíO.OD'm') 
= 0.314 Webers. 
(b) The induced emf is 
Δ ί 
Δ ί 
^ (10)((1000 - 200) webers/m^)(7r(0.01)W) 
1 sec 
= 2.51 volts. 
Example 2. A long solenoid of cross-sectional 
area 10 cm^ is wound with 10(X) turns per meter. The 
windings carry a current of 0.50 amperes. A secon-
dary winding of 10 turns encircles the solenoid. 
When a switch in the circuit providing the current 
to the solenoid is opened, the magnetic field of the 
solenoid becomes zero in 0.10 seconds. What is the 
average induced emf in the secondary? 
SOLUTION 
The change in field through the solenoid when the 
switch is opened is 
Δ β = μ ο η 7 - 0 
= (47Γ X 10' nt/amp')(1000 m-')(0.50 amp) 
= 6.28x 10"'webers/m'. 
The induced emf in the secondary is 
/6.28xl0-'webers/m'\ 
^=N.A|f=(10)(10-m^)(^ 
= 6.28x10' volts. 
(0.10 sec) 
Example 3. An inductor of inductance 2 henrys, 
and resistance 20 ohms, is connected to a 6 volt bat-
tery of zero internal resistance. Find: 
(a) the initial rate of current increase in the cir-
cuit. 
(b) the rate of increase of current at the instant 
when the current is 0.2 amperes, 
(c) the current 0.1 seconds after the circuit is 
closed, and 
(d) the final constant current. 
SOLUTION 
(a) Apply Eq. (33-18) in the form 
dt' 
V 
L" 
m 
L' 
At 
and 
dl 
dt' 
i = 0 , 
1 = 0, 
V 
6 volts 
L 
2 henrys 
(b) At / = 0.2 amperes. 
= 3 amp/sec. 
dl^V 
IR^( 
6vohs \ 
["(0.2amp)(20ohms) 
dt 
L 
L 
\2 henrys/ 
L 
2 henrys 
= 1 amp/sec. 
(c) The current 0.1 seconds after the switch is 
closed is obtained by applying Eq. (33-22). 
6 volts 
2 ohms (l-e 
-[(20 
ohmsXO.l sec)/2 henrys]\ 
= 3^1 - 
amp = 1.90 amp. 
(d) The final constant current is obtained by sub-
stituting ί = 00 into Eq. (33-22), which gives 
_ 
V 
6 volts 
^ = Ä = 2 ^ = ^^"^P-
P R O B L E M S 
1. A proton of mass = 1.67 x 10"^' kg, charge = 1.60 x 10'^ coul, moves in a ckcular path of 12 cm radius 
perpendicular to a uniform magnetic field. The velocity of the proton is 1 x 10' m/sec. What is the total 
magnetic flux enckcled by the circular path of the proton? 
2. Derive an expression for the magnetic flux density inside a long, straight conductor of ch-cular cross 
section, having a radius JR. The current is uniformly distributed over the cross-sectional area of the 
conductor so that the current density j = HTTR^. 
3. (a) Derive an equation for Β  at the center of a circular coil of radius a. 
(b) What is the value of Β  at the center of a circular coil of radius 4 cm containing 20 turns and carrying 
a current of 10 amperes? 

Problems 
301 
4. A coil of 5 turns has dimensions of 7 cm x 9 cm. It rotates at an angular velocity of SOTT rad/sec in a field 
of 0.4 webers/m\ What is the maximum emf induced in the coil if the axis of rotation is perpendicular to 
the field? 
5. Two coils placed close together have a mutual inductance of 25 millihenrys. The current in coil A is 
increasing at the rate of 2 amp/sec. What is the emf generated in coil Β by this changing current? 
6. A particle of mass 1 gm and charge 5 microcoulombs is moving in a circle in a solenoid of 50 turns and 
length 10 cm carrying a current of 10 amperes. The radius of the circle is 10 cm. Find the linear speed. 
7. How much energy is stored in the magnetic field of a coil whose self-inductance is 10 henrys, when it is 
carrying a steady current of 4 amperes? 
8. A flat square coil of 10 turns has sides of length 12 cm. The coil rotates in a magnetic field whose flux 
density is 0.025 webers/m\ 
(a) What is the angular velocity of the coil if the maximum emf produced is 20 millivolts? 
(b) What is the average emf at this velocity? 
9. What is the time constant of an R-L circuit containing a pure resistance of 200 ohms in series with a 5 
henry inductor whose resistance is 100 ohms? 
10. A 500 ohm resistance and a 5 henry inductance are connected in series with a 20 volt battery. 
(a) What is the value of the time constant of this circuit? 
(b) How much energy is stored in the inductor after the switch has been closed for a period of time 
equal to the time constant? 
11. An inductor of inductance L and a resistor of resistance R are connected in series with a battery that 
has set up a steady current Jo in the circuit. At time ί = 0, the battery is shorted out so that the series 
circuit consists of only an inductor and the resistor. Set up the differential equation relating L, R, I, and 
t, and integrate this equation to obtain the ciurent through the resistor as a function of time. 
12. An electron 
= 1.60 x 10"^^ coul) circulates about a proton in an orbit of 0.52 x 10"'° m radius at a speed 
of 2.5 X 10^ m/sec. 
(a) To what current is this equivalent? 
(b) What is the magnetic field at the nucleus due to the orbital current? 
13. (a) Show that self-inductance over resistance (L/R) and resistance times capacitance (RC) both have 
the dimensions of time, 
(b) Show that 1 weber per second equals 1 volt. 
14. A closely wound rectangular coil of 50 turns has dimensions of 12 cm x 25 cm. The plane of the coil is 
rotated from a position where it makes an angle of 45° with a magnetic field of flux density 2 webers/m^ 
to a position perpendicular to the field in time ί = 0.1 seconds. What is the average emf induced in the 
coil? 
15. (a) If the current increases uniformly from 0 to 2 amperes in 0.40 seconds, in a coil whose coefficient of 
self-induction is 0.60 henrys, what is the magnitude of the emf produced? 
(b) With the aid of a diagram, specify its direction. 
16. A uniform field of magnetic flux density Β  is changing in magnitude at a constant rate dB/dt. Given a 
mass m of conducting material (density d, resistivity p) drawn into a wire of radius r and formed into a 
circular loop of radius R, show that the induced current in the loop when placed normally in the above 
field does not depend upon the size of the wire or the loop, and is given by 
UiTodAdtr 

302 
Magnetic Field of a Current 
17. In Figure 33-13, V is 120 volts, L is 120 henrys, and R is 60 ohms. 
(a) To what value does the current build up 
after the switch is closed? 
(b) How much energy will be stored in the in-
ductor when the current has reached its 
final value? 
(c) Calculate the current and the rate of change 
of the current when the potential difference 
across the resistor is 50 volts. 
Figure 33-13 
18. In Figure 33-14, V = 6volts, 1?, =0.1 ohms, 
R2 = 0.9 ohms, and L = 1 henry. Find the po-
tential across R2 after switch S has been closed 
for 0.4 seconds. What is the power dissipated in 
the circuit 0.4 seconds after switch S has been 
closed? 
Figure 33-14 
19. An inductor of inductance 4 henrys and resistance 4 ohms is connected to a battery of terminal voltage 
12 volts. Find 
(a) the initial rate of increase of current in the circuit, 
(b) the rate of increase of current at the instant when the current is 1 ampere, 
(c) the current 0.4 seconds after the circuit is closed, and 
(d) the final steady current. 
20. Derive the expression 
NA 
for a search coil used with a galvanometer to measure a magnetic field, where R is the total resistance of 
the coil and galvanometer, Q is the total charge passing through R , Ν  is the number of turns of wire in 
the coil, and A is the area of the coil. The operation of a search coil is as follows: Originally the coil is 
placed in the magnetic field with the plane of the coil perpendicular to the field. It is then quickly 
removed from the field region. This induces a current in the coil which flows through the galvanometer 
attached to the coil. 

34 
Magnetic and 
Thermoelectric 
Properties of Matter 
34-1 
Origins of Magnetism 
303 
34-2 
Magnetization, Magnetic Intensity, 
Susceptibility, and Permeability 
304 
34-3 
Diamagnetism 
304 
34-4 
Paramagnetism 
305 
34-5 
Ferromagnetism 
305 
34-6 
Thermoelectricity 
306 
34-1 
O R I G I N S OF M A G N E T I S M 
The earliest guess as to the origin of magnetism 
in magnets was that the material consisted of a 
number of particles each being a small magnet. 
These small magnets are oriented at random when 
the material is unmagnetized. When the material is 
magnetized they are aligned with unlike poles adja-
cent to one another within the material, thus leaving 
unlike poles at each end of the material, as illus-
trated in Figure 34-1. 
After the discovery that a small coil carrying a 
current behaves like a small magnet. Ampere 
suggested the following: (1) magnetism is due to 
small circulating currents associated with each 
atom; (2) each atom possesses a magnetic dipole 
moment; and (3) the magnetic moment of any sub-
stance is equal to the vector sum of the magnetic di-
pole moments of its constituent atoms. 
The magnetic moment of a current loop enclosing 
an area A is defined by 
m = nIA, 
(34-1) 
where π  is a unit vector in the direction of m, which 
is normal to the plane of the loop and is in the 
direction a right-hand screw projects if turned in 
the direction of the current. Let us now list some of 
the properties of the magnetic moment m of the 
current associated with the orbiting electrons in an 
atom, which have been verified experimentally. 
(1) The resultant magnetic moment of an atom is 
related to the vector sum of the individual 
magnetic moments of its electrons. 
IS N| |S 
N| is 
N| IS Nl |S 
Nj 
Is 
N| Is 
N| Is N| Is N| Is N| 
(a) Unmagnetized 
(b) Magnetized 
Hgure 34-1 Materials in unmagnetized and magnetized states. 
303 

304 
Magnetic and Tliermoelectric Properties of Matter 
(2) An atom, ion, or molecule has zero perma-
nent magnetic moment if the total angular 
momentum of the electrons is zero. If the 
total angular momentum is not zero, the par-
ticle has a permanent magnetic moment. 
(3) Most free atoms have a permanent magnetic 
moment, while molecules nearly always have 
a zero permanent magnetic moment. 
It should be clear to the reader that the above 
properties of the magnetic moment indeed support 
Ampere's suggestions. 
34-2 
M A G N E T I Z A T I O N , M A G N E T I C 
INTENSITY, SUSCEPTIBILITY, A N D 
P E R M E A B I L I T Y 
If we place a material substance within a coil car-
rying a ciurent, it is quite likely that the value of Β  
will be different than when the substance is not pre-
sent. It is possible to consider the resultant Β  within 
the coil material as the sum of two quantities Bo due 
to the current in the coil and a quantity μ ο Μ  from 
the magnetic properties of the substance in the coil. 
We may write 
Β = Β ο  + μ ο Μ
, 
(34-2) 
where μ ο  is referred to as the permeability of free 
space and Μ  is called the magnetization of the sub-
stance. The magnetization Μ  is the magnetic mo-
ment per unit volume and is analogous to the elec-
tric polarization P. Also, it is convenient to define 
another vector Η  called the magnetizing force or 
magnetic intensity just as in electrostatics it was 
convenient to define the vector D . Η  is defined by 
the relationship 
μ ο  
(34-3) 
This allows us to write Ampere's law [Eq. (33-2)] as 
> H d l = NJ, 
(34-4) 
for any substance placed within a coil of Ν  turns. 
In MKS units, Η  is measured in ampere turns per 
meter. 
In an isotropic substance, Μ  is directly propor-
tional to Η  and parallel to it, so that we may write 
M = Xn^H, 
(34-5) 
where Xm is known as the magnetic susceptibility 
and is a dimensionless quantity. If we substitute 
Eq. (34-3) into Eq. (34-2) for Bo, we get 
Β = μ ο (Η  + Μ
) = μ ο (ΐ + ^ ) Η . 
(34-6) 
Substituting Eq. (34-5) into Eq. (34-6), gives 
Β = μ ο (1 + ;^..)Η  
(34-7) 
Β = μ Η , 
(34-8) 
or 
where 
μ =μ ο (1 
+ χ Λ 
(34-9) 
The ratio μ / μ ο  is called the relative permeability μ ^. 
It follows from Eq. (34-9) that 
μ . = 1+;^„ 
or 
χ „ ,=μ τ-1. 
(34-10) 
The susceptibility of a vacuum equals zero, since 
μ Γ = 1 in a vacuum. 
Since Μ  can be interpreted as the magnetic mo-
ment per unit volume, Xm is a quantity whose 
interpretation is in terms of a unit volume of the 
material. For comparison between experiment and 
theory, it is convenient to define mass susceptibility 
or specific susceptibility Xms, 
^Xrn 
p ' 
(34-11) 
where ρ is the density of the material. Another 
useful quantity is obtained by referring χ „  to a 
particular number of molecules of the material. We 
therefore define the term molar susceptibility by 
the relation 
_ M 
Xm mole ~ ρ X"^* 
(34-12) 
where Μ  is the molecular mass of the material. 
Real materials are usually characterized by the 
sign of their susceptibility. In the next three sec-
tions we discuss the three types of magnetic 
materials. 
34-3 
D I A M A G N E T I S M 
Diamagnetic substances are those for which Xm is 
negative and μ r is less than unity. In diamagnetic 
substances, χ „  is practically independent of temp-
erature and field strength, and is approximately 
constant for a particular type of atom or ion. 
The explanation of diamagnetism follows from 
Faraday's law and Lenz's law of electromagnetic 
induction. Atoms can be visualized as consisting of 
electrons moving in orbits around the nucleus. This 

Ferromagnetism 
305 
orbital motion of the electrons constitutes a tiny 
current loop. For certain atoms, those which make 
up diamagnetic materials, the net magnetic moment 
for the orbital motion of the electrons is zero. If a 
magnetic field is applied to the substance, each 
electron experiences a force, and its orbital motion 
is altered so that it acquires an angular momentum 
and hence a magnetic moment. The change in the 
orbits will produce a net flux for the atom, which 
will oppose the flux due to Bo. Thus, the net flux 
through the material will be less than if the material 
were not present, causing the resultant Β  to be less 
than Bo, or for the material to have a negative Xm. 
Diamagnetism is present in all substances, but its 
presence is masked in substances where the atoms 
have a net magnetic moment. Superconductors, 
which are substances that have zero electrical resis-
tance at very low temperatures, are 
highly 
diamagnetic when in the superconducting state. 
When a superconductor is placed in a magnetic field 
low enough not to destroy its superconductivity, 
currents are produced in the surface layers and the 
magnetic field in the interior of the sample remains 
zero. Thus, we may say that superconductors are 
perfectly diamagnetic. 
34-4 
P A R A M A G N E T I S M 
Paramagnetic substances are those for which χ η , 
is positive and μ , is slightly greater than unity. 
Paramagnetism occurs in substances that possess a 
permanent magnetic moment. In the absence of an 
external magnetic field, the atomic dipoles point in 
random directions, and there is no resultant mag-
netization. When a field is applied to the substance, 
the atomic dipoles orient themselves parallel to the 
field. This gives a net magnetization parallel to the 
field and a positive contribution to the susceptibil-
ity. This positive contribution is affected by temp-
erature, since thermal agitations affect the random-
ness of the orientations of the dipoles. At a low 
temperature, there is less thermal agitation, thus 
giving less random orientation, which permits 
greater magnetization for a given field. The temper-
ature dependence of χ „  for paramagnetic sub-
stances is given by an empirical relation known as 
Curie's law, 
C 
ature. This relation is good for paramagnetic gases 
at ordinary field strengths and temperatures, where 
the interaction of the molecules is negligible. 
For liquids and solids, where the interaction of 
the molecules may be large, a modified Curie law 
called the Curie-Weiss law is used. 
Xm 
= Τ -θ' 
(34-14) 
Xm 
rp. 
(34-13) 
where C is a constant and Τ  is the absolute temper-
where θ is the Weiss temperature and is charac-
teristic of the substance. It may be positive or nega-
tive. The equation only holds for Τ  > Θ, and for 
many substances no single equation holds over a 
wide temperature range. Of the conmion gases, 
only oxygen and nitric oxide are paramagnetic. 
Most solids have filled electron inner shells and are 
diamagnetic. Exceptions are compounds of transi-
tion elements, which have unfilled electron shells. It 
should be mentioned again that diamagnetism is 
also present in paramagnetic substances, but its 
presence is masked by the stronger paramagnetic 
effect. The magnetic behavior of metals is even 
more complex. In metals, the bound electrons are 
attached to positive ions, which (except in the 
transition group) have filled electron shells and are, 
therefore, diamagnetic. The conduction electrons 
give rise to a diamagnetic and paramagnetic effect, 
both of the same order of magnitude and both inde-
pendent of temperature. Exceptions are ferro-
magnetic metals such as iron, nickel, cobalt, along 
with many alloys. 
34-5 
F E R R O M A G N E T I S M 
Ferromagnetic substances are those for which Xm 
is positive and large, and 
is much greater than 
unity. In ferromagnetic materials, the permeability 
is not a constant but is a function of temperature 
and magnetic field. Ferromagnetic substances are 
all solids. Their magnetic properties change ab-
ruptly at a certain temperature called the "Curie 
point." Above the Curie point, χ ^ is independent of 
field strength and follows the Curie-Weiss law ap-
proximately. Below the Curie point, very large val-
ues of magnetization are produced by quite small 
fields, and the magnetization varies nonlinearly 
with field strength, which means that the permeabil-
ity varies with field strength. Figure 34-2 is a typical 
magnetization curve for a ferromagnetic material. 
This curve is not reversible. If the material is in-
itially unmagnetized, and an increasing field Η  is 

306 
Magnetic and Ttiermoelectric Properties of Matter 
Figure 34-2 
material. 
Magnetization curve for a ferromagnetic 
applied, Β  follows the solid curve 05 and reaches a 
saturation value B,. If the field Η  is now reduced, Β  
follows the broken curve SS'. When Η  = 0, Β  stiU 
has a finite value, known as residual induction. As 
the field is increased in the reverse direction, Β  
decreases and finally becomes zero. The value of 
this reversed field to make Β  = 0 is called the co-
ercive force. If we continue to increase the field Η  in 
the reverse dúection, a reverse induction Β  is set 
up, which reaches a saturation value BÍ = - B,. If 
the field is now again increased in the positive di-
rection, Β  will follow the broken curve S'S, and we 
have what is called a hysteresis loop or cycle traced 
out by the broken curve. 
A detailed explanation of ferromagnetism is 
beyond the scope of this book, so only a brief qual-
itative description will be attempted here. Fer-
romagnetic substances contain atoms with perma-
nent magnetic dipoles. Groups of these atomic di-
poles point in the same direction. These groups are 
called "domains." In the unmagnetized material, 
the domains are oriented at random, so there is no 
resultant magnetic moment in any direction. When 
a magnetic field is applied, the domains where the 
magnetization is parallel or at a small angle with the 
dh-ection of the field grow at the expense of those 
where the magnetization is anti-parallel, and the 
boundaries of the domains are displaced. It is pos-
sible to map out domain boundaries and to detect 
the shifting of domain orientation experimentally. 
There are two classes of ferromagnetic materials: 
(1) magnetically soft materials, which have a high 
permeability and are easily magnetized and demag-
netized, and (2) magnetically hard materials, which 
have a relatively low permeability and are difficult 
to magnetize and demagnetize. 
34-6 
T H E R M O E L E C T R I C I T Y 
In 1821, Seebeck discovered that a current flows 
in a simple cucuit composed of two different 
metals, when the junctions are maintained at differ-
ent temperatures. A thermocouple, a temperatiu-e 
measiuing device that utilizes this effect, is illus-
trated in Figure 34-3. The extended junctions are 
temperature probes composed of metals A and Β 
twisted together. Since a steady current flows in the 
ch-cuit, an emf must exist. This emf is called a 
thermal emf because it is produced by a difference 
in temperature between the two junctions. The 
effect is called the Seebeck effect. 
The Seebeck current is due to the fact that the 
density of free electrons in a metal depends on the 
temperature and is different for different metals. 
Therefore, when two different metals are con-
nected to form two junctions and the junctions are 
at different temperatures, electron diffusion at the 
junctions takes place at different rates, and there is 
a net flow of electrons from one junction to the 
other. In terms of energy considerations, energy is 
absorbed from the heat source at the hot junction, 
which is in part liberated to the surroundings at the 
cold junction, the remainder being converted into 
electrical energy in the circuit. 
In 1834, Peltier discovered that if a current from 
an external source is sent through a cúcuit com-
posed of two different metals (similar to a ther-
mocouple), one junction warms up and the other 
cools, which means that heat is absorbed at one 
junction and liberated at the other. The heat 
absorbed or liberated at a junction between two 
metals depends on the direction of the current. If 
the Seebeck emf is from metal A to metal Β  at a 
Temperature 
probe . 
Metal A 
Metal a 
Temperature 
probe 
Figure 34-3 Thermocouple of metals A and β  with 
junctions 1 and 2 at temperatures t,, and f, (t,, >t,). 

Problems 
307 
hot junction, an external emf applied in this direc-
tion produces a cooling at this junction and a heat-
ing at the other junction. The heat absorbed or lib-
erated at a junction is dependent on the quantity of 
charge that crosses the junction, on the tempera-
ture of the junction, and on the nature of the two 
metals. 
It was discovered by Sir William Thomson (Lord 
Kelvin) that a thermal emf is set up in a metal hav-
ing a temperature gradient. The emf in a section of 
the conductor is directly proportional to the differ-
ence in temperature between its ends. It is possible 
to explain this effect in terms of our electron gas 
model of conducting materials. The density of free 
electrons at the hot end of the conductor will be 
greater than at the cold end. This will cause the 
electrons to migrate from the hot end to the cold 
end of the conductor, leaving the hot end positive 
and the cold end negative. As a result, an emf 
should be produced in the conductor directed from 
the cold to the hot end. This is the case for most 
metals, but there are some metals in which the emf 
is in the opposite direction, an effect for which 
there is no simple explanation. 
The Seebeck, Peltier, and Thomson effects are all 
completely reversible. The Peltier emf, called the 
Peltier coefficient (π AB) of a junction of metals A 
and B, is defined as the quantity of heat absorbed or 
liberated per unit electric charge crossing the junc-
tion. Thus, 
H p 
TT A B —  ' 
(34-15) 
where Hp is the Peltier heat. 
The Thomson emf of an infinitesimal length of 
wire that has a temperature difference dt may be 
written crdi, where σ  is called the Thomson coeffi-
cient and is defined as the heat absorbed or liber-
ated in this length of wire per unit electric charge 
transferred. Thus, 
adt=^. 
(34-16) 
where Η τ is the Thomson heat. The Seebeck emf in 
a circuit like the thermocouple circuit (Figure 34-3) 
is the resultant of two Peltier emf's and two Thom-
son emf's. Referring to this thermocouple circuit. 
^ A B = Í'TTABK 
•^('TTBA)t, 
+ 
aBdt. 
(34-17) 
In this equation, the direction of the emf is taken so 
that the current flows from JB to A at the hot junc-
tion. The above equation is the "fundamental ther-
mocouple equation." Since the emf is a function of 
the th and i/, it is possible to use a thermocouple to 
measure temperature if either th or ti is fixed. 
Example 1 . A toroid with a core of relative per-
meability 500 is 1 m in circumference. It has 1000 
turns of wire carrying a current of 0.5 amperes. 
(a) What is the magnetic intensity? 
(b) What is the flux density in the core? 
SOLUTION 
(a) The magnetic intensity is given by Eq. (34-4), 
^ N 7 ^ 
(1000X0.5 amp) ^ 3 ^ 
/ 
1 m 
(b) The flux density in the core is given by Eq. 
(34-8), 
Β = μ Η = (500 nt/amp')(500 amp/m) 
= 25 X 10^ webers/m^ 
P R O B L E M S 
1. What is the magnetic moment of a flat coil of 20 turns, having an area of 5 cm^ and carrying a current of 
8 amperes? 
2. Consider a cylindrical rod of iron, length L = 60 cm, cross-sectional area A = 5 cm^ which is bent into a 
ring with a small air gap / = 0.1 cm between the ends. If 400 turns of wire, carrying 6 amperes, are wound 
around the iron ring, and if the relative permeability of the iron is 5000, what is the magnetic flux inside 
the iron? 
3. An iron ring of relative permeability 1200 has a mean circumference of 40 cm and an area of 5 cm^ The 
ring is wound with 350 turns of wire and carries a current of 0.2 amperes. 
(a) What is the magnetic intensity in the ring? 
(b) What is the flux density in the ring? 

308 
Magnetic and Ttiermoelectric Properties of Matter 
4. A metal ring having 800 turns of wire and a mean diameter of 16 cm carries a current of 0.3 amperes. The 
relative permeability of the core is 650. 
(a) What is the flux density in the core? 
(b) What part of the flux density is due to electronic loop currents in the core? 
5. The area of a hystersis loop is proportional to an energy loss. By the consideration of the physical 
dimensions of Β  and H, show that the area of the loop has the units joules/m\ 
6. An iron rod of cross-sectional area 4 cm^ and relative permeability 1000 is bent into a circle with inside 
radius 9 cm and outside radius 11 cm. It is wound with 1000 turns of wire per meter and a current of 2 
amperes is sent through the wire. Determine the magnetic flux in the iron. 
7. A solenoid has a cross-sectional area of 6cm^ and the magnetic flux density (B) in its air core is 
10"^ webers/m^ What is the value of Β  and flux φ  in the solenoid if an iron core of relative permeability 
1000 replaces the air core? 
8. The iron rod of Problem 6 is an alloy of relative permeability 10,000 and is wound with 1000 turns of 
wire carrying 1 ampere. Find H, B, and φ , 
9. The hysteresis loops for four ferromagnetic alloys are shown in Figure 34-4, all to the same scale. Which 
of the materials would make the best permanent magnet? Explain your answer. 
Figure 34-4 
10. The current in the windings of a coil of 500 turns wound around an iron ring of relative permeability 
1000 and circumference 50 cm is 1 ampere. Calculate: 
(a) the magnetic intensity, 
(b) the flux density, 
(c) the magnetization, and 
(d) the magnetic susceptibility. 
11. A magnet has a coercive force of 6 x 10' amp/m. It is inserted in a solenoid 10 cm long, having 50 turns. 
What current must be carried in the solenoid in order to demagnetize it? 
12. Why is Eq. (34-17) the result of two Peltier emf's and two Thomson emf's? 
13. A closed thermocouple circuit consists of the two thermal junctions and a low resistance galvanometer 
all in series. The galvanometer has a resistance of 9 ohms and reads directly in millivolts across its 
terminals. The resistance of the rest of the circuit is 1 ohm, and the thermocouple develops an emf of 
20 microvolts per degree difference of temperature between the junctions. When one junction is 
maintained at 0°C and the other is placed in a molten metal, the galvanometer reading is 5.40 millivolts. 
What is the temperatiue of the molten metal? 

35 
Alternating 
Currents 
35-1 
Introduction 
309 
35-7 
Parallel Circuit with Resistances, 
35-2 
Properties and Effects of 
Capacitance, and Inductance 
313 
Alternating Currents and EMF's 
309 
35-8 
Power in Alternating Current 
35-3 Circuit with Resistance Only 
310 
Circuits 
314 
35-4 
Circuit with Capacitance Only 
311 
35-9 
Resonant Circuits 
315 
35-5 
Circuit with Inductance Only 
312 
35-10 The Transformer 
315 
35-6 
Series Circuit with Resistance, 
Capacitance, and Inductance 
312 
35-1 
INTRODUCTION 
An alternating current is one that flows alter-
nately in opposite directions in a circuit. When the 
current rises from zero to a maximum, returns to 
zero, increases to a maximum in the opposite direc-
tion, and ñnally returns to zero again, it is said to 
have completed a cycle. The time requh-ed for the 
completion of one cycle is called the period (r). 
The number of cycles per second is called the fre-
quency (/) of alternation of the current. The maxi-
mum instantaneous current in either direction is 
called the amplitude. 
Figure 35-1 is a graph of an alternating current 
versus time, where positive values of the instan-
taneous current i represent currents in one direc-
tion and negative values in the other direction. This 
graph can be represented by the equation 
i = IM sin 
(2τφ -^φ ), 
(35-1) 
Where IM is the maximum value of the current and / 
is the frequency of alternation of the current, φ  is a 
constant angle called the phase angle, which fixes 
the value of i at time ί = 0 relative to its maximum 
value. 
27Γ ft 
Figure 35-1 Graph of sinusoidal alternating current. 
35-2 
P R O P E R T I E S A N D E F F E C T S O F 
A L T E R N A T I N G C U R R E N T S A N D 
E M F ' S 
The effects of alternating currents and emf's are 
quite different from those due to direct currents and 
emf's. In general, it is possible for the applied emf 
and the resulting current not to be in phase. That is, 
the current and emf do not have maximum or 
309 

310 
Alternating Currents 
minimum values at the same time. Alternating cur-
rents can flow continuously in a capacitor, while di-
rect currents cannot. In an alternating current cir-
cuit, the simple form of Ohm's law must be modi-
fied to take care of capacitance and inductance. 
Since an alternating current varies from zero to a 
maximum value, we may ask what is the effective 
value of this current, or, in other words, what is 
meant by an alternating current of 1 ampere. Exper-
iment has shown that the heating effect of an elec-
tric current is independent of current direction. As 
a result, the heating effect of an alternating current 
is taken as the basis for the definition of an ampere. 
An alternating current is said to have an effec-
tive value of 1 ampere when it will develop the 
same amount of heat in a given resistance as 
1 ampere direct current at the same time. If we des-
ignate the effective alternating current by J, then 
RI^ = average value of Ri^ 
or 
/ = (average of i')''' 
= [average of JM' sin' (2τφ  + φ )]''\ 
The average value of sin' θ and cos' θ per cycle 
must be the same since both have the same shape, 
where θ = (2nft + φ ). Since sin' θ + cos' θ = 1, the 
average value of both sin'ö  and cos' θ is i There-
fore, 
voltage V as 
/ = : ^ = 0.707JM. 
(35-2) 
V = ; ^ = 0.707 VM, 
(35-3) 
where VM is the maximum or peak voltage. Alter-
nating currents and voltages are almost always ex-
pressed in terms of their effective values, which are 
sometimes called their "root-mean-square" values. 
35-3 
C I R C U I T WITH R E S I S T A N C E 
O N L Y 
In the case of an alternating current circuit that 
has only a pure resistance, the instantaneous cur-
rent and voltage are connected by Ohm's law. 
V 
'=R' 
(35-4) 
In a similar way, we define the effective value of the 
Therefore, if an alternating voltage ν  = VM sin 27rfi 
is across the resistance, the instantaneous current is 
i = ^ ^ ^ ^ = I . s m 2 ^ , 
(35-5) 
and the current and voltage are both in phase since 
they both vary with sin27rfit. This is illustrated 
graphically in Figure 35-2(a). The instantaneous 
current and voltage values may be represented as 
the projection of vectors on the vertical axis, which 
are rotating about a conunon origin with an angular 
velocity ω  = 277/. Figure 35-2(b) is such a vector 
representation showing current and voltage in 
phase with one another. Since the effective current 
and voltage are each given by J = 0.701 IM and V = 
(a) 
Reference 
direction 
(b) 
Figure 35-2 

Circuit Witt) Capacitance Only 
311 
0.707 VM, respectively, the relation 
(35-6) 
is also true. 
35-4 
CIRCUIT WITH C A P A C I T A N C E 
O N L Y 
It was mentioned at the beginning of the chapter 
that an alternating current can flow continuously in 
a capacitor. It should be pointed out that no current 
flows through the dielectric between the plates of a 
capacitor. The current in the circuit is due to the 
charging and discharging of the capacitor in oppo-
site directions. The current through the circuit is 
equal to the time rate of change of charge on the 
capacitor, since the current flowing to or from 
either plate is the amount of charge transferred per 
second. Therefore, if an alternating-current voltage 
is applied to the capacitor, the instantaneous charge 
on the capacitor is 
q = Cv = CVM sin Ζττ/ί, 
and the current i is 
i=^ 
= j¡{CVMsm27Tft) 
= ITÍCVM 
C O S lirft. 
The maximum current is obviously 
and 
7M=27rfCVM 
(35-7) 
i = 7M C O S 27rfi. 
(35-8) 
Ohm's law in terms of the effective voltage and cur-
rent is 
27r/C 
(35-9) 
The quantity XllirfC is called the capacitive reac-
tance (Xc) of the capacitor and has the dimensions 
of resistance. Then 
1 = 
Xc' 
(35-10) 
Figure 35-3(a) is a graphical representation of the 
instantaneous voltage and current equations, and 
Figure 35-3(b) shows the phase difference by a 
vector representation. 
The current and the voltage are out of phase by 
90°. The current reaches its maximum one-quarter 
period ahead of the voltage. Thus, it may be said 
that in a capacitor the current leads the voltage by 
90°. 
(a) 
Figure 35-3 
Reference 
2 ^ ft 
direction 
(b) 

312 
Alternating 
Currents 
Reference 
direction 
(a) 
(b) 
Figure 35-4 
35-5 
CIRCUIT WITH I N D U C T A N C E 
O N L Y 
If an alternating-current voltage is applied to a 
pure inductance L, the instantaneous potential 
difference across it is given by 
di 
and 
V = Vm sinlirft 
= 
di=j^sm2nftdt. 
Integration of both sides, gives 
The maximum current is obviously 
Vm 
Im 
= lirfL 
and 
i = - Im c o s 2τφ . 
(35-11) 
(35-12) 
Figure 35-4(a) is a graphical representation of the 
instantaneous voltage and current equations and 
Figure 35-4(b) shows the phase difference by a 
vector diagram. 
The current and the voltage are out of phase by 
90°. The voltage reaches its maximum one-quarter 
period ahead of the current. Thus, it may be said 
that, in an inductance, the voltage leads the current 
by 90°. 
35-6 
S E R I E S CIRCUIT WITH R E S I S T A N C E , 
C A P A C I T A N C E , A N D I N D U C T A N C E 
We shall now consider a circuit with a resistance, 
capacitance^nd inductance in series, as illustrated 
in Figure 35-5. The instantaneous potential differ-
ence V between the terminals a and b is equal to 
the sum of the potential differences across each cir-
cuit element. If we use the voltage-current relations 
Ohm's law in terms of effective voltage and current 
is 
V 
ί = 277fL* 
(35-13) 
The quantity 27rfL is called the inductive reactance 
Xl of the inductance. Then, 
J = 
Xl' 
(35-14) 
Figure 35-5 A series R-L-C circuit. 

Parallel Circuit with Resistance, Capacitance, and Inductance 
313 
for each element, and use the vector diagram for 
this type of circuit as illustrated in Figure 35-6 as a 
guide, we can obtain a voltage-current relation for 
this circuit. Referring to the vector diagram, we see 
that 
V = 
lVR'-^(XL-Xcy 
or 
V 
. V 
ζ  
We have introduced the quantity 
roduced the quantity 
(35-15) 
(35-16) 
(35-17) 
which is called the impedance of the circuit. 
The voltage in this circuit can either lead or lag 
the current, depending on the relative values of the 
inductance and capacitance. From Figure 35-6 the 
phase angle is given by the relation 
tan φ  = 
=— 
(35-18) 
Figure 35-6 Vector diagram for a series R-L-C circuit. 
Example 1. A series circuit connected across a 
110 volt, 60 cycle line consists of a coil with 
100 ohms resistance and 0.20 henrys inductance, 
and a capacitor of capacitance 100 microfarads. 
Calculate: 
(a) the current in the circuit, 
(b) the phase angle between the current and sup-
ply voltage, and 
(c) sketch the vector diagram for this circuit. 
SOLUTION 
(a) 
XL = lirfL = (27τ)(60 cps)(0.20 henrys) 
= 75.4 ohms. 
1 
1 
Xc = lirfC 
(27τ)(60 cps)(100 x 1 0 ' farads) 
26.6 ohms 
Z = 
V^T{X[-XcY 
= V(lOOn)' + [(75.4 - 26.6)Ω ]' 
= 111 ohms 
V 
110 volts = 0.991 amp. 
Ζ 
111Ω  
Π.Λ  
t a « ^ XL-XC 
75.4Ω -26.6Ω  
(b) 
t a n c i > = — ^ = j
^
^
^ 
ψ =26° 
Voltage leads the current by 26°. 
(c) 
= 0.488 
IXl-IXc' 
IXc-
35-7 
P A R A L L E L CIRCUIT 
WITH 
R E S I S T A N C E , C A P A C I T A N C E , 
A N D 
I N D U C T A N C E 
Another circuit we will examine is one with resis-
tance, capacitance, and inductance in parallel, as 
illustrated in Figure 35-7. In this circuit, the poten-
tial difference across each branch is the same, but 
Figure 35-7 A parallel R-L-C circuit. 

314 
Alternating Currents 
the current through each branch is generally differ-
ent. If V is the effective value of the applied emf, 
then the ciurent through each branch of the circuit 
is 
- 
:— 
- 
-TT 
V 
lirfC 
lirfL 
Xl ' 
The vector diagram for this cucuit is shown in 
Figure 35-8. The total effective current in the cúcuit 
is obtained by the vector addition of the component 
currents. The magnitude of this current is 
1 / 1 
1 
and the impedance is given by the relation 
Z = ( ¿ , + 
( 2 ^ C - ¿ ) ) " ' " 
1 
(35-19) 
(35-20) 
The phase angles (φ ) may be computed with the 
aid of the diagram. 
tan φ  = 
\ 
Xc —4r 
2'TrfC-
1 
XL 
R 
R 
lirfL 
(35-21) 
icy 
1 
Reference 
^ 
direction 
Figure 35-8 Vector diagram for a parallel R-L-C 
circuit. 
It should be noted that the signs of Xc and Xl are 
reversed in comparison with theu use in the series 
circuit. Why? 
35-8 
P O W E R 
IN 
A L T E R N A T I N G 
C U R R E N T 
C I R C U I T S 
The instantaneous power in an alternating cur-
rent circuit is 
p=vi 
= VMIM sin lirft sin (lirft - φ ), 
(35-22) 
where φ  is the phase angle between the voltage 
and the current, which may be either positive or 
negative. The quantity of interest is not the instan-
taneous power, but the average power. We may 
transform Eq. (35-22) by the use of the trigonomet-
ric identity 
sin a sinb = |[cos (a-b)- 
cos (a + b)], 
and obtain the equation 
ρ = I VMIM[cos Φ  - cos (Airft - Φ ) ] . 
(35-23) 
In this equation, cos (47r/i - φ ) is periodic, alternat-
ing between positive and negative values, so that its 
average value is zero. Thus, the average power in 
the ch-cuit is given by 
Ρ  = | V M I M cos φ . 
(35-24) 
In terms of the effective current / and voltage V, 
the average power may be expressed as 
Ρ  = Vicos φ . 
(35-25) 
The term cos φ  is called the power factor of the 
circuit. For a pure resistance circuit, φ  = 0 and the 
average power is Ρ  = VL For a pure inductance or 
capacitance, φ  = 90° and Ρ  = 0. 
Equation (35-23) may be put in the form 
ρ = V/[(l - cos 4τ φ ) 
cos φ  + sin 47rft sin φ ], 
(35-26) 
by expanding the variable term and replacing VM 
and IM by V and J. The first term in the square brac-
kets represents the fluctuating power dissipated 
into heat, for this term is non-negative and exists 
alone if the component has a pure resistance. The 
second term represents the power alternately 
stored in a cucuit component and released to the 
cucuit, for this term changes sign and is the only 

The Transformer 
315 
term present if the component has a pure reactance. 
An ideal capacitor stores energy in the form of an 
electric field and an ideal inductor—in the form of a 
magnetic field. The peak power involved in energy 
storage is 
Ps = VI sin φ . 
(35-27) 
This is called reactive power. To help distinguish 
reactive power from instantaneous or average 
power, different units are used. For instantaneous 
or average power, the unit is the watt if the current 
is in amperes and the electromotive force or poten-
tial difference is in volts, while, by convention, 
reactive power is left in volt-amperes. 
35-9 
R E S O N A N T 
C I R C U I T S 
Equation (35-17) is the expression for the impe-
dance of a series AC circuit. It is evident that for a 
given value of the resistance R, the impedance Ζ is 
a minimum when 
27rfL=; 1 
27r/C-
This happens when the frequency is 
1 
(35-28) 
where 
is called the resonant frequency of the cir-
cuit. We then have what is called a resonant circuit. 
For this particular frequency, the impedance re-
duces simply to the resistance, i and ν  are in phase, 
and the current-voltage relationship is as given by 
Ohm's law for a dkect current circuit. Also, at the 
resonant frequency, the current is a maximum, and 
the voltages across the capacitance and inductance 
are equal and opposite, so the phase angle is zero. 
It is also possible to have a resonant parallel cir-
cuit. Equation (35-20) is the expression for the im-
pedance of a parallel AC circuit. It is evident that, 
for a given value of the resistance R, the impedance 
Ζ is a minimum when 
27r/C=; ^ 
which leads again to 
fR = 
lirfL' 
1 
27rVLC' 
(35-29) 
It should be noted from Eqs. (35-19) and (35-20) 
that in a resonant parallel circuit the impedance is a 
maximum and the current is a minimum. 
Since the capacitance and inductance of a circuit 
can be varied, the resonant frequency can also be 
varied. Resonant circuits are of great importance in 
radio and television. They provide a means by 
which one particular frequency can be selected so 
as to respond to the signal of a desired station. 
35-10 
T H E T R A N S F O R M E R 
An alternating current has the advantage over di-
rect current in that its voltage can be changed in 
any desired ratio by means of a transformer. This is 
the reason why alternating current is used almost 
universally for power transmission. 
A transformer consists of two separate coils 
wound on opposite sides of a common iron core as 
illustrated in Figure 35-9. The coil to which the cur-
rent is supplied is called the primary coil, and the 
coil from which the current is dehvered is called the 
secondary coil. An alternating current is induced in 
the secondary coil by the changing magnetic flux in 
the primary coil due to the impressed alternating 
current in the primary coil. If we let Np equal the 
number of turns in the primary coil and 
the 
number of turns in the secondary coil, respectively, 
and assume that the magnetic flux (φ ) is the same 
throughout the iron core, then the primary flux link-
age is Ν ρφ  and the secondary flux linkage is Ν ,φ . 
Since the flux is varying, an induced emf is pro-
duced in both coils that is proportional to the rate of 
change of flux, and we have 
V 
— É L 
^'dt 
(35-30) 
The rate of change of flux will be the same in both 
coils, since the change in both coils is due to the 
same source, an alternating current at a definite fre-
Figure 35-9 Schematic diagram of a transformer. 

316 
Alternating Currents 
quency. Therefore, Eq. (35-30) becomes 
Thus, the ratio of the secondary voltage to the 
primary voltage can be controlled by the choice of 
the number of turns on each coil. 
The power loss in a transformer is usually quite 
small, and it can be shown that the current and volt-
age are nearly in phase. We can therefore arrive at 
the following approximate relations, namely. 
Vsis 
= Vp/p 
and 
V. 
Ν / 
(35-32) 
Equation (35-32) points out that the voltage ratio is 
the same as the ratio of the number of turns, 
while the current ratio is the inverse ratio of the 
number of turns. 
Example 2. A series circuit includes a resistor of 
resistance 20 ohms, a capacitor of capacitance 30 
microfarads, 
and an inductor 
of 
inductance 
0.30 henrys. Impressed upon this is a 60 cycle AC 
voltage whose effective value is 120 volts. Calcu-
late; 
(a) the inductive reactance, 
(b) the capacitive reactance, 
(c) the impedance, 
(d) the effective current in the circuit, 
(e) the phase angle, 
(f) the power factor, 
(g) the power dissipation in the circuit, 
(h) the maximum instantaneous voltage applied 
to the circuit, 
(i) if the frequency of the supply voltage could 
be changed, the resonant frequency, 
(j) when in resonance the current in the circuit, 
and 
(k) the voltage drop across the capacitor at res-
onance. 
SOLUTION 
(a) Xl = iTTfL = (27γ)(60 cps)(0.30 henrys) 
= 113.1 ohms. 
(b) Xc 
1 
1 
" lirfC 
(27γ)(60 cps)(30 x 1 0 ' farad) 
88.3 ohms. 
(c) Z c = V ä ' + ( X l - X c ) ' 
= ν (20Ω )' + (113.1Ω  - 88.3Ω )' 
= 32 ohms. 
V 
120 volts 
(d) 
/ = i = - 32Ω  - = 3.75 amp. 
(e) 
^ = t a n - ^ ^ ^ = t a n - ^ = 51.r. 
(f) 
c o s ^ = cos 51. r = 0.628. 
(g) P = V J c o s ^ 
= (120 volts)(3.75 amp)(0.628) = 282 watts. 
(h) 
= V2 V = (1.414)(120 vohs) = 169.5 voUs. 
(i) 
ÍR = 
^ 
27rVIC 
1 
277 V(0.30 henrys)(30 x 1 0 ' farad) 
1 
6ir X 10 
-5 = 53 sec-\ 
^ 
V 
120 volts 
^ 
^> 
^ = : R = - 2 o í T - = ^ " " ^ P -
(k) 
V = JXc = 
V27TÍ,C 
6 amp 
V(27r)(53 cps)(30 X 1 0 ' farad) 
= 60 volts. 
P R O B L E M S 
1. In a series R-L-C circuit, jR = 80 ohms, Xl = 100 ohms, and Xc = 40 ohms. What is the impedance of 
the ch-cuit? 
2. An AC circuit has a resistance of 500 ohms, an inductor of inductance 5 henrys, and a capacitor of 
capacitance 20 microfarads all in series. 
(a) At a frequency of 60 cps, what will be the power supplied to this ch*cuit if Vm = 220 volts? 
(b) At what frequency will resonance occur? 

Problems 
317 
3. A resistor of resistance 8 ohms is connected in a series circuit to an inductor of inductive reactance 
8 ohms. 
(a) What is the impedance of this circuit? 
(b) Does the current lag or lead the voltage? 
(c) What is the phase angle? 
4. Every inductor has a resistance due to the wire in its windings. With reference to Figure 35-10. 
(a) Calculate the effective current in the cir-
cuit. 
(b) Calculate the average power in the circuit. 
Figure 35-10 
R=40 ohms 
V = 28 volts 
400 cps 
L =0.1 henry 
/?L=20 ohms 
5. A series circuit consists of an 8 ohm resistor, a 12 ohm inductor, and a capacitor. A 60 cycle AC is 
impressed across this circuit. 
(a) What capacitance will cause this circuit to be in resonance? 
(b) What will be the phase angle between the current and the voltage when the circuit is in resonance? 
6. A series AC circuit has an impedance of 50 ohms and a power factor of 0.60 at 60 cps, the voltage 
- 
lagging the current. 
(a) Should an inductor or a capacitor be placed in series with the circuit to raise its power factor? 
(b) What size element will raise the power factor to unity? 
7. (a) An AC series circuit contains a resistance of 200 ohms, an inductance of 10 henrys, and a 
capacitance of 2.5 microfarads in series with a variable frequency generator. What is the resonant 
frequency of the circuit? 
(b) If the applied voltage in part (a) has a peak value of 200 volts, what is the effective current at 
resonance? 
8. An AC series circuit consists of a generator, which provides an effective voltage of 500 volts at 
2000 rad/sec to a resistor with a resistance of 150 ohms, an inductor of inductance 0.4 henrys, and a 
capacitor of capacitance 0.5 microfarads. Calculate: 
(a) the reactance of the circuit, 
(b) the impedance of the circuit, 
(c) the effective current and current amplitude, 
(d) the power factor and average power dissipation, and 
(e) draw the vector diagram of the circuit roughly to scale. 
9. A resistance of 80 ohms, an inductance of 0.4 henrys, and a capacitance of 100 microfarads are 
connected in series with an AC source whose angular frequency is 250 rad/sec. The effective current is 
1 ampere. 
(a) Calculate the reactances XL and Xc of the circuit. 
(b) Calculate the effective voltage and the voltage amplitude. 
(c) Draw the vector diagram of the circuit roughly to scale. 
10. The mutual inductance of a transformer is 0.50 henrys. Find the expression for the voltage induced in 
the secondary coil when the primary coil carries an alternating current Ip = 0.2 sin a>t. 

318 
Alternating 
Currents 
11. A series circuit includes a resistor of resistance 5 ohms, a capacitor of capacitance 20 ntícrofarads, and 
an inductor of inductance 0.50 henrys. Impressed on this is an AC voltage whose effective value is 
120 volts and whose frequency is 60 cps. Calculate: 
(a) the impedance, 
(b) the effective current, 
(c) the power factor, and 
(d) the power dissipation in the circuit. 
12. For the circuit shown in Figure 35-11, it is de-
sued that the current h shall lead the current J, 
by 30°. What must be the value of l?2 if L, = 
12 henrys, 
L 2 = 7 henrys, l?i = 3ohms, / 
is 
1000 cps, and VM is 68 volts? 
Figure 35-11 
13. A resistance of 400 ohms, a pure inductance of 0.55 henrys, and a capacitance of 4 microfarads are 
connected in series with an AC source whose effective voltage is 100 volts. The angular frequency is 
1000 rad/sec. Calculate: 
(a) the reactance of the circuit, 
(b) the impedance of the ch-cuit, 
(c) the effective and peak currents, 
(d) the voltage across the resistor, 
(e) the voltage across the capacitor, and 
(f) the power dissipated in the circuit. 
14. A resistance of 100 ohms, a capacitance of 10 microfarads, and a coil having a resistance of 200 ohms 
and an inductance of 0.5 henrys are connected in parallel to a 110 volt 60 cps power line. Calculate: 
(a) the current in the circuit and 
(b) the power supplied by the line. 
15. A series cúcuit connected across a 200 volt 60 cps line consists of a capacitor of capacitive reactance 
30 ohms, a resistor of 44 ohms, and a coil of inductive reactance 90 ohms and resistance 36 ohms. Calcu-
late: 
(a) the current in the cucuit and 
(b) the potential difference across each unit. 
16. If the circuit in Problem 15 consisted of all the units in parallel, what would be the current in the circuit? 
17. A step-down transformer operates on a 2200 volt line and supplies a load of 20 amperes. The ratio of the 
secondary to primary windings is 1:20. Determine: 
(a) the secondary voltage, 
(b) the primary current, and 
(c) the power output. 
18. A toroidal coil has a 1000 turn primary and a 5 turn secondary wound on an iron ring. The rate of change 
of flux is 0.5 webers/sec in the core. Certain information is missing; hence, you cannot calculate all of 
the following quantities: 
(a) voltage across the secondary, 
(b) voltage across the primary, and 
(c) permeability of the core. 
For each quantity, calculate the value or list the missing data, and show how the calculation could be 
made had all the data been given. 

36 
Maxwell's Equations 
and Electro-
magnetic Waves 
36-1 
Introduction 
319 
36-2 
Maxwell's Equations 
319 
36-3 The Wave Equation 
321 
36-4 
The Nature and Energy of 
Electromagnetic Waves 
322 
36-5 
The Electromagnetic Spectrum 
323 
36-1 
INTRODUCTION 
Clerk Maxwell condensed all of electromagnetic 
field theory into four field equations. In recognition 
of this outstanding contribution to electromagnetic 
theory, they are called Maxwell's equations, and 
they 
govern 
all 
macroscopic 
electromagnetic 
phenomena. Maxwell developed these equations by 
correlating the expressions for electric and magne-
tic fields based on Gauss's, Ampere's, and Fara-
day's laws, which we have previously discussed. 
The equations used in either the differential or in-
tegral form are applicable to time dependent as well 
as constant fields, and lead to expressions for elec-
tromagnetic waves, which we shall discuss later in 
this chapter. 
36-2 
M A X W E L L ' S E Q U A T I O N S t 
Gauss's law was written in Chapter 30 in the fol-
lowing form: 
Ε · ndA 
which may be written 
= 5 3 
(36-1) 
tThe vector relationships required in this chapter are 
discussed in Appendix I. 
Ε 
ndA 
(36-2) 
where Q = l^q. This is one of Maxwell's equations 
in integral form. For a continuous charge distribu-
tion, either Eq. (36-1) or (36-2) may be written 
¿ Ε ndA = - ¿ φ
, 
(36-3) 
JA 
€ OJv 
where ρ is the volume charge density inside the 
volume ν  enclosed by the Gaussian surface. 
The divergence theorem in vector analysis states 
that, for any vector function F, 
j 
F - n d A = ^ V'Fdv, 
(36-4) 
where dv is any volume element enclosed by the 
surface area A. If we apply this theorem to Eq. 
(36-3), it becomes 
¿ V'Edv=-¿ 
pdv. 
(36-5) 
JA 
€ OJv 
Since we are integrating over the same volume, the 
integrands are equal, and we obtain the equation 
or 
V . E = - ^ 
€ o 
div Ε = -^, 
€ o 
(36-6) 
319 

320 
Maxwell's Equations and Electromagnetic Waves 
which is the differential form of Maxwell's Eq. 
(36-2). 
Ampere's law for a complete circuit was given by 
Eq. (32-4) of Chapter 32 as 
4π 7 r ' 
(36-7) 
which tells us that the lines representing Β  are 
cueles about the axis of a current element. In Chap-
ter 33, we saw that 
B-ndA =0, 
(36-8) 
since magnetic induction lines are closed lines and 
every line entering a closed surface must leave it. 
Equation (36-8) is another of Maxwell's equations 
in integral form. Since, from the divergence 
theorem. 
j> B-ndA =j> divBdr, 
thus 
or 
divB = 0 
V B = 0, 
(36-9) 
which is another of Maxwell's equations, Eq. 
(36-8), in differential form. 
In Chapter 33, we discussed Faraday's law, 
which is expressed in analytical form in that chap-
ter by Eq. (33-4); namely. 
di' 
(36-10) 
where ^ is the induced emf and φ  is the magnetic 
flux. Also 
Φ  - i 
B n d A 
and 
= * E d s , 
(36-11) 
(36-12) 
Eq. (33-1) and Eq. (29-8), respectively. Substituting 
Eqs. (36-11) and (36-12) into Eq. (36-10), gives 
JE'ds 
= -^j 
B-ndA, 
(36-13) 
which is another one of Maxwell's equations in in-
tegral form. By Stoke's theorem. 
E - d s = φ  curlE-ndA. 
(36-14) 
Furthermore, the time and space coordinates are in-
dependent variables. Therefore, Eq. (36-13) can be 
rewritten in the form 
j c u r i E - n d A = - ^ 
^-ndA. 
(36-15) 
Since this equation holds for any surface area, the 
integrands must be equal, giving the differential 
form of Maxwell's equation [Eq. (36-13)] 
curi Ε = - dB 
or 
V x E = -
dt 
dB 
dt ' 
(36-16) 
We will now derive the remaining Maxwell's 
equation. In Chapter 33, we derived Ampere's 
law—namely. 
<pB-ds = MoI, 
(36-17) 
where I is the total current passing through any sur-
face bounded by the path. In general, / is the sum 
of the true or conduction current Ic and the dis-
placement current ID. Since Β  = μ ο Η  for free space, 
we may write Eq. (36-17) as 
Η  · ds = Jc + ID, 
(36-18) 
which is the ñnal Maxwell equation in integral 
form. It may be written in terms of current densities 
as 
^ H - d s = £ j - n d A + £ ^ - n d A , 
(36-19) 
where j is the conduction current density and d D/di 
is the displacement current density. By Stoke's 
theorem, Eq. (36-19) can be written 
j curiH · ndA = ^ j - ndA 4 - ^ ^ 
- ndA. 
(36-20) 
Since this equation holds for any surface, the integ-
rands must be equal, giving the differential form of 
the above Maxwell equation 
or 
curiH = i + 
V x H = j + 
dD 
dt 
dD 
dt' 
(36-21) 
Equations (36-19) and (36-21) tell us that the dis-
placement current density dD/di has the same 

The Wave Equation 
321 
effect in producing a magnetic field as does the con-
duction current density j . For steady currents, Eqs. 
(36-19) and (36-21) reduce to 
and 
j>H'ds = j>^ i-ndA 
(36-22) 
curlH = j , 
(36-23) 
respectively. Let us consider the space between the 
plates of a capacitor which is being charged. Here 
there is no conduction current between the plates 
but, since the voltage between the plates is changing 
with time, there is a displacement current between 
them; Eqs. (36-19) and (36-21) become 
ndA 
(36-24) 
and 
respectively. 
curl Η  = dO 
dt ' 
(36-25) 
36-3 
THE W A V E 
EQUATION 
One of the most important applications of Max-
well's equations is the derivation of the elec-
tromagnetic wave equation. It is possible to derive 
a wave equation for each of the field vectors E, H, 
B, and D. Let us derive the wave equation for the 
field vector E. We first take the curl of Maxwell's 
equation for the field vector Ε—namely, curl Ε = 
-dB/dt, 
This gives 
curl curl Ε = - ^ curl B. 
By the use of the theorem 
V^A = grad div A - curl curl A, 
we can write 
curl curl Ε = grad div Ε - V'E. 
(36-27) 
Substituting Eq. (36-27), Maxwell's equation 
curi Η  = j + (dOldt) and Β  = μ  Η  into Eq. (36-26), we 
obtain 
graddivE-V'E = - μ
^ - μ
^ . 
(36-28) 
If the field is uniform, the medium obeys Ohm's 
law, and the susceptibility is independent of the 
field intensity, then we can show that 
div Ε = 0, 
j = σ Ε , 
and 
D = eE. 
By substitution of these equations into Eq. (36-28), 
we get 
^ Ε - μ ί ^ - μ σ ^ = 0. 
(36-29) 
This is the three-dimensional wave equation for Ε 
in an absorbing medium. It will be left as a problem 
to derive the wave equation for each of H, B, and D. 
These equations are similar to Eq. (36-29). All four 
equations can be written collectively as 
) S 
(36-30) 
D 
In a medium in which the absorption or power 
loss to the medium is zero, the conductivity (σ ) is 
also zero, and Eq. (36-29) may be written 
Let us introduce a quantity υ such that 
-,2_ 1 
(36-31) 
and 
IF 
μ €  
= ϋ V E . 
(36-32) 
Dimensionally, Eq. (36-32) in MKS units is 
volts 
(36-26) 
3o that 
meters seconds 
2 volts 
meter? 
υ = 
meters 
second' 
Thus, it appears that ν  has the dimensions of speed. 
This speed is characteristic of the medium, being 
dependent on the constants μ  and e. We may, 
therefore, write 
1 
V = V μ €  
(36-33) 
The index of refraction η  of a medium, as indicated 
in Eq. (10-26), is 
V μ ο € ο  
(36-34) 
where c is the speed of electromagnetic radiation in 
space, μ r is the relative permeability, and Cr—the 
relative permittivity of the medium. 

322 
Maxwell's Equations and Electromagnetic Waves 
We have shown in this section that the elec-
tromagnetic field quantities have the attributes of 
wave phenomena. That is, they satisfy a wave equ-
ation and their speed of propagation is directly re-
lated to the medium in which they propagate. In the 
next section, we discuss in more detail the nature 
and energy of electromagnetic waves. 
Example 1. Polystyrene has a relative permea-
bility (μ ^) of 1 and a relative permittivity (Cr) of 
2.70. Find the index of refraction for polystyrene, 
and the speed of electromagnetic waves in poly-
styrene. 
SOLUTION 
η  = ν μ 7 . = V(2.70)= 1.65 
cy 
. 
c 
3 X10* m/sec 
Speed = —== = 
• — 
VJi^r 
1.65 
= 1.82xltf m/sec. 
36-4 
T H E N A T U R E A N D E N E R G Y O F 
E L E C T R O M A G N E T I C 
W A V E S 
A system of traveling electric and magnetic fields 
comprises what is called an electromagnetic wave. 
There is no microscopic vibration of a material sub-
stance taking place, but only an alternation of the 
electric and magnetic field. An electromagnetic 
wave can travel through empty space, that is, 
energy can be transmitted through empty space. 
This energy transmission through space by an elec-
tromagnetic wave must be related to the energy as-
sociated with electric and magnetic fields, which we 
described previously. We saw in Chapters 30 and 33 
that the energy per unit volume stored in an electric 
and magnetic field in empty space is equal to \€ oE^. 
and (l/2μ o)ß^ = l μ o H ^ respectively. The simplest 
form of an electromagnetic wave is one in which 
the electric field Ε and the magnetic field Η  vary 
sinusoidally in space and time, are at right angles to 
one another, have the same frequency, and are in 
phase. This kind of electromagnetic wave is called a 
plane polarized wave. Here, the total energy per 
unit volume, WIv, at any point is given by 
W / l ; = | € o E ^ + | μ o H ^ 
The Ε vector oscillates in one plane and the Η  
vector in a plane perpendicular to it, as illustrated in 
Figure 36-1. Therefore, the energy flow associated 
with the electromagnetic wave also oscillates and is 
Envelope of electric 
intensity vectors 
/ 
S 
Wave motion 
Envelope of magnetic induction vectors 
Figure 36-1 A simple electromagnetic wave. 
best represented by a vector (S in Figure 36-1). The 
sense of this vector gives the dh*ection of energy 
flow, and the length of the vector gives the mag-
nitude of the energy flow per unit area, per unit 
time, across a plane perpendicular to the vector. 
Although we will not prove it here, the rate at which 
this energy per unit area is transported is given by 
S = EXH. 
The vector S is called the Poynting vector. We see 
that, for energy to be transported in an elec-
tromagnetic field, we must have an electric and a 
magnetic field not parallel to one another. Why? It 
is worth repeating that Ε and Η  are not independent 
of one another. The change in Ε gives rise to H, and 
the change in Η  gives rise to E. In electrostatics, the 
Poynting vector is zero since a static charge does 
not produce a magnetic field. 
Example 2. If the earth has a downward electric 
field of 100 volts/m at the Equator and a northward 
magnetic field of 40 amperes per meter, what is the 
Poynting vector? 
SOLUTION 
Since Ε is perpendicular to H, the magnitude of S 
is EH. Therefore, 
S = EH= 100 volts/m x 40 amp/m 
= 4000watts/m'. 
The duection of S by the right-hand screw rule for 
vector products applied to the diagram following is 
east. 

The Electromagnetic Spectrum 
323 
36-5 
THE E L E C T R O M A G N E T I C S P E C T R U M 
Electromagnetic waves exist with all possible 
wavelengths, from the longest radio waves of about 
10'm to the shortest gamma rays of about IO'"* m. 
The shorter wavelengths are usually measured in 
either angstroms (1 Ä = 10~^° m) or X-ray units 
(1 XU = 10"'^ m). The electromagnetic spectrum 
has no definite upper or lower limit, and the spec-
trum regions overlap. Table 36-1 gives the approxi-
mate wavelength limits of the different types of 
electromagnetic waves and how each is produced. 
detected, or used. Light is defined as the portion of 
the electromagnetic spectrum that is capable of 
affecting the eye's sense of vision and ranges from 
about 4300 Ä to about 6900 Ä. 
We have previously discussed light in some de-
tail. However, we have said little on the other por-
tions of the electromagnetic spectrum. At the long 
wavelength end of the spectrum, we have radio 
waves. An antenna is used to produce the waves. 
The size of the antenna depends upon the 
wavelength to be transmitted. Antenna designers 
have found that antennas about one-quarter the 
wavelength give efficient transmission. Thus, anten-
nas can vary in size from 100 m in length for long 
wave radio transmission to a few centimeters for 
some radar sets operating in the microwave region. 
Thermal radiation is produced by heating a solid. 
When the solid is heated, it first appears a dull red, 
and its spectrum is almost wholly in the red. As its 
temperature is increased, its spectrum extends to 
include all the spectral colors from red to violet. 
The shorter the wavelength or the higher the fre-
quency of the radiation emitted, the greater the 
energy. Figure 36-2 is a distribution of the energy in 
a continuous spectrum of thermal radiation for 
different temperatures of the source. 
Infrared radiation is given out by most ordinary 
light sources. About 95% of the radiation given out 
by an ordinary incandescent lamp is in the infrared 
Table 36-1 The Electromagnetic Spectrum. 
Approximate 
How 
Kind of 
units of 
How 
detected 
radiation 
wavelength 
produced 
or used 
Radio waves 
ΙΟ ^-ΙΟ 'Ίη  
Oscillating electronic 
Radio, radar, television 
circuit 
Thermal radiation 
10-^-8 X 10-'m 
Hot bodies 
Heat effects, thermopile 
Infrared light 
5xlO-'-8xlO-'m 
Light sources 
Heat effects, photography 
Visible light 
7000-4000Ä 
Light sources 
Eye, photography, photo-
electric cell 
Ultraviolet light 
4000-50Ä 
Electric spark, gas dis-
Photography, photo-
charge, the sun 
electric cell, ionization. 
fluorescence 
X-rays 
50,000-100 XU 
Electron impact on solid 
Photography, ionization. 
bodies 
fluorescence 
7-rays 
100-6 XU 
Radioactive materials 
Ionization, photography 
Cosmic rays 
0.7-0.4 XU 
Bombardment of earth's 
Ionization, nuclear 
atmosphere by high 
disintegration 
energy particles from 
outer space 

324 
Maxwell's Equations and Electromagnetic Waves 
Figure 36-2 Distribution of energy in the continuous 
spectrum of thermal radiation, for different tempera-
tures of the source. 
region. The spectrum of both thermal radiation and 
infrared radiation can be studied by means of a 
thermopile. A thermopile consists of a large 
number of copper-bismuth thermocouple junctions 
in series, one set of junctions being exposed to the 
radiation, while the other set is protected. The cur-
rent produced is proportional to the radiant energy 
received by the instrument. The spectrum of in-
frared radiation can also be studied by means of 
specially prepared photographic plates and coarse-
ruled diffraction gratings. 
The greatest source of ultraviolet radiation ob-
served on the Earth is the Sun. Ultraviolet radiation 
in ordinary light has an important effect on animal 
health. When absorbed by the skin, it causes chemi-
cal reactions that are usually beneficial to the 
health. One of these reactions is the generation of 
vitamin D in both animal and vegetable tissue. 
However, overexposure to high frequency ul-
traviolet radiation can cause bums to the tissue, 
especially such sensitive tissue as that of the eye. 
As mentioned in Table 36-1, the ultraviolet portion 
of the spectrum can be studied by means of photo-
graphy, ionization, and fluorescence. In fluores-
cence, the emitted wavelength is longer than the 
Figure 36-3 X-ray tube. 

Problems 
325 
wavelength absorbed, so that when ultraviolet light 
falls on a fluorescent material, visible light is emit-
ted permitting the detection and study of the ul-
traviolet radiation. 
X-rays are produced by accelerating electrons in 
a vacuum tube. Figure 36-3 shows an X-ray tube. 
Electrons are emitted by the hot filament F, accel-
erated across the tube striking a target T, usually 
made of some heavy metal. The electrons are accel-
erated by a constant high potential placed across 
the tube. The high energy electrons, on hitting the 
target, cause it to radiate X-rays. X-rays affect a 
photographic plate, ionize gases, cause some mater-
ials to fluoresce, and can be detected and studied 
because of these properties. For the lighter ele-
ments, the absorption of X-rays is directly propor-
tional to the electron density in the absorbing ma-
terial. It is this property of materials that makes 
possible the successful application of X-rays to 
medicine and industry. For example, if a part of the 
body is placed between a beam of X-rays and a 
photographic plate, the photographic plate will 
show shadows of the bones, since the bones ab-
sorb X-ray radiation to a greater extent than the 
flesh. 
Gamma (γ-) rays are emitted by radioactive ma-
terials and have a greater penetrating power than 
ordinary X-rays. They affect a photographic plate 
and will ionize gases like X-rays. The γ-rays of 
radioactive elements are very often used in treat-
ment of cancer, although the use of very short 
wavelength X-rays for such treatments has in-
creased. 
Cosmic rays are present all over the Earth and 
come from outer space. In spite of many experi-
mental investigations, the origin of cosmic rays is 
still not completely understood. When they reach 
the Earth's surface, the rays consist of elec-
tromagnetic radiation of shorter wavelength than 
γ-rays along with high energy charged particles. 
They can be detected by such devices as ionization 
chambers and Geiger counters. 
P R O B L E M S 
1. A parallel plate capacitor of area A is connected to a voltage source that increases slowly and linearly 
with time. Find an expression for 
(a) the displacement current density and 
(b) the displacement current at a point between the plates. 
2. Starting from Maxwell's equations, show that the electric field intensity due to a point charge q is given 
by 
1 
E = 
47Γ € ο  r 
3. Paraffin has a relative permittivity € r of 2 and a relative permeability μ Γ of 1. Find: 
(a) the index of refraction for paraffin and 
(b) the speed of electromagnetic waves in paraffin. 
4. Write out Maxwell's equations in terms of Ε and Η  only for a non-homogeneous medium in which e and 
μ  are functions of the coordinates. 
5. Develop the wave equation in Ε for a plane wave traveling in the y-direction with Ε in the ζ -direction. 
6. Given a plane monochromatic wave traveling in a linear, isotropic homogeneous medium, show that the 
electric and magnetic energy densities are equal. 
7. A plane electromagnetic wave, having an electric field of magnitude Ε = 100 volts/m, is traveling in a 
non-conducting medium of relative permittivity 9 and relative permeability unity. Find: 
(a) the velocity of the wave, 
(b) the index of refraction of the medium, and 
(c) the energy density of the wave in the medium. Hint: Refer to Problem 6. 
8. A low frequency radio wave is incident upon a body of distilled water of relative permittivity 9 and 
relative permeability unity. Find: 
(a) the speed of the wave in the water and 
(b) the index of refraction of the water. 

326 
Maxwell's Equations and Electromagnetic Waves 
9. Derive Snell's law for light refraction through the consideration of polarized plane light waves incident 
obliquely at a plane dielectric interface. 
10. A plane radio wave travels in the jc-direction and is plane-polarized with its electric field vector in the 
z-direction. Its frequency is 10' cps. The average power propagation by the wave is 16 watts/m^ Find: 
(a) the wavelength of the wave and 
(b) the effective values of Ε and Η  for this wave. 
11. Calculate the Poynting vector for a plane wave in free space having an effective electric field of 50 
micro volts/m. 
12. A wire of diameter 2 nun, resistivity 10"^ ohm/m, carries a current of 5 amp. Calculate the component of 
the Poynting vector (S) perpendicular to the wire. 
13. Lines of magnetic induction are always closed. They neither originate nor terminate at any point, and as 
many flux lines enter a volume as leave it. Assuming the preceding statement is true, prove Maxwell's 
equation 
V - B = 0. 
14. A radar transmitter emits a 10 kilowatt pulse lasting for 10"' seconds. At a certain distance from the 
transmitter, the area of the wavefront is 0.5 km\ Find the average energy density within the pulse. 

Appendices 
Appendix I Vector Analysis 
327 
Appendix II Unit Conversion Factors 
335 
Appendix III 
Fundamental Constants 
335 
Appendix IV Mathematical Tables 
336 
A P P E N D I X I 
V E C T O R 
A N A L Y S I S 
1 
Vector Quantities and Vectors 
Vector quantities have magnitude and direction, 
for example, force, displacement, electric field, etc. 
The complete specification of a vector quantity re-
quues (1) a unit for the quantity, (2) a number stat-
ing how many times the unit is contained in the 
quantity, and (3) a statement of direction. 
Vectors are duected line segments or arrows 
used to represent vector quantities. The length of 
the line segment gives the magnitude and its sense 
the direction of the vector quantity. 
2 
Addition and Subtraction of Vectors 
Vectors may be added by two methods: the 
polygon method and the method of components, 
which will be discussed briefly in the next section. 
In the polygon method, the tail of the second vector 
is joined to the head of the first vector, the tail of 
the third to the head of the second, and so on, 
keeping their direction as given. The sum of the 
vectors is a vector drawn from the tail of the first 
vector to the head of the last vector. Figure 1 
represents the addition of two vectors A and B. C is 
the sum of the two vectors 
or 
A + B = C 
B + A = C. 
(1) 
Figure 1 Vector addition. 
Therefore, the sum of the two vectors is commuta-
tive. 
Subtraction of the vector Β  from the vector A is 
defined as the addition of the negative vector - Β  to 
A. Thus, 
A - B = A + (-B). 
The sum of any number of vectors is associative. 
That is, vectors may be added in any desired 
manner. 
R = (A + B)-fC = A + (B-hC) = (A + C) + B. 
(2) 
3 
Components of a Vector 
Vectors may be conveniently represented in Car-
tesian coordinates by means of the unit vector con-
cept. A unit vector is one of unit magnitude. 
327 

328 
Appendices 
Let the vectors i, j , k be unit vectors along with 
x-,y-, 
ζ -axes, respectively. A vector A can be con-
sidered to be the sum of the vectors iA„  jAy, and 
kA,, each of which lies along one of these axes. 
A = iA, +jAy +kA„ 
(3) 
where A„  Ay, and A^ are known as the Cartesian 
components of A. The sum of vectors A, B, C in 
terms of their components can be written 
A + B + C = i(A. + 
+ G ) 
j(Ay -f By + Cy) 
+ k ( A . + B . + C ) , 
(4) 
since the components along the x-axis add directly, 
as do those along the y- and ζ -axis. The magnitude 
of the length of a vector A is 
A = VÄ7+Ä7TÄ7. 
(5) 
Proof: 
Refer to Figure 2. 
A' 
= (OPf 
+ 
A^ 
(OPf 
= A.' + 
Ay\ 
A ' = A.' + Ay' + A / 
A = V Ä 7 + Ä 7 + Ä ? . 
or 
4 
The Products of Two Vectors 
(a) Scalar (Dot) Product. The scalar (dot) pro-
duct is the product of the magnitude of one vector 
by the magnitude of the projection of another vec-
tor upon it. The result is a scalar, and the process is 
commutative; that is. 
A Β  = Β  A. 
(6) 
Consider the scalar product of the two vectors A 
and Β  as shown in Figure 3. 
A Β  = Β  A = A B cos a 
(7) 
If A · Β  = 0, then A = 0, Β  = 0, or A is perpendicular 
to B. If A · Β  = AB, then A and Β  are parallel. 
The scalar products of the unit Cartesian vectors 
obey the relations 
i j = j k = k i = 0 
i i = j j = k 
k = l . 
(8) 
The scalar product of two vectors is the sum of 
the products of the corresponding components. 
Consider the two vectors A and Β  written in terms 
of their components: 
A = iA,+jAy+kAz 
and 
Β  = IB, + jBy + kB,. 
Multiplying out in the usual way, and then sub-
stituting the values of the products of the unit vec-
tors, the scalar product becomes 
A · Β  = Α ,Β . + AyBy + Λ Β . . 
(9) 
The scalar product is distributive, that is, 
Α · ( Β 4 - 0 = Α ·Β - Κ Α ·€ , 
(10) 
which we shall not prove here since its proof is 
available in any book on vector analysis. 
Figure 3 Scalar product of two vectors. 
(b) Vector (Cross) Product. The vector, or 
cross product, of two vectors A and Β  is written A x 
B. It is defined as the vector perpendicular to the 
plane determined by A and B, as illustrated in Fig-
ure 4, and is equal in magnitude to AB sin Θ, where 
θ is the angle between A and B, that is. 
V = A X Β  = π Α Β  sin θ. 
(11) 
Figure 2 Cartesian components of a vector. 
where η  is a unit vector in the direction of V. It is so 
directed that a 
right-handed 
rotation about V 
through an angle θ carries A into B. A x Β  is equal 
in magnitude to the area of the parallelogram. A x Β  

\/ector Analysis 
329 
Figure 4 Vector product of two vectors. 
is non-commutative 
A x B = - B x A , 
(12) 
since rotation from Β  to A is opposite to that from A 
to B. 
If A and Β  are parallel, 0 = 0° or 180° and A x Β  = 
0. Conversely, if the vector product is zero, one of 
the vectors is zero or else the two are parallel. If A 
and Β  are peφ endicular, A x B = nAB, in which 
case the two vectors and their product are mutually 
at right angles. 
The vector products of the unit Cartesian vectors 
are seen to obey the following relations: 
iXi = j X J = k X k = 0, 
i x j = - j x i = k, 
j χ  k = - k X j = i, 
k x i = - i x k = i. 
(13) 
The vector (cross) product of A and Β  may be 
expressed in terms of their components and the unit 
vectors. 
A X Β  = (iA, + jAy + kAJ X (IBx + jBy + kBJ, 
= (AyB, 
- ABy)i 
+ (A.a - A.BJj 
+ 
(AxBy-AyBx)k. 
As a determinant, this can be written 
A x B = 
ί 
j 
k 
Α χ  
Ay 
Az 
Bx 
By 
Bz 
(14) 
(15) 
The vector product, like the scalar product, is dis-
tributive; that is. 
5 
The Products of Three Vectors 
(a) Scalar Triple Product [A · (B χ  C)]. A (B χ  
C) is a scalar and represents the volume of a paral-
lelepiped as illustrated in Figure 5. Any face of the 
solid may be taken as the base. The three equival-
ent expresssions for the volume are: 
Α ·(Β  X C) = B-(C x A) = C -(A x B). 
(17) 
Since the order of the terms in a scalar product is 
immaterial, these relations are equivalent to 
(BxC) A = (CXA) B = (AXB) C 
(A X B)-C = (B X O - A = (C X Α )·Β . 
(18) 
or 
Therefore, in the scalar triple product, the dot and 
cross can be interchanged without changing the 
value of the result. 
The sign of the scalar triple product is unchanged 
as long as the cyclic order of the factors is un-
changed. For every change in cyclic order, a minus 
sign is introduced, for example, 
A (BXC) = - A (CXB). 
Therefore, three vectors have six identical scalar 
triple products. 
The scalar triple product may be expressed in 
determinant form. Consider the product A · (B x C). 
B x C = 
i 
j 
k 
Bx 
By 
Bz 
Cx 
Cy 
Cz 
A. = 
iA,+iA,+kA„  
A - ( 8 X 0 = 
A, 
B. 
Ay 
Az 
By 
Bz 
Cy 
Cz 
(19) 
When three vectors are in a plane, the volume of 
the parallelopiped formed by them is zero. Hence, 
Ax(BH-C) = A x B + A x C . 
(16) 
Figure 5 The scalar triple product. 

330 
Appendices 
the condition for three vectors to be coplanar is that 
their scalar triple product equals zero. 
(b) Vector Triple Product [Ax(BxC)]. 
Ax 
(B X C) is a vector. The sign of the product changes 
every time the order of the factors A and Β  x C is 
changed in A x (B x C) or whenever the order of the 
factors Β  and C is changed in (B x C). The vector 
Β  X C is perpendicular to the plane containing Β  and 
C. The vector A x (B x C) is perpendicular to the 
plane containing A and (B x C) and is, therefore, in 
the same plane as Β  and C, as illustrated in Figure 6. 
The product A x (B x C) can be reduced to a sim-
pler form as follows: Let the x-axis be along vector 
B, the y-axis in the plane Β  and C, and z-axis along 
Β  X C, as in Figure 6. 
A = iA.+jBy+kß„ 
B = ÍB., 
C = i G + J G , 
B x C = kB.G, 
A χ  (Β  χ  C) = (ΙΑ . + ¡Ay + kA.) Χ  (kB.Cy) 
= 
ÍAyB,Cy-}AB,Cy. 
But 
and 
A · C = (ÍA. + iAy + kA.) · (ÍG + }Cy) 
AxCx ~l~ A.yCy 
A · Β  = (Μ , 
+jAy 
+kA.) 
· 
(¡B.) 
= A A . 
From Eqs. (24), (21), and (22), 
(20) 
(21) 
(22) 
(23) 
(24) 
(25) 
(26) 
^ - ^ ' " ^ 
Ax(BxC) 
A X (B X C) = AyCyB 
- jA.B.G 
= (AyG + A . G ) B - Í A . B . G - j A A G 
= (A.G 
+ AyCy 
)B - A.B. (ÍG 
+ j G 
). 
From Eqs. (22), (25), and (26), 
A X (B X C) = (A · C)B - (A · B)C. 
(27) 
Each term of the product involves the external fac-
tor in a scalar product first with the extreme and 
then with the middle factor. 
6 
Vector Integration 
(a) Line Integral. Suppose ds is an element of 
length along a curve, where s = s(i) is the equation 
of a curve. Let V denote a vector at an angle θ with 
that of the length element, as illustrated in Figure 7. 
It is possible to form the integrals 
| v d s = J VcosÖds 
and 
(28) 
V x d s = n 
V sin θ ds. 
(29) 
each being a line integral along the curve C. Such 
integrals occur frequently, for example, if V is a 
force on a particle and ds an element of its dis-
placement, Eq. (28) denotes the work done on the 
particle; while if V is a force on an extended body 
free to rotate about a given axis and ds is an ele-
ment of its moment arm, Eq. (29) denotes the tor-
que on the body. Also, the integral / Sds could be 
formed where S is a scalar function. 
(b) Surface and Volume integrals. Consider a 
surface A drawn in any vector field, as illustrated in 
Figure 8. Let V be the value of the vector at an 
Figure 6 The vector triple product. 
Figure 7 Tangential line integral of a vector. 

Vector Analysis 
331 
Figure 8 Normal surface integral of a vector. 
element of surface dA, η  a unit outward vector nor-
mal to dA, and θ the angle between π  and V. The 
total outward flux is then given by 
φ =j^\'ndA=j^V 
cos 
edA. 
(30) 
The outward flux or flow is positive if it is in the 
direction of the outward unit normal vector n, and 
negative—if in the opposite duection. Other sur-
face integrals that may be formed are 
f SdA, 
I SndA 
and 
( \xndA, 
(31) 
JA 
JA 
JA 
where S is a scalar function. 
Consider a closed surface enclosing a volume v. 
The following volume integrals may be formed 
I. 
\dv 
and Jsdt;, 
(32) 
where V and S are vector and scalar ñelds, respec-
tively. 
7 
Differentiation 
(a) Differentiation of Vectors. If V is a vector 
function of a scalar variable t, then when t changes 
from ί to ί +Δ ί, V becomes V + AV, and Δ Υ /Δ ί = 
average rate of change of V with t. Also, 
y i ? Δ ί 
dt' 
which is the derivative of V with respect to t. When 
V is expressed in rectangular coordinates, 
V = iV.+jV,+kV.. 
Since Vx, Vy, Vz are functions of t and i, j , k are unit 
vectors. 
dt ' d í ^ d í 
" d f 
and similarly for higher derivatives. 
(33) 
(b) Differentiation of S u m s and Products. If 
V = A + B, which are all functions of i, 
- ^ - ^ ( A + B ) - ^ + - ^ , 
(34) 
so we can say that differentiation of vector sums is 
distributive. 
Consider the scalar product 
and 
S = A B 
. ^ _ d A 
- - ^ • B + A 
dS 
dt 
dB 
dt' 
(35) 
Since the order of the factors can be interchanged 
in Eq. (35), the differentiation of scalar products 
gives a result that is commutative. 
Consider the vector product 
and 
V = A x B 
dl^dX 
dB 
dt - dt ^ B + A X - ^ . 
(36) 
Since a change in order of the factors changes the 
sign of the result, the differentiation of vector pro-
ducts gives a result that is non-commutative. 
(c) Partial Differentiation. Consider a vector as 
a function of more than one scalar independent var-
iable. Let a vector V be a function of the Cartesian 
coordinates x, y, and z. If y and ζ  remain constant 
while X changes, the partial derivative d\/dx 
de-
notes the rate of change of V with respect to x. 
If X, y, and ζ  change simultaneously by dx, dy, 
and dz, the total change or total differential is 
d V = — dx + - 7 - dy + — dz, 
dx 
dy 
^ 
dz 
' 
which may be written 
dV = 
V. 
(37) 
If r = ix Η -jy +kz is the radius vector from the 
origin, then 
dr = idx +jdy -hkdz. 
(38) 

332 
Appendices 
We define an operator V by 
(39) 
By the use of this operator, Eq. (39), and Eq. (38), 
Eq. (37) may be rewritten as 
dV = ( V d r ) V . 
(40) 
The vector differential operator V is of great 
physical significance in vector analysis. It is useful 
in defining three important quantities—namely, the 
gradient, divergence, and curl, which will now be 
discussed. 
8 
Gradient, Divergence, and Curl 
(a) The Gradient of a Scalar Field. In studying 
the field of some scalar function S we sometimes 
need to know the rate of change of S in going from 
one point to another. This information is obtained 
with the aid of a vector called the gradient of the 
field. 
When S(x, y,z) is a scalar point function, then 
the vector 
dx 
* dy 
dz 
is called the gradient of S, grad S, or VS. Symboli-
cally, in Cartesian coordinates. 
dx ' dy 
dz 
(41) 
Gradient S or VS is a vector perpendicular to a 
level or equipotential surface at the point x, y, z, a 
vector which is equal in magnitude to the fastest 
rate of increase of S with distance and points in the 
direction of the fastest rate of increase of S. 
(b) The Divergence of a Vector Field. If V 
represents a vector field whose components, to-
gether with their partial derivatives, are continuous, 
a useful scalar field can be derived from it by the 
scalar multiplication of V and V. The scalar quan-
tity resulting from V · V is known as the divergence 
of V or div V. Symbolically, in Cartesian coordi-
nates. 
or 
dx 
dy 
dz ' 
(42) 
The divergence of a vector field at a point gives 
the rate of flow per unit volume into or out of a 
volume element. If the divergence is positive, the 
flow is away from the point; if it is negative, the 
flow is toward the point. In the case of thermal, 
electric, or magnetic fields, the existence of a di-
vergence means the presence of a source or sink of 
flux at the point. When the divergence is zero 
everywhere, there are no sources or sinks and the 
flux leaving equals the flux entering any volume ele-
ment. 
(c) The Curl of a Vector Field. The curi of a 
vector field V is the vector multiplication of V and 
V. The vector quantity resulting from V x V is 
known as the curl of V or curl V. Symbolically, in 
Cartesian coordinates. 
+ k 
1 
j 
k 
d 
d 
d 
dx 
dy 
dz 
or 
\dy 
dzj 
'\dz 
dx) 
+ k idVy_dV\ 
\ dx 
dy ) ' 
(43) 
In order to discuss the physical meaning of the 
curl of a vector, we shall first define the mathemati-
cal quantity circulation. Circulation is the line in-
tegral of a vector field along a given path. If the 
vector field is V, its circulation about a closed path 
is 
V d s , 
where ds is an element of length along the path. 
The curl of a vector field at a point is the limit ap-
proached by the ratio 
circulation about a small closed path 
area of surface bounded by that path 
as the path is allowed to shrink to a point. The curl 
is thus the limiting value of circulation per unit 
area. 

Vector Ar}alysis 
333 
9 
More Vector Operators 
(a) Div Grad. If 5 is a scalar function of position 
in space, 
div grad S= V· VS 
and the operation 
divgrad = V^ = |l 
+ ^ 
+ ¿, 
(44) 
is known as Laplace's operator. 
can also oper-
ate on a vector, if 
V = i V . + j V , + k V , 
v^v = vH Vx + ν  j V, + V'k V. 
This result is important in mechanics and in elec-
tricity and magnetism. 
(b) Curl Grad. If S is a scalar point function of 
position in space, 
is a vector, and it is possible to calculate its curl, 
curl grad S = V x VS 
i 
j 
k 
d 
d 
d 
dx 
dy 
dz 
dS 
dS 
dS 
dx 
dy 
dz 
= 0. 
(45) 
The result illustrates the fact that the line integral of 
any vector resulting from the gradient of a scalar 
(for example, grad S) around a closed path is zero. 
(c) Grad Div. If V is a vector field, div V is a 
scalar field which, therefore, has a gradient. 
grad div V = V(VV) 
Vdx 
'dy"dzJXdx 
^ dy 
dz) 
\ dx^ 
dxdy 
dxdz) 
\dxdz 
dydz 
dz'r 
(d) Div Curl 
(46) 
curl V = 
i 
j 
k 
d 
d 
d 
av; 
dx 
dy 
dz "Uy" az> 
Vx 
Vy 
V 
/av._m 
/av,_avA 
\dz 
dxP^Kdx 
dyn 
and 
,
^
j
d 
/ a v 
aVy\ 
a /av, 
aVA 
div curi V = - ( — - — ) 4- - 
( — - — j 
a 
idVy 
dVx\ 
\dx 
dy) 
dz 
d'Vz 
d'Vy 
, 
d'Vx 
dxdy 
dxdz 
dydX 
a^V 
^ d'Vy 
d'Vx^^ 
dydX 
dzdX 
dxdy 
(e) Curl Curl. 
curi V = 
ί 
J 
k 
a 
a 
a 
aVy 
dx 
ay 
dz 
- ^ a y 
dz 
Vx 
Vy 
V 
.(dV^_dV,\ 
(dV,_dV\ 
Adz 
dx)^\dx 
dy)' 

334 
Appendices 
curl curl V = 
_d_ 
dx 
_d_ 
dy 
_d_ 
dz 
(dV_dVy\ 
(dV_dV\ 
(dVy_dVÄ  
\dy 
dz)\dz 
dx)\dx 
dy) 
dxdy 
dy^ 
dzdx 
4- analogous terms for j and k 
= grad div - V\ 
so that the operation 
curl curl = grad div - V^ 
(f) Other Formulas Involving V. If A and Β  are 
differentiable vector functions, and φ  and φ  are 
differentiable scalar function of position x, y, z. 
then 
ν (φ  + ψ) = ν φ  + νψ  
ν ·(Α  + Β ) = ν · Α  + ν -Β  
Vx(A + B) = V x A + V x B 
V · (ψ Α ) = (ν φ )· Α  + φ (ν · A) 
ν χ (ψ Α ) = (ν φ )χ Α +φ (ν χ Α ) 
ν(φ ψ) = (ν<ί>)ψ  + <ί>(νψ) 
ν ·(Α χ Β ) = B ( V x A)-A-(VxB) 
ν χ (Α χ Β ) = (B-V)A-B(VA) 
- ( Α · V)B + A(VB) 
V(AB) = (Β  V)A + (A V)B + B 
x(VxA) + Ax(VxB). 
10 
G a u s s ' s Divergence Theorem and Stokes' 
Theorem 
The total flux through a closed surface A is given 
by 
φ  = 
^\'η ά Α, 
where V is a vector field and η  is a unit outward 
vector normal to dA. Also, the total flux diverging 
from the volume ν  enclosed by the surface A is 
given by 
Therefore, 
φ =ο  
V'\dv. 
Jv 
£ v n d A = £ ν · Vdü , 
(47) 
which is Gauss's divergence theorem. 
Stokes' theorem states that the tangential line in-
tegral of a vector function V around a closed path is 
equal to the normal surface integral of curl V over 
the enclosed surface or cap enclosed by the path. 
In symbols. 
η  · curl VdA, 
(48) 
where η  is a unit vector normal to the element of 
surface. 
Proof: 
Let the surface (Figure 9) be divided into in-
finitesimal surface elements. Consider the shaded 
area of the vector area ndA. At the middle of dA, 
the vector field has a curl. From the definition of the 
curl (namely, that it is the limiting value of the cir-
culation per unit area), we may write 
< ) V d s = n-curi VdA 
for this element of area. A similar process is applied 
to all the elements. The line integral along the com-
mon sides of the elements cancel, thus leaving only 
the contributions along the surface boundary. 
Therefore, 
^ V d s = £ η  curi VdA, 
which is Stokes' theorem. 
Figure 9 Stokes's theorem. 

Unit Conversion Factors/Fundamental 
Constants 
335 
A P P E N D I X II 
UNIT C O N V E R S I O N 
F A C T O R S 
1 
Length 
1 meter 
= 39.37 inches 
1 kilometer= 0.6214 miles 
1 angstrom = 10"'° meters 
1 micron 
= 10"^ meters 
1 foot 
= 0.3048 meters 
1 inch 
= 0.0254 meters 
1 mile 
= 1.609 kilometers 
2 
Force 
1 newton = 10' dynes = 0.2248 pounds 
1 kilogram-force = 9.807 newtons = 2.205 pounds 
1 pound = 0.4536 kilogram-force 
3 
M a s s 
1 kilogram = 0.0685 slugs 
1 slug = 14.59 kilograms 
Pressure 
atmosphere = 1.013 x 10' nt/m^ 
= 1.013x10'dynes/cm' 
= 14.7 lb/in' 
5 
Energy 
1 joule = 
1 kilocalorie = 
1 electron volt = 
1 kilowatt-hour = 
1 foot-pound = 
1 B.T.U. = 
10' ergs = 0-7376 foot-pounds 
4185 joules = 3.%8 B.T.U. 
3086 foot-pounds 
1.60x10"'' ergs 
= 1.60x10" joules 
= 2.655 X 10' foot-pounds 
= 1.356 joules 
= 778 foot-pound 
= 0.252 kilocalories 
Power 
1 kilowatt = 737.6 foot-pounds/sec 
= 1.341 horsepower 
1 horsepower = 550 foot-pounds/sec 
= 0.7457 kilowatts 
A P P E N D I X ill 
F U N D A M E N T A L 
C O N S T A N T S 
Name of quantity 
Symbol 
Value 
Velocity of light in vacuum 
c 
2.9979x 10« m/sec 
Charge of electron 
Qe 
-1.602x10·'coul 
Rest mass of electron 
rrie 
9.109x10" kg 
Planck's constant 
h 
6.625 x lO"'''* joules-sec 
Boltzmann's constant 
1.380 X 10"" joules/°K 
Avogadro's number 
N o 
6.025 X 10^' molecules/mole 
Universal gas constant (chemical 
scale) 
R 
8.314 joules/mole-°K 
Mechanical equivalent of heat 
J 
4.185 joules/cal 
Standard atmospheric pressure 
1 atm 
1.013x10'nt/m' 
Volume of ideal gas at 0°C and 
1 atm 
22.415 liters/mole 
Absolute zero of temperature 
0°K 
-273.15°C 
Acceleration due to gravhy (sea 
8 
level, at equator) 
9.78049 m/sec' 
Universal gravitational constant 
G 
6.673 X 10"·'nt-m'/kg' 
Mass of earth 
niE 
5.975 xlO'" kg 
Mean radius of earth 
6.371 X 10* m = 3959 miles 
Coulomb's law constant 
Κ 
8.98X lO'nt-m'/coul' 
Permittivity of free space 
€ o 
8.85X10--"^"";,' 
coup 
Permeability of free space 
μ ο  
47Γ X 10"^ nt/amp' 

336 
Appendices 
A P P E N D I X IV 
M A T H E M A T I C A L 
T A B L E S 
1 
Natural Trigonometric Functions 
Angle 
Angle 
Angle 
De-
Ra-
Co-
Tan-
De-
Ra-
Co-
Tan-
De-
Ra-
Co-
Tan-
gree dian 
Sine 
sine 
gent 
gree dian Sine 
sine 
gent 
gree dian 
Sine 
sine 
gent 
0° 
.000 
0.000 
1.000 
0.000 
r .017 
.018 
1.000 
.018 
31° 
.541 
.515 
.857 
.601 
61° 
1.065 
.875 
.485 
1.804 
2° 
.035 
.035 
0.999 
.035 
32° 
.559 
.530 
.848 
.625 
62° 
1.082 
.883 
.470 
1.881 
30 
.052 
.052 
.999 
.052 
33° 
.576 
.545 
.839 
.649 
63° 
1.100 
.891 
.454 
l.%3 
4° 
.070 
.070 
.998 
.070 
34° 
.593 
.559 
.829 
.675 
64° 
1.117 
.899 
.438 
2.050 
5° 
.087 
.087 
.9% 
.088 
35° 
.611 
.574 
.819 
.700 
65° 
1.134 
.906 
.423 
2.145 
6° 
.105 
.105 
.995 
.105 
36° 
.628 
.588 
.809 
.727 
66° 
1.152 
.914 
.407 
2.246 
7° 
.122 
.122 
.993 
.123 
37° 
.646 
.602 
.799 
.754 
67° 
1.169 
.921 
.391 
2.356 
8° 
.140 
.139 
.990 
.141 
38° 
.663 
.616 
.788 
.781 
68° 
1.187 
.927 
.375 
2.475 
9 0 
.157 
.156 
.988 
.158 
39° 
.681 
.629 
.777 
.810 
69° 
1.204 
.934 
.358 
2.605 
10° 
.175 
.174 
.985 
.176 
40° 
.698 
.643 
.766 
.839 
70° 
1.222 
.940 
.342 
2.747 
11° 
.192 
.191 
.982 
.194 
41° 
.716 
.658 
.755 
.869 
71° 
1.239 
.946 
.326 
2.904 
12° 
.209 
.208 
.978 
.213 
42° 
.733 
.669 
.743 
.900 
72° 
1.257 
.951 
.309 
3.078 
13° 
.227 
.225 
.974 
.231 
43° 
.751 
.682 
.731 
.933 
73° 
1.274 
.956 
.292 
3.271 
14° 
.244 
.242 
.970 
.249 
44° 
.768 
.695 
.719 
.966 
74° 
1.292 
.%1 
.276 
3.487 
15° 
.262 
.259 
.966 
.268 
45° 
.785 
.707 
.707 
1.000 
75° 
1.309 
.966 
.259 
3.732 
16° 
.279 
.276 
.961 
.287 
46° 0.803 .719 
.695 
1.036 
76° 
1.326 
.970 
.242 
4.011 
17° 
.297 
.292 
.956 
.306 
47° 
.820 
.731 
.682 
1.072 
77° 
1.344 
.974 
.225 
4.331 
18° 
.314 
.309 
.951 
.325 
48° 
.838 
.743 
.669 
1.111 
78° 
1.361 
.978 
.208 
4.705 
19° 
.332 
.326 
.946 
.344 
49° 
.855 
.755 
.656 
1.150 
79° 
1.379 
.982 
.191 
5.145 
20° 
.349 
.342 
.940 
.364 
50° 
.873 
.766 
.643 
1.192 
80° 
1.3% .985 
.174 
5.671 
21° 
.367 
.358 
.934 
.384 
51° 
.890 
.777 
.629 
1.235 
81° 
1.414 
.988 
.156 
6.314 
22° 
.384 
.375 
.927 
.404 
52° 
.908 
.788 
.616 
1.280 
82° 
1.431 
.990 
.139 
7.115 
23° 
.401 
.391 
.921 
.425 
53° 
.925 
.799 
.602 
1.327 
83° 
1.449 
.993 
.122 
8.144 
24° 
.419 
.407 
.914 
.445 
54° 
.942 
.809 
.588 
1.376 
84° 
1.466 
.995 
.105 
9.514 
25° 
.436 
.423 
.906 
.466 
55° 
.960 
.819 
.574 
1.428 
85° 
1.484 
.996 
.087 
11.43 
26° 
.454 
.438 
.899 
.488 
56° 
.977 
.829 
.559 
1.483 
86° 
1.501 
.998 
.070 
14.30 
27° 
.471 
.454 
.891 
.510 
57° 
.995 
.839 
.545 
1.540 
87° 
1.518 
.999 
.052 
19.08 
28° 
.489 
.470 
.883 
.532 
58° 
1.012 
.848 
.530 
1.600 
88° 
1.536 
.999 
.035 
28.64 
29° 
.506 
.485 
.875 
.554 
59° 
1.030 
.857 
.515 
1.664 
89° 
1.553 1.000 
.018 
57.29 
30° 
.524 
.500 
.866 
.577 
60° 
1.047 
.866 
.500 
1.732 
90° 
1.571 1.000 
.000 
00 

2 Table of Logarithms to B a s e 10 
Ν  
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
1 
2 
P.P. 
3 
4 
5 
10 0000 0043 0086 0128 0170 0212 0253 0204 0334 0374 4 
8 
12 17 21 
11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 
8 
11 15 19 
12 0792 0828 0864 0899 0934 0%9 1004 1038 1072 1106 3 
7 
10 14 17 
13 1139 1173 1206 1239 1271 1303 1335 1307 1399 1430 
3 
6 
10 13 16 
14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 
3 
6 
9 12 15 
15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 
3 
6 
8 11 14 
16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 
3 
5 
8 11 13 
17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 2 
5 
7 10 12 
18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 
2 
5 
7 
9 12 
19 2788 2810 2833 2856 2878 2900 2923 2945 2%7 2989 2 
4 
7 
9 11 
20 3010 3032 3054 3075 30% 3118 3139 3160 3181 3201 
2 
4 
6 
8 11 
21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 2 
4 
6 
8 10 
22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 
2 
4 
6 
8 10 
23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 
2 
4 
5 
7 
9 
24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3%2 
2 
4 
5 
7 
9 
25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 
2 
3 
5 
7 
9 
26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 2 
3 
5 
7 
8 
27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 
2 
3 
5 
6 
8 
28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 2 
3 
5 
6 
8 
29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 
1 
3 
4 
6 
7 
30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 
1 
3 
4 
6 
7 
31 4914 4928 4942 4955 4%9 4983 4997 5011 5024 5038 
1 
3 
4 
6 
7 
32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 
1 
3 
4 
5 
7 
33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 
1 
3 
4 
5 
6 
34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 
1 
3 
4 
5 
6 
35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 
1 
2 
4 
5 
6 
36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 
1 
2 
4 
5 
6 
37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 
1 
2 
3 
5 
6 
38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 
1 
2 
3 
5 
6 
39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 
1 
2 
3 
4 
6 
40 6021 6031 6042 6053 6064 6075 6085 60% 6107 6117 
1 
2 
3 
4 
5 
41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 
1 
2 
3 
4 
5 
42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 
1 
2 
3 
4 
5 
43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 
1 
2 
3 
4 
5 
44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 
1 
2 
3 
4 
5 
45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 
1 
2 
3 
4 
5 
46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 
1 
2 
3 
4 
5 
47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 
1 
2 
3 
4 
5 
48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6803 
1 
2 
3 
4 
4 
49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 
1 
2 
3 
4 
4 
50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 
1 
2 
3 
3 
4 
51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 
1 
2 
3 
3 
4 
52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 
1 
2 
2 
3 
4 
53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 
1 
2 
2 
3 
4 
54 7324 7332 7340 7348 7356 7364 7372 7380 7388 73% 
1 
2 
2 
3 
4 
337 

Table of Logarithms to B a s e 10 (cont'd) 
Ν  
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
1 
2 
p.p. 
3 
4 
5 
55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 
1 
2 
2 
3 
4 
56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 
1 
2 
2 
3 
4 
57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 
1 
2 
2 
3 
4 
58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 
1 
1 
2 
3 
4 
59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 
1 
2 
3 
4 
60 7782 7789 77% 7803 7810 7818 7825 7832 7839 7846 
1 
1 
2 
3 
4 
61 7853 7860 7868 7875 7882 7889 78% 7903 7910 7917 
1 
1 
2 
3 
4 
62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 
1 
1 
2 
3 
3 
63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 
1 
1 
2 
3 
3 
64 8062 8069 8075 8082 8089 80% 8102 8109 8116 8122 
1 
2 
3 
3 
65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 
1 
1 
2 
3 
3 
66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 
1 
1 
2 
3 
3 
67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 
1 
1 
2 
3 
3 
68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 
1 
1 
2 
3 
3 
69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 
1 
2 
3 
3 
70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 
1 
1 
2 
2 
3 
71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 
1 
1 
2 
2 
3 
72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 
1 
1 
2 
2 
3 
73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 
1 
1 
2 
2 
3 
74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 
1 
2 
2 
3 
75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 
1 
1 
2 
2 
3 
76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 
1 
1 
2 
2 
3 
77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 
1 
1 
2 
2 
3 
78 8921 8927 8932 8938 8943 8949 8954 8960 8%5 8971 
1 
1 
2 
2 
3 
79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 
1 
2 
2 
3 
80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 
1 
1 
2 
2 
3 
81 9085 9090 90% 9101 9106 9112 9117 9122 9128 9133 
1 
1 
2 
2 
3 
82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 
1 
1 
2 
2 
3 
83 9191 91% 9201 9206 9212 9217 9222 9227 9232 9238 
1 
1 
2 
2 
3 
84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 
1 
2 
2 
3 
85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 
1 
1 
2 
2 
3 
86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 
1 
1 
2 
2 
3 
87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 0 
1 
1 
2 
2 
88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 
0 
1 
1 
2 
2 
89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 0 
^ 
2 
2 
90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 0 
1 
1 
2 
2 
91 9590 9595 9600 9605 9609 %14 %19 %24 %28 %33 
0 
1 
1 
2 
2 
92 %38 9643 9647 %52 %57 9661 9666 %71 %75 9680 0 
1 
1 
2 
2 
93 9685 9689 %94 %99 9703 9708 9713 9717 9722 9727 0 
1 
1 
2 
2 
94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 0 
2 
2 
95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 
0 
1 
1 
2 
2 
96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 0 
1 
1 
2 
2 
97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 0 
1 
1 
2 
2 
98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 0 
1 
1 
2 
2 
99 9956 9%1 9%5 9%9 9974 9978 9983 9987 9991 99% 
0 
2 
2 
338 

3 Table of Exponentials 
χ  
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
0.0 
1.000 
1.010 
1.020 
1.031 
1.041 
1.051 
1.062 
1.073 
1.083 
1.094 
0.1 
1.105 
1.116 
1.127 
1.139 
1.150 
1.162 
1.174 
1.185 
1.197 
1.209 
0.2 
1.221 
1.234 
1.246 
1.259 
1.271 
1.284 
1.297 
1.310 
1.323 
1.336 
0.3 
1.350 
1.363 
1.377 
1.391 
1.405 
1.419 
1.433 
1.448 
1.462 
1.477 
0.4 
1.492 
1.507 
1.522 
1.537 
1.553 
1.568 
1.584 
1.600 
1.616 
1.632 
0.5 
1.649 
1.665 
1.682 
1.699 
1.716 
1.733 
1.751 
1.768 
1.786 
1.804 
0.6 
1.822 
1.840 
1.859 
1.878 
1.896 
1.916 
1.935 
1.954 
1.974 
1.994 
0.7 
2.014 
2.034 
2.054 
2.075 
2.0% 
2.117 
2.138 
2.160 
2.181 
2.203 
0.8 
2.226 
2.248 
2.270 
2.293 
2.316 
2.340 
2.363 
2.387 
2.411 
2.435 
0.9 
2.460 
2.484 
2.509 
2.535 
2.560 
2.586 
2.612 
2.638 
2.664 
2.691 
1.0 
2.718 
2.746 
2.773 
2.801 
2.829 
2.858 
2.886 
2.915 
2.945 
2.974 
1.1 
3.004 
3,034 
3.065 
3.0% 
3.127 
3.158 
3.190 
3.222 
3.254 
3.287 
1.2 
3.320 
3.353 
3.387 
3.421 
3.456 
3.490 
3.525 
3.561 
3.597 
3.633 
1.3 
3.669 
3.706 
3.743 
3.781 
3.819 
3.857 
3.8% 
3.935 
3.975 
4.015 
1.4 
4.055 
4.0% 
4.137 
4.179 
4.221 
4.263 
4.306 
4.349 
4.393 
4.437 
1.5 
4.482 
4.527 
4.572 
4.618 
4.665 
4.712 
4.759 
4.807 
4.855 
4.904 
1.6 
4.953 
5.003 
5.053 
5.104 
5.155 
5.207 
5.259 
5.312 
5.366 
5.419 
1.7 
5.474 
5.529 
5.585 
5.641 
5.697 
5.755 
5.812 
5.871 
5.930 
5.989 
1.8 
6.050 
6.110 
6.172 
6.234 
6.297 
6.360 
6.424 
6.488 
6.554 
6.619 
1.9 
6.686 
6.753 
6.821 
6.890 
6.959 
7.029 
7.099 
7.171 
7.243 
7.316 
2.0 7.389 
7.463 
7.538 
7.614 
7.691 
7.768 
7.846 
7.925 
8.004 
8.085 
2.1 
8.166 
8.248 
8.331 
8.415 
8.499 
8.585 
8.671 
8.758 
8.846 
8.935 
2.2 9.025 
9.116 
9.207 
9.300 
9.393 
9.488 
9.583 
9.679 
9.777 
9.875 
2.3 9.974 
10.07 
10.18 
10.28 
10.38 
10.49 
10.59 
10.70 
10.80 
10.91 
2.4 
11.02 
11.13 
11.25 
11.36 
11.47 
11.59 
11.70 
11.82 
11.94 
12.06 
2.5 
12.18 
12.30 
12.43 
12.55 
12.68 
12.81 
12.94 
13.07 
13.20 
13.33 
2.6 
13.46 
13.60 
13.74 
13.87 
14.01 
14.15 
14.30 
14.44 
14.59 
14.73 
2.7 
14.88 
15.03 
15.18 
15.33 
15.49 
15.64 
15.80 
15.% 
16.12 
16.28 
2.8 
16.44 
16.61 
16.78 
16.95 
17.12 
17.29 
17.46 
17.64 
17.81 
17.99 
2.9 
18.17 
18.36 
18.54 
18.73 
18.92 
19.11 
19.30 
19.49 
19.69 
19.89 
3.0 
20.09 
20.29 
20.49 
20.70 
20.91 
21.12 
21.33 
21.54 
21.76 
21.98 
3.1 
22.20 
22.42 
22.65 
22.87 
23.10 
23.34 
23.57 
23.81 
24.05 
24.29 
3.2 
24.53 
24.78 
25.03 
25.28 
25.53 
25.79 
26.05 
26.31 
26.58 
26.84 
3.3 
27.11 
27.39 
27.66 
27.94 
28.22 
28.50 
28.79 
29.08 
29.37 
29.67 
3.4 
29.% 
30.27 
30.57 
30.88 
31.19 
31.50 
31.82 
32.14 
32.46 
32.79 
χ  
.0 
.1 
.2 
.3 
.4 
.5 
.6 
.7 
.8 
.9 
3 
20.09 
22.20 
24.53 
27.11 
29.% 
33.12 
36.60 
40.45 
44.70 
49.40 
4 
54.60 
60.34 
66.69 
73.70 
81.45 
90.02 
99.48 
109.9 
121.5 
134.3 
5 
148.4 
m.o 
181.3 
200.3 
221.4 
244.7 
270.4 
298.9 
330.3 
365.0 
6 
403.4 
445.9 
492.7 
544.6 
601.8 
665.1 
735.1 
812.4 
897.8 
992.3 
7 
1097 
un 
1339 
1480 
1636 
1808 
1998 
2208 
2441 
2697 
8 
2981 
3295 
3641 
4024 
4447 
4915 
5432 
6003 
6634 
7332 
9 
8103 
8955 
9897 
10938 
12088 
13360 
14765 
16318 
18034 
19930 
log.oß" = χ  log.o« = 0.43429 χ  
339 

Table of Exponentials (contd) 
X 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
0.0 
1.000 
.9900 
.9802 
.9704 
.9608 
.9512 
.9418 
.9324 
.9231 
.9139 
0.1 
.9048 
.8958 
.8869 
.8781 
.8694 
.8607 
.8521 
.8437 
.8353 
.8270 
0.2 
.8187 
.8106 
.8025 
.7945 
.7866 
.7788 
.7711 
.7634 
.7558 
.7483 
0.3 
.7408 
.7334 
.7261 
.7189 
.7118 
.7047 
.6977 
.6907 
.6839 
.6771 
0.4 
.6703 
.6637 
.6570 
.6505 
.6440 
.6376 
.6313 
.6250 
.6188 
.6126 
0.5 
.6065 
.6005 
.5945 
.5886 
.5827 
.5769 
.5712 
.5655 
.5599 
.5543 
0.6 
.5488 
.5434 
.5379 
.5326 
.5273 
.5220 
.5169 
.5117 
.5066 
.5016 
0.7 
.4966 
.4916 
.4868 
.4819 
.4771 
.4724 
.4677 
.4630 
.4584 
.4538 
0.8 
.4493 
.4449 
.4404 
.4360 
.4317 
.4274 
.4232 
.4190 
.4148 
.4107 
0.9 
.4066 
.4025 
.3985 
.3946 
.3906 
.3867 
.3829 
.3791 
.3753 
.3716 
1.0 
.3679 
.3642 
.3606 
.3570 
.3535 
.3499 
.3465 
.3430 
.33% 
.3362 
1.1 
.3329 
.32% 
.3263 
.3230 
.3198 
.3166 
.3135 
.3104 
.3073 
.3042 
1.2 
.3012 
.2982 
.2952 
.2923 
.2894 
.2865 
.2837 
.2808 
.2780 
.2753 
1.3 
.2725 
.2698 
.2671 
.2645 
.2618 
.2592 
.2567 
.2541 
.2516 
.2491 
1.4 
.2466 
.2441 
.2417 
.2393 
.2369 
.2346 
.2322 
.2299 
.2276 
.2254 
1.5 
.2231 
.2209 
.2187 
.2165 
.2144 
.2122 
.2101 
.2080 
.2060 
.2039 
1.6 
.2019 
.1999 
.1979 
.1959 
.1940 
.1920 
.1901 
.1882 
.1864 
.1845 
1.7 
.1827 
.1809 
.1791 
.1773 
.1755 
.1738 
.1720 
.1703 
.1686 
.1670 
1.8 
.1653 
.1637 
.1620 
.1604 
.1588 
.1572 
.1557 
.1541 
.1526 
.1511 
1.9 
.14% 
.1481 
.1466 
.1451 
.1437 
.1423 
.1409 
.1395 
.1381 
.1367 
2.0 
.1353 
.1340 
.1327 
.1313 
.1300 
.1287 
.1275 
.1262 
.1249 
.1237 
2.1 
.1225 
.1212 
.1200 
.1188 
.1177 
.1165 
.1153 
.1142 
.1130 
.1119 
2.2 
.1108 
.1097 
.1086 
.1075 
.1065 
1054 
.1043 
.1033 
.1023 
.1013 
2.3 
.1003 
•9926 
*9827 
*9730 
*%33 
*9537 
*9442 
*9348 
*9255 
*9163 
2.4 0.0 
9072 
8982 
8892 
8804 
8716 
8629 
8544 
8458 
8374 
8291 
2.5 0.0 
8208 
8127 
8046 
7966 
7887 
7808 
7730 
7654 
7577 
7502 
2.6 0.0 
7427 
7353 
7280 
7208 
7136 
7065 
6995 
6925 
6856 
6788 
2.7 0.0 
6721 
6654 
6587 
6522 
6457 
6393 
6329 
6266 
620a 
6142 
2.8 0.0 
6081 
6020 
5%1 
5901 
5843 
5784 
5727 
5670 
5613 
5558 
2.9 0.0 
5502 
5448 
5393 
5340 
5287 
5234 
5182 
5130 
5079 
5029 
3.0 0.0 
4979 
4929 
4880 
4832 
4783 
4736 
4689 
4642 
45% 
4550 
3.1 0.0 
4505 
4460 
4416 
4372 
4328 
4285 
4243 
4200 
4159 
4117 
3.2 0.0 
4076 
4036 
39% 
3956 
3916 
3877 
3839 
3801 
3763 
3725 
3.3 0.0 
3688 
3652 
3615 
3579 
3544 
3508 
3474 
3439 
3405 
3371 
3.4 0.0 
3337 
3304 
3271 
3239 
3206 
3175 
3143 
3112 
3081 
3050 
X 
.0 
.1 
.2 
.3 
.4 
.5 
.6 
.7 
.8 
.9 
3 0.0 
4979 
4505 
4076 
3688 
3337 
3020 
2732 
2472 
2237 
2024 
4 0.0 
1832 
1657 
1500 
1357 
1228 
n i l 
1005 
•9095 
*8230 
*7447 
5 0.00 
6738 
6097 
5517 
4992 
4517 
4087 
3698 
3346 
3028 
2739 
6 0.00 
2479 
2243 
2029 
1836 
1662 
1503 
1360 
1231 
1114 
1008 
7 0.000 
9119 
8251 
7466 
6755 
6112 
5531 
5004 
4528 
4097 
3707 
8 0.000 
3355 
3035 
2747 
2485 
2240 
2035 
1841 
1666 
1507 
1364 
9 0.000 
1234 
1117 
1010 
•9142 
*8272 
•7485 
*6773 
*6128 
*5545 
*5017 
10 0.0000 
4540 
4108 
3717 
3363 
3043 
2754 
2492 
2254 
2040 
1846 
logio^"* = -xlogjo^ = - 0.43429 X 
340 

Answer Section 
A N S W E R S T O O D D - N U M B E R E D P R O B L E M S 
Chapter 1 
1. 
25°C 
3. 25°C 
5. 80°F 
7. 
122°F, 50°C 
Chapter 2 
1. 
23.3°C 
3. (a) lost in the form of vapor or steam 
(b) 100°C 
5. 0.57 cal/gm °C 
7. 7200 cal 
9. 543 cal/gm 
11. 0.10 cal/gm °C 
13. (a) 5.9°C (b) 0°C 
15. 
lOkcal/gm 
Chapter 3 
1. 
340T 
5. 8.64cm^ 
7. 83.5°C 
9. 0.291cm' 
11. d = - ^'^ 
13. 
/(steel) = 21.7 cm,/(brass) = 13.7 cm 
15. 369 cm' 
17. 9.65 x 1 0 " ' 1 9 . 
27.8 cm' 
21. 
(a) 2.67x10'"C* 
(b) 18x 10' 
Chapter 4 
1. 
190°C 
3. 31.6kcal/sec-m^ 
5. 0.12 
7. 3700 
9. 5920°K 
11. (a) 640 kcal 
(b) 19,500 seconds 
13. 100.5°C 
15. (a) -2.5°C 
(b) -6.7°C 
Chapter 5 
1. 
(a) yes 
(b) yes 
(c) positive 
3. 
13.5 mm 
5. 5°C, 51% 
7. 10"'atm 
9. O.OrC 
Chapter 6 
5. 
D = 3.23 cm 
11. δ  = 52.4° 
Chapter 7 
5. 
ρ = 7.5 cm 
7. h = 15.1 m 
9. 
\.6R from the front of the bowl 
11. 
Image is erect, real, twice original size and located yR to right of spherical surface 
Chapter 8 
1. 
Correct: final image is 3.41? behind front surface; Incorrect: final image is 1.51? behind 
front surface 
3. / = - 60 cm 
7. ± 12 cm, ± 60 cm 
9. 60 cm behind the diverging lens 
11. 
( a ) / = - 7 5 cm 
(b) 100 cm 
13. 2.5,16.7 cm 
17. //4 
Chapter 9 
1. 
i = 2 x l 0 " ' s e c 
3. c = 3.01 x 10'm/sec 
5. c = 2.998 x 10'm/sec 
341 

342 
Answer Section 
Chapter 10 
3. 
4.3 m 
5. y =2sin[27Γ(0.2 
JC - 4 ί ) ] ; / = 4cycles/sec 
9. (a) ττ  radians 
(b) 0 (c) 0 
(d) 7Γ/2 radians 
11. /, = 523 or 501 cycles/sec 
13. 
η λ = 2 7 Γ Γ 
13. 
η λ = 2 π Γ 
Chapter 11 
1. 
λ=6800Α  
3. í = | x l O " ' m 
7. ( a ) y = 8 . 1 m m 
(b) η  = 3 for 4500Ä; η  = 8 for 6000Ä 
9. 
η  = 182 
11. 
( N a - N v a c ) « 9 4 
Chapter 12 
5. 
(a) r« = 4.01 mm 
(b)r8=1.26cm 
9. (a) Ö, = 11.5°; Ö 2 = 20.5° (b) Ay =8.5 cm 
17. 
(c) λ, =4920 Ä 
(d) ft =22° 
Chapter 13 
1. 
(a) Xtotai = V3 Λ  
sin 
3. 
(a) 
L ^ ( 2 - i ; i ) + 60° 
(b) Xtotal — Via 
sin [ft+70.9°] 
9. 
(a) |A + B| = 23.2 
(b) α = 72.9° 
| A - B | = 30.5 
β  =126° 
|2A + B| = 30.1 
7 = 61.9° 
11. 
(a) Ö„ax = 45° (b)J/Jo = l 
15. 
(a) 53° (b) vertical 
17. 
i = 8.56xlO-'m 
19. 
(b) All light transmitted by the first polaroid film will be transmitted 
Chapter 14 
1. 
11.5 ft/sec' 
3. 142 ft 
5. -1.33 ft/sec' 
7. (a) 56 ft 
(b) - 36 ft/sec, 36 ft from ground 
9. 
(a) 3000 ft west 
(b) 157 ft/sec 
11. (a) 2 m/sec 
(b) 15 m, 5 m/sec south 
13. (a) 36 ft 
(b) 0.5 sec 
(c) 48 ft/sec 
(d) - 3 2 ft/sec' 
Chapter 15 
5. 
L /Lo« 1 - 1 X 10"' 
7. 
AÍ2 = 5.77 x 10"' sec 
9. 
Ü = 0.6 x 10' m/sec toward S, 
Chapter 16 
1. 
F = l n t 
5. s « 2 . 1 x l 0 - ' ' m 
7. 
JC = ^ 1 ? £ M 
11. g = go/d + Ä / Ä E ) ' 
13. 
ί < 12 X 10' sec or « I day 
15. (a) F21 = é x 10"'' nt toward m, 
(b) F23 = J x 10"" nt toward ms 
(c) ay = é X 10"'' m/sec' up; a, = έ  x 10"" m/sec' to right 
(d) a = 1.45 x 10"" m/sec'; Ö « 29° 
Chapter 17 
1. 
(b) ω /(2sin <p) (c) 30° 
3. (a) ' m/sec' 
(b) 2.1 m/sec'; 6.3 nt 
5. 
16101b 
7. (a) 20 cm/sec' 
(b) 0.384 nt = 3.84 x 10'dynes 
11. (a) 19.6nt 
(b) 2.1kg 
13. 
(a) 1701b 
(b) 1501b 
(c) 0 ("free fall") 

Answer Section 
343 
Chapter 18 
1. 
(a) 4 ft/sec' 
(b) 1.41b 
3. 4.25 m/sec' 
5. (a) 3.92 m/sec' 
(b) 16.2nt 
(c) 2.75 kg 
7. 
(a) 5.201b 
(b) 20.51b 
(c) 25.51b 
9. (a) 42 ft, 1.5 seconds 
(b) 259 ft 
11. (a) 86,900 ft 
(b) x = 23,000 ft, y =53,400 ft 
13. (a) 30 seconds 
(b) 14,400 ft 
(c) 96,000 ft 
15. 
(a) 15.8 seconds 
(b) 500 ft 
17. 0.318 rev/sec 
Chapter 19 
1. 
49,000 cm/sec 
3, 14 cm/sec 
5. 3.1 ft/sec 
7. 
1000 m 
9. (a) 1 ft/sec 
(b) 40,000 lb sec 
11. (a) 20 ft/sec 
(b) 2 ft/sec', 0.5 lb 
(c) 0.062 
13. 40 m/sec, 0.2 seconds 
15. 
(a) 188 slug ft/sec 
(b) 3761b 
Chapter 20 
I. 
(a) 1000ft lb 
(b) 211.51b 
(c) 45.3ft lb 
(d) 743.2ft lb 
3. (a) 100ft lb 
(b) 0.50 
5. 
(a) 10* joules; 1.% x 10' joules 
(b) 3.6 x 10' joules; 6.42 x 10' joules 
(c) 1 x 10' joules, 0 
7. 
(a) 1.06 hp 
(b) 300 watts 
9. (a) 2700 ft lb 
(b) 1800 ft lb 
(c) 624 ft lb 
(d) 276 ft/lb 
11. 
0.30 hp, 224 watts 
13. (a) 1501b 
(b) 1.14 
15. (a) 41.71b 
(b) 834 ft lb 
(c) 0.152 
Chapter 21 
1. 
(a) 50 rad/sec 
(b) 10 rad/sec' 
(c) 100 rad/sec 
(d) 100 cm/sec' 
(e) 1000 cm/sec 
(f) 10'cm/sec' 
3. (a) 21b ft 
(b) 4.5 rad/sec' 
(c) 27 rad/sec 
(d) 162ft-lb 
(e) 54 ft 
5. 
16 rad/sec 
7. (a) 15.8 seconds 
(b) 500 ft 
9. (a) 1.33 rad/sec 
(b) 0.89 rad/sec' 
11. 
(a) 1220 ft 
(b) 12.51b 
13. 14.3 rad/sec 
15. 32 ft/sec 
Chapter 22 
1. 
(a) 0.314 m/sec 
(b) -0.49 m/sec' 
(c) 0.99 nt/m 
3. (a) 1.15 m/sec 
(b) 0.82 m 
5. 
0.10 
7. (a) 1.75 s e c ' 
(b) 0.061 joules 
(c) 1.32 m/sec 
(d) 14.5 m/sec' 
9. (a) 0.40nt 
(b) y = Λ  sin 27Γ (τ -ίς ^^ 
11. 80.4 cm/sec; 64,700 cm/sec' 
\0.05 sec 
2 cm/ 
Chapter 23 
1. 
1.5 m 
3 . 0.026 cm 
5. (a) 0.112 in. 
(b) 2.08 x 10'ft-lb 
7. 0.24 joules 
9. 0.50 joules 
I I . 
10'nt 
13. 0.46 in' 
15. 17.3 lb/in' 
Chapter 24 
1. 
7.45x10'nt/m' 
3. 1000 nt 
5. 4 m/sec 
7. 2.5gm/cm'; 1.2gm/cm' 
9. (a) 10.0166gm 
(b) 0.16% 
11. 28lb/in' 
13. (a) 30ft/sec and 40ft/sec 
(b) Oft and 10.8ft 
15. 
3.96 lb/in' 
Chapter 25 
1. 
60.6 joules; 191 joules 
5. (a) 10'joules 
(b) 6.93 joules 
(c) 0 
9. (a) 2.50 x 10'joules 
(b) 6.93x10'joules 
(c) 1.50 x 10'joules 
11. 298°C 
13. (a) -9.05 kcal 
(b) 7.56 kcal 
(c) 0 (adiabatic) 
15. (a) 2833 cm' 
(b) 287 (c) 140 
17. 200 m 
19. -78.5kcal/kg 
Chapter 26 
1. 
(a) 320°K 
(b) 20% 
3. (a) 33.4 joules 
(b) 57% (c) 2.33 
5. 
540cal/gm 
7. 
600 joules 
9. (a) 11.4 kg (b) 3.68 x 10'joules 
11. 1.25 
13. 0.62kcal/°K 
Chapter 27 
1. 
(a) true 
(b) false 
5. (a) 1170 joules 
(b) 2600 m/sec 
7. 
16.4 kcal; 11.7 kcal; 4.7 kcal; 
11.7 kcal 
11. 94,400 cm/sec 

344 
Answer Section 
Chapter 28 
1. 
1.95 X 10" nt/m' 
3. (a) 5030 watts 
(b) 1060 cycles/sec 
5. 20 db 
7. 310 cps 
9. 29 seconds 
11. 0.20watts/m' 
13. 19.4 m/sec 
Chapter 29 
1. 
(a) 1.73x10" nt toward Λ  
(b) 13.8 x 10'nt/coul 
(c) 0 
3. (a) 9 x 10'volts 
(b) 7.07 X 10' volts/m at an angle of 38.2° with the + JC direction 
(c) 0.45 joules 
5. (b) 20 vohs 
(c) 3.2x lO'nt downward 
(d) 3.5 x 10''m/sec' downward 
7. (b) 100 volts 
(c) 8 x 1 0 ' ' n t downward 
(d) 8.8 x 10" m/sec'downward 
9. (a) 1.45 x 10'm/sec 
(b) 2.76 X 1 0 ' sec 
(c) 9.60 x 1 0 " joules 
11. 8 x 10' volts; 0 
1 3 . - 4 volts/m; 4 volts/m 
15. (a) Ε = - c 
(b) Ε = - gc(d, - dz) (c) volts/m 
Chapter 30 
' · 
^ - ^ ^ " - é ' » ? 
, . , . , 0 
0 . ) - ^ 
(c) 
^ 
(a) 8 X 10"'coul 
(b) 0.16 joules 
11. (a) 133 volts 
(b) 0.053 joules 
(c) polarized oil dielectric 
15. (a) 2 micromicrofrads 
(b) 2x 10"'coul 
(c) 667voks and 333 
volts 
(d) 10"'joules 
17. (a) 13.3 microcoul/m' 
(b) 4.5 microcoul/m' 
(c) 4.5 microcoul/m' 
(d) 13.3 microcoul/m' 
(e) 6.6 joules/m' 
19. Ε = y - ^ ; Ρ  = τ ^ ^ ; D = 
Chapter 31 
1. 
6.05 ohms 
3. 0.0025/°C 
5. 6 ohms 
7. (a) 0.30 ohms 
(b) 12 volts 
9, 2.1 volts 
11. 
(a) 25 watts 
(b) 4500 joules 
13. 600 ohms 
15. (a) 27 volts 
(b) 27 watts 
(c) 4 ohms 
(d) 12 volts 
(e) 2 amp 
17. 0.5 seconds 
Chapter 32 
1. 
300 nt 
3. 12 nt-m 
5. (a) 0 (b) 0.0125 amp-m' 
7. (a) 0.80 amp-m'/sec along negative 
z-axis 
(b) 0.173 nt-m parallel to y-axis 
9. (a) 25 amp 
(b) 1.6 amp 
11. 2.3 amp 
13. 2450 ohms 
15. 1.69 ohms 
17. (a) 0.303 ohms 
(b) 14,970 ohms 
19. 5 x 10"'nt/amp-m 
vertically downward 
Chapter 33 
1. 
0.04 webers 
3. 3.14 x 10"'webers/m' 
5. 0.050 volts 
7. 80 joules 
9. 
16.7millisec 
15. 3 volts 
17. (a) 2 amp 
(b) 240 joules 
(c) i amp; π  amp/sec 
19. (a) 3 amp/sec 
(b) 2 amp/sec 
(c) 1 amp 
(d) 3 amp 
Chapter 34 
1. 
0.08 amp-m' 
3. (a) 175 amp-turns/m 
(b) 0.264 webers/m' 
7. (a) 1 (b) 6 x 10"' 
9. a 
11. 
12 amp 
13. 300°C 
Chapter 35 
1. 
100 ohms 
3. (a) 11.31 ohms 
(b) lag (c) 45 deg 
5. (a) 221 microfarads 
(b) Odeg 
7. (a) 200 rad/sec 
(b) 0.707 amp 
9. (a) 100 ohms; 40 ohms 
(b) 100 volts; 141 volts 
11. 
(a) 56.2 ohms 
(b) 2.14 amp 
(c) 0.09 (d) 23 watts 
13. (a) 300 ohms 
(b) 500 ohms 
(c) 0.20 amp and 0.283 amp 
(d) 80 volts 
(e) 50 volts 
(f) 16 watts 
15. (a) 2 amp 
(b) V c = 60volts; VR= 88volts; Vcoii = 194volts 
17. (a) 110volts 
(b) lamp 
(c) 2200watts 
Chapter 36 
1. 
(a) aD/ai=curlH 
(b) Displacement current = Λ  curl Η  
3. (a) 1.41 
(b) 2.13 
7. 
(a) I X 10'm/sec 
(b) 3 (c) 8 x 10"'joules/m' 
11. 6.65 x 10"" watts/m' 

Index 
Aberration 
chromatic 
59 
of Hght 64 
spherical 
44 
Absolute 
humidity 
33 
pressure 
216 
scales of temperature 
3 
temperature 
3 
thermodynamic temperature scale 
3, 232, 233 
zero 
2, 233 
Absorber, perfect 
25 
Absorption 
factor 
25 
of light 
24-26 
of radiation 
251 
AC 
309-316 
circuits 
309-316 
parallel 
313-314 
power relations 
314 
series 
312-313 
vector diagrams 
310-314 
generator 
297-298 
motors 
297-299 
Accelerated motion 
109-112, 126 
Acceleration 
108 
angular 
176 
average 
110 
centripetal 
147 
constant 
110 
instantaneous 
110 
of center of gravity 
185-186 
of gravity 
127 
variations 
127 
of rockets 
159-160 
radial 
148-149 
tangential 
149 
Accommodation 
56 
Acoustics (see also Sound) 
245-249 
t i l i refers to a table. 
Action and reaction 
137-138 
Adiabatic 
compression 
222 
expansion 
224-225 
processes 
222, 224-225 
in sound waves 
245-246 
Alternating current 
see AC 
Ammeter 
289 
Amorphous 
solid 
241 
substance 
29 
Ampere 
271 
Ampere's law 
293-294 
Amperian currents 
303 
Amplitude 
of AC 
309 
of longitudinal wave 
67 
of simple harmonic motion 
191, 193 
of transverse wave 
67 
Analogues, rotational 
180 
Analyzer 
101 
Angle 
critical 
40 
of incidence 
38 
of reflection 
38 
of refraction 
39 
of shear 
201 
phase 
69, 192 
plane 
45, 177; Units and conversion factors 
45, 177 
solid 
260; Units 
260 
Angstrom 
322 
Angular 
acceleration 
176 
displacement 
176 
impulse 
180 
momentum 
180, 184 
conservation of 
184, 185 
velocity 
176 
Antenna 
323 
Aperture 
58 
Aqueous humor 
56 
345 

346 
Index 
Archimedes principie 
211 
Armature 
see Generator and Motor 
Astigmatism 
56 
Atmospheric 
humidity 
33 
pressure 
211, 212 
Atom 
251 
concept 
251 
nuclear model 
251 
Atwood's machine 
140 
Audibility 
247 
Avogadro's 
law 
240 
number 
239; Value 
239 
Axis 
optic 
104 
Balance equal arm 
127 
Barometer 
212 
Battery 
274, 275 
Beam, light 
37 
Beat frequency 
70 
Beats 
70 
Bel 
247 
Bernoulli's equation 
213-216 
Bimetallic thermometer 
15 
Biot-Savart law 
285-286 
Birefringence 
104 
Black body 
25 
radiation 
25 
Boiling 
point curve 
31 
points 
10, Iii, 31 
Boltzmann's constant 
239 
Bomb calorimeter 
11 
Bound charges 
265 
Boyle's law 
16 
Bradley, James 
64 
Brewster's angle and law 
103 
British-engineering system 
126 
British thermal unit (BTU) 
275 
Brown, Robert 
241 
Brownian motion 
241 
Bulk modulus 
201, 204i 
Buoyant forces 
211 
Calorie 
7 
Calorimeter 
8 
bomb 
11 
continuous flow 11 
Calorimetry 
8 
Camera 
58, 59 
Capacitance 
262-264; Units 
262 
Capacitive circuits 
voltage relations in 
311 
Capacitive reactance 
311 
Capacitor 
262 
charge 
262 
cylindrical 
267 
discharge 
277-278 
energy of 
262-263 
in parallel 
263 
in series 
263 
parallel plate 
262 
Carnot 
cycle 
229-231 
efficiency 
231 
engine 
230-233 
performance coefficient 
231 
refrigerator 
231 
theorem 
231 
Cavendish experiment 
129 
Cavendish, Henry 
129 
Center 
of curvature of mirror 
43 
of gravity 
178 
of mass 
185 
Centigrade scale 
2 
Central force 
151 
Centrifugal force 
149 
Centripetal 
acceleration 
147 
force (see also Central force) 
155 
Charge, 
electric 
(see 
also 
Electric 
251-256; Units 
252 
capacitor 
262 
on conductors 
260-262 
Charles' law 
16 
Chemical energy 
274-275 
source of emf 
274-275 
Chromatic aberration 
59 
Circuits 
see AC and DC 
Circular motion 
146-149 
Clausius 
229 
Coated optics 
79 
Coaxial conducting cylinders 
267 
Coefficient of 
area expansion 
14 
friction 
143 
linear expansion 
13, 14i 
thermal conductivity 
21, 22i 
charge) 

Index 
347 
Coefficient of (contd.) 
volume expansion 
14, 15i 
Coercive force 
306 
Cohen, E. R. 
65 
Coherence 
77-78 
Collisions 
158-159 
elastic 
158 
inelastic 
158 
Combustion, heat of 
10, Hi 
Compass needle 
283 
Components 
rectangular 
327-328 
vector 
99-100 
Compound microscope 
57 
Compressibility 
202 
Compression of a gas 
adiabatic 
222 
isothermal 
223 
Concave mirrors 
43-47 
Concurrent forces 
175 
Condensation 
32 
Conducting coaxial cylinder 
267 
Conduction 
of electricity 
271-272 
of heat 
21 
Conductivity 
electrical (see also Resistivity) 
27 
thermal 
21, 22i 
Conductors, electrical 
251 
charges on 
260-262 
Cones (receptors) 
56 
Conservation of 
angular momentum 
184-185 
energy 
158, 169, 171 
mechanical energy 
169, 171 
momentum 
156-158 
Conservative and non-conservative forces 
166 
Constants, fundamental 
335 
Continuity equation 
214 
Continuous flow calorimetry 
11 
Contraction Lorentz 
120 
Convection 
21, 23, 24 
coefficient 
24 
Converging lens 
53 
Conversion factors, 335i 
Convex mirrors 
43 
Cooling curve 
9 
Cooling, Newton's law of 
24 
Coordinate systems 
112-113 
moving 
112-114 
Cornea 
56 
Cosines, 336i 
Cosmic rays 
325 
Coulomb 
252 
Coulomb's law 
252 
determination of constant in 
252 
Crest 
71 
Critical 
angle 
40 
constants, 31 ί 
point 
30 
pressure 
30 
temperature 
30 
Cross (vector) product 
328 
Crystalline 
lens of eye 
56 
solids 
241 
Curie 
law 
305 
point 
305 
Curie-Weiss law 
305 
Curi 
332 
Current, electric (see also AC and DC) 
271-278; 
Units 
271 
decay in inductive circuit 
299 
density 
271 
element 
284-285 
growth in inductive circuit 
299 
lagging 
313 
leading 
313 
Curvature, center of 
43 
Cycle 
191 
Carnot 
229-231 
electrical 
309 
Cyclotron 
287 
Dalton's law of partial pressures 
240 
Damped oscillations 
194-195 
DC 
271-278 
circuit 
273-278 
generator 
297-299 
motor 
297-299 
Decay time 
195 
Decibel 
247 
Deñnite integral 
111-112 
Deformation 
199-200 
Degradation of energy principle 
234 
Degree 
see Angle and Temperature 
Density 
current 
271 
magnetic flux 285 
of matter, definition 
210, 210i 

348 
Index 
Deviation angle 
41 
Dew point 
33 
Diamagnetic materials 
304 
Dichroism 
104 
Dielectric 
264-266 
constant 
266 
permittivity 
266 
polarization 
264-266 
Difference of potential 
254 
Differentiation 
109 
Diffraction 
85-92 
gratings 
91-92 
of light 
85-92 
of waves 
73-74, 85 
Diffuse reflection 
30 
Dipole 
264 
Dipole moment 
electric 
264 
magnetic 
288 
Direct currents 
see DC 
Dispersion 
40-41 
by diffraction 
91 
by prism 
40-41 
by refraction 
40 
of light 
40-41, 91 
Displacement 
108 
angular 
176 
electrical 
266 
rate of change of 
109 
Dissipation of mechanical energy 
172 
Divergence 
332 
Diverging 
lens 
54 
mirror 
43 
Domain, magnetic 
306 
Doppler effect 
248-249 
Dot (scalar) product 
163, 328 
Double refraction 
104 
Double slit diffraction 
89-91 
Du Mond, J. W. M. 
65 
Dynamic equilibrium 
133 
Dynamics 
107 
of fluids 213-216 
of particle 
125-137 
of rigid body 
179 
of rotational motion 
183 
Dyne 
335 
Ear 
247 
Earth rotation 
148-149 
Effective value of current and voltage 
310 
Efliciency 
Carnot 
231 
of heat engine 
231 
Einstein, Albert 
117 
Einstein's relativity theory 
119 
Elastic 
body 
199 
collisions 
158 
constants (moduli) 
200, 204i 
relations among 
202 
materials 
199 
potential energy 
170 
restoring force 
170 
waves 
205-206 
Elasticity 
199-203 
moduli of, 204Í 
Electric, Electrical 
charge 
251-256; Units 
252 
negative 
251 
positive 
251 
circuits 
see AC and DC 
conductivity 
272 
conductors 
251 
current 
see also AC and DC 
alternating 
309-316 
direct 
271-278 
element 
284-285 
magnetic effects 
283-300 
dipole 
264 
displacement 
266 
energy 
274 
field 
253-254 
strength 
253-254; Units 
254 
force 
252 
generator 
297 
insulator 
251 
intensity 
254 
lines 
253 
meters 
288-290 
permittivity 
252-253 
potential 
254-255; Units 
255 
power 
274 
resistance 
272-274; Units 
272 
resistivity 
272-273; Units 
272 
structure of matter 
251 
susceptibility 
265-266 
Electromagnetic 
induction 
294-296 
radiation 
323-324 
spectrum 
323, 332i 

Index 
349 
Electromagnetic (contd.) 
theory of light 
323-325 
waves 
322-325 
Electromotive force (see emf) 
274 
Electron 
251 
charge on 
256 
mass 
256 
Electrostatics 
251-267 
emf 
induced 
294 
thermal 
306 
Energy; Units and conversion factors 
164-165 
capacitor 
262-263 
conservation 
158, 169, 171 
elastic 
170 
electric 
274 
internal 
219 
kinetic 
158, 165, 179, 180 
mechanical 
see Kinetic and Potential 
dissipation of 
172 
transformation of 
221 
of a capacitor 
262-263 
of a rolling body 
185-186 
of self inductance 
296-297 
potential 
169-171 
elastic 
170 
relations in harmonic motion 
193-194 
rotational 
179-180 
states 
171 
thermal 
1 
total 
171 
Engine 
see Heat engine 
Enthalpy 
223 
Entropy 
233-235 
Equal-arm balance 
127 
Equation of state of a gas 
16, 17, 32 
Equilibrium 
7, 133 
static and dynamic 
133 
thermal 
7 
Erecting lens 
58 
Erg 
164 
Ether (luminiferous), 
velocity of 
117-118 
Evaporation 
30 
Expansion 
adiabatic free 
223 
isothermal 
222 
thermal 
13-16 
Exponentials 
339i, 340i 
External work 
219 
Extraordinary ray 
104 
Eye 
56^59 
defects of 
56 
far point of 
56 
near point of 
56 
Eyepiece 
57 
Fahrenheit scale 
2 
Farad 
262 
Faraday, Michael 
117 
Faraday's law 
294 
Far point of eye 
56 
Fermi, Enrico 
161 
Ferromagnetic materials 
305-306 
Ferromagnetism 
305-306 
Fields 
see Electric and Magnetic 
First law of motion (see also Newton's) 
136 
First law of thermodynamics 
221 
Fitzgerald, George 
117 
Fizeau's toothed wheel experiment 
64 
Flatness, interference tests for 
80 
Flow of fluid 213-216 
Fluid 
209 
dynamics 
213, 216 
ideal 
209 
pressure 
209-210 
real 
209 
statics 
209-212 
Fluorescence 
324-325 
Flux, magnetic 
293 
Focal length 
of lens 
54 
of mirror 
45 
Focal point 
44, 54 
Foot, definition 
126 
Foot-pound 
165, 335 
Force definition 
125; Units and conversion fac-
tors 
126 
buoyant 
211 
central 
151 
centrifugal 
149 
centripetal (same as Central force) 
155 
constant of a spring 
192 
electric 
252 
frictional 
143 
gravitational 
128 
magnetic 
282-286 
moment of 
175-178 
normal 
136, 143 
nuclear 
125 
of gravity 
127 
restoring 
192 

350 
Index 
Force definition (contd.) 
types 
125 
Forced oscillations 
196 
Forces, systems of 
concurrent 
175 
conservative and nonconservative 
166 
equilibrium of 
133 
fundamental 
125 
interatomic 
241 
Foucault, Jean 
64 
Frames of reference 
112 
Fraunhofer diffraction 
88-91 
Free body diagram 
138 
Free expansion 
223 
Freezing 
see Fusion 
Frequency 
68, 192, 246, 309 
Fresnel diffraction 
86-88 
Friction 
142-144 
coefficient of 
143, 144i 
kinetic 
143 
on inclined plane 
152, 153 
static 
143 
work against 
167, 168 
Fundamental 
constants 
335 
forces 
125 
frequency 
71, 247 
quantities 
see Introduction 
units 
see Introduction 
Fusion 
10, 31 
curve 
31 
latent heat of 
10, Hi 
/-value 
59 
/-value of lens 
59 
g 
127 
G 
128-130 
Galilean transformation 
112-114 
Galvanometer 
288, 289 
element in voltmeter and ammeter 
289 
Gamma rays 
325 
Gas 
adiabatic processes 
222, 224 
external work 
219 
internal energy 
221 
isothermal processes 
223 
kinetic theory 
237-240 
laws 
16-18 
liquefaction 
32 
real 
240 
specific heats 
224, 224i 
thermal energy 
1 
thermal properties 
16, 17 
thermometer 
4 
universal constant 
17 
Gauge pressure 
216 
Gauss's divergence theorem 
334 
Gauss's law 
259-262 
General gas law 
17 
Generator 
297-299 
Geometrical optics 
37 
Gilbert, William 
283 
Gradient 
operator 
255, 332 
potential 
255 
temperature 
22 
Grating 
diffraction 
91 
Gravitation 
Cavendish experiment 
129 
constant 
130 
Newton's law of 
128 
universal 
128-130 
Gravity 
acceleration of 
127 
variations 
127 
center of 
178 
force of 
127 
Gyration, radius of 
183, 183i 
Harmonic motion 
191-197 
Harmonics 
71, 247 
Heat 
1 
capacity 
7 
conduction 
21-23 
coefficient of 
21, 22i 
convective transfer of 
23-24 
engine 
230-233 
efficiency of 
231 
latent 
10-11 
of combustion 
10, Hi 
quantity 
7 
radiative transfer of 
24-26 
specific 
7-10, 9i 
transfer 
21-26 
units 
8 
Henry 
296 
Hertz, Heinrich 
117 
Hooke, Robert 
192 
Hooke's law 
192 
Horsepower 
165 
Human audible range 
245 

Index 
351 
Humidity 
33 
absolute 
33 
relative 
33 
Huygens, Christian 
63 
Huygens' principle 
73 
Hygrometer (wet and dry bulb) 
34 
Hygrometry 
33 
Hypermetropic eye 
56 
Hysteresis 
306 
Ideal 
fluid 
209 
gas 
224 
Image 
44 
distance 
44 
formation 
44 
real 
47 
virtual 
47 
Impedance 
313 
Impulse 
155-156 
Incidence, angle of 
38 
Inclined plane 
motion on 
152-153 
Index of refraction 
39, 40i, 321 
in eyeball, 56i 
Induced emf 
294 
Inductance 
296-300, Units 
296 
mutual 
296 
self 
296 
energy of 
296-297 
Induction 
294-297 
electromagnetic 
294-296 
residual 
306 
Inductive circuits 
current growth and decay 
299 
voltage relations in 
311 
Inductive reactance 
312 
Inelastic 
collisions 
158 
properties of solids 
204-205 
Inertia 
125 
law of 
137 
moment of 
177, 181-183 
rotational 
177 
Inertial reference frame 
137 
Infrared radiation 
323 
Infrasonic waves 
245 
Insulators 
251 
thermal 
22 
Integral 
definite 
111-112 
line 
164 
Intensity 
electric 
254 
magnetic 
304 
of light beam 
102 
of sound 
246-247 
Interatomic forces 
241 
Interference 
69, 77-82 
fringes 
82 
in thin films 79-80, 82 
of light waves 
77-79 
Interferometer 
81, 118 
Internal 
energy 
220-221 
of gas 
221-225 
reflection 
40 
resistance 
274-275 
work 
219 
Iris 
56 
Irreversible process 
234 
Isobaric processes 
222 
Isothermal processes 
222 
Isothermals of real gases 
32 
Isotopes 
251 
Joule 
164 
Jupher, satellites of 
63 
Kelvin, Lord (William Thomson) 
307 
Kelvin temperature scale 
3, 232 
Kepler's laws 
150-151 
Kilocalorie 
7 
Kilogram, definition 
126 
Kilowatt-hour 
165 
Kinematics 
107-115, 180 
Kinetic 
energy 
1, 158, 165 
friction 
143 
theory of gases 
237-240 
theory of liquids 
241 
theory of solids 
241-242 
Kirchhoff's law of radiation 
25 
Kirchhoff's rules of electric circuits 
275 
Land, E. H. 
104 
Laser 
77-78 
Latent heat (see also Fusion and Vaporization) 
nt 
fusion 
10 
vaporization 
10 
Lawrence, Ernest O, 
287 

352 
Index 
LeChatelier's principle 
29 
Length 
126 
Lens 
53-59 
axis 
54 
combinations 
55 
converging 
53 
diverging 
54 
equation 
54 
focal length 
54 
focal point 
54 
/-value 
59 
maker's equation 
54 
nonreflecting 
79 
thick and thin 
53 
Lenz's law 
294-296 
Light 
aberration 
64 
absorption 
24-26 
diffraction 
62 
dispersion 
40-42, 62 
interference 
63, 77-79 
nature of 
61-65 
particle theory of 
62 
point source of 
37 
polarization 
101-104 
ray 
37 
rectilinear propagation of 
37, 62 
reflection 
38-39, 62 
refraction 
39-40, 62 
speed 
63-65 
waves 
63, 67 
Linear 
expansion 
13 
polarized waves 
101 
Lines 
electric 
253 
magnetic 
293 
Liquefaction of gases 
32 
Liquids 
kinetic theory of 
241 
thermal properties 
15-16 
Lloyd's mirror experiment 
79 
Lodestone 
283 
Logarithms 
337-338Í 
Longitudinal wave 
67, 97 
Loops 
in electrical networks 
275 
Lorentz contraction 
120 
Lorentz force law 
286 
Lorentz, Hendrick 
117 
Lorentz transformation 
120 
Magnet 
283 
external field of 
283 
magnetic moment of 
303 
Magnetic 
deflection of charged particles 
286-287 
dipole moments 
288 
domain 
306 
field 
283 
of coil 
297 
of currents 
293-300 
of solenoid 
297 
torque on loop in 
288 
vector 
253 
flux. Units 
293 
density (induction) 
284-286; Units 
285,293 
in solenoid 
297 
force 
on conductors 
283-284 
on current element 
284 
on moving charge 
286 
hysteresis 
306 
intensity 
304; Units 
304 
lines 
293 
materials 
304-306 
diamagnetic 
304-305 
ferromagnetic 
305-306 
paramagnetic 
305 
real 
304 
moment 
303 
per unit volume 
304 
permeability constant 
285 
permeability relative 
304 
properties of matter 
303-306 
susceptibility 
304 
Magnetization 
304 
curve 
306 
intensity of 
304; Units 
304 
saturation 
306 
Magnetizing force 
304; Units 
304 
Magnification 
47 
Magnifier 
57 
Magnifying power 
56-58 
of microscope 
57 
of telescope 
58 
Malus law 
102 
Mascons 
127 
Mass 
125-130; definition 
126; Units and conver-
sion factors 
126 
inertial and gravitational 
127-128 
rest 
161 
standard 
126 

Index 
353 
Mass (contd.) 
variation with speed 
161 
Mathematical tables 
336-340 
Matter 
electrical structure of 
251 
Maxwell, James Clerk 
117, 319 
Maxwell's electromagnetic equations 
319-321 
Mechanical 
energy 
see also Energy 
conservation 
169, 171 
dissipation 
172 
equivalent of heat 
221 
units, systems of 
164-165 
Melting (see Fusion) 
10 
point 
Hi, 29 
pressure effect 
30 
Meter 
126 
in terms of standard wavelengths 
81 
Meters 
electrical 
288-290 
Venturi 
215 
Michelson 
interferometer 
81, 118 
measurement of speed of light 
65-66 
Michelson-Morley experiment 
117 
Micron 
335 
Microscope 
57 
compound 
57 
simple 
57 
Mirror 
center of curvature 
43 
concave 
43-47 
convex 
43-47 
equation 
46 
focal length 
45 
focal point 
44 
parabolic 
45 
paraxial rays 
45 
plane 
38 
radius of curvature 
44 
spherical 
45 
MKS system of units 
126 
Modulus, elastic 
bulk 
201 
shear 
201 
Young's 
201 
Molecular mass 
239 
Moment 
arm 
175 
magnetic 
203 
of force 
175-178 
of inertia 
177, 181-183, 183i 
parallel axis theorem 
181 
perpendicular axis theorem 
182 
Momentum 
angular 
180, 184 
conservation of 
184-185 
linear 
156 
conservation 
156-158 
Morley, Edward 
117 
Motion 
see also Dynamics and Kinematics 
accelerated 
110-112, 126 
Brownian 
241 
in a circle 
146-149 
non-equilibrium 
134-136 
of a point in a rotating body 
177-178 
periodic 
191 
planetary 
149-151 
projectile 
144-146 
rolling 
185, 186 
rotational 
179-186 
systems with variable mass 
159-160 
wave 
67 
Motor 
297-299 
AC 
297-299 
DC 
297-299 
Musical instruments 
247-248 
Mutual inductance 
296 
Myopia 
56 
Near point of eye 
56 
Negative charge 
251 
Neutrino 
161 
Neutron 
251 
Newton, Sir Isaac 
62 
Newton (nt) 
126 
Newton's 
first law 
136 
law of cooling 
24 
law of gravitation 
128 
laws of motion (or of mechanics) 
136 
rotational analogues 
180 
particle theory of light 
62 
rings 
80 
second law 
136 
in terms of momentum 
159 
rotational analogue 
183 
third law 
136 
Nicol prism 
104 
Noise 
247 
Nonequilibrium motion 
134-136 
Nonreflecting lenses 
79 

354 
Index 
Normal force 
136, 143 
Nuclear 
forces 
125 
model of atom 
251 
Nucleus 
251 
Object (optical) 
44 
distance 
44 
real 
47 
virtual 
47 
Objective 
57-58 
of microscope 
57 
of telescope 
57 
Ocular 
57 
Ohm 
272 
Ohm's law 
272 
Optic 
axis 
104 
nerve 
56 
Optical 
instruments 
56-59 
path length 
78-82 
Optics coated 
79 
Ordinary ray 
104 
Oscillations 
damped 
194 
forced 
196 
of pendulum 
193 
Overtones 
71 
Parabolic mirror 
45 
Parallel 
AC circuit 
313-314 
-axis theorem 
181 
connections of capacitors 
263 
connections of resistors 
274 
-plate capacitor 
262 
Paramagnetic materials 
305 
Paraxial rays 
45 
Partial derivative 
205 
Partial pressure 
204 
Particle 
107 
accelerators 
287 
dynamics of 
107 
theory of light 
62 
Pascal, Blaise 
213 
Pascal's principle 
213 
Peltier 
effect 
306-307 
emf 
307 
heat 
307 
Pendulum simple 
193 
Performance coefficient of refrigerator 
231 
Period 
68, 191, 309 
Periodic motion 
191 
Permeability constant 
285 
Permittivity 
dielectric 
266 
electrical 
252-253; Units 
253 
Perpendicular 
axis theorem 
182 
Phase angle 
69, 192 
AC 309 
Phases of matter 
30-31 
Pitch 
of sound 
247 
Plane mirror 
38 
Planetary motion 
149-151 
Point charge 
252 
Poisson's ratio 
203 
Polarization 
dielectric 
264-266 
of light 
101-104 
Polarizer 
101 
Polaroid 
102 
Position vector 
108 
Positive charge 
251 
Potential 
difference 
255 
electric 
254-255 
energy 
169-171 
gradient 
255 
Potentiometer 
276-277 
Pound 
126 
Power 
165; Units and conversion factors 
165 
average 
165 
factor 
314 
in electric circuits 
274 
instantaneous 
165 
reactive 
315 
relations in AC 
314 
Poynting vector 
322 
Presbyopia of eye 
56 
Pressure 
absolute 
216 
atmospheric 
211, 212 
critical 
30, 30i 
effect on melting 
30 
fluid 
209-210 
gauge 
216 
gradient 
210 
partial 
240 
temperature diagram 
31 

Index 
355 
Pressure (contd.) 
vapor 
30 
volume diagram 
32 
Prevost's theory of exchanges 
24 
Principle of degradation of energy 
234 
Principle of equipartition of energy 
239 
Prism 
deviation of light by 
40-42 
dispersion 
40-42 
Problem solving 
141-142 
Projectile motion 
144-146 
Proton 
251 
Psychological effects of sound waves 
247 
Pupil (of eye) 
56 
Quality of sound 
247 
Quantity 
of electricity 
see Charge 
of heat 
7 
scalar 
99 
Radar 
323 
Radial acceleration 
148-149 
Radian 
45, 177 
Radiant energy 
see Radiation 
Radiation (see also Electromagnetic) 
21 
absorption 
24-26 
antenna 
323 
black-body 
25 
electromagnetic 
24 
infrared 
323 
Kirchhoff's law 
25 
radio-frequency 
323 
Stefan-Boltzmann law 
26 
thermal 
24-26,323-324 
ultraviolet 
323-324 
visible 
see Light 
Radiator perfect 
25 
Radio waves 
323 
Radius 
of curvature 
44 
of gyration 
183 
Range of projectile 
145 
Rankine temperature scale 
3 
Ratio of specific heats of gas 
244, 239, 240 
Rays, light 
37 
extraordinary 
104 
ordinary 
104 
paraxial 
45 
R-C series circuit 
277-278 
Reactance 
311-313 
capacitive 
311 
inductive 
312 
Reaction forces 
137 
Reactive power 
315 
Real 
fluid 
209 
gases 
240 
image 
47 
magnetic materials 
304 
Rectilinear propagation of light 
37 
Reference frame 
112 
inertial 
137 
Reflection 
38 
by plane mirror 
38 
by spherical mirrors 
44-47 
diffuse 
30 
factor 
25 
laws of 
38 
of waves 
71 
polarization by 
103-104 
specular 
38 
total internal 
40 
Refraction 
39 
dispersion by 
40 
double 
104 
index 
39, 40-4li, 321 
laws of 
39 
of light 
39-42 
of superconductors 
273 
of waves 
71 
Refractive index 
40, 40-4li 
Refrigerator 
231 
performance coefficient 
231 
work done on 
231 
Regelation 
30 
Relative humidity 
33 
Relative permeability 
304 
Relative velocity 
113, 118 
Relativistic expressions 
addition of velocities 
121-122 
mass 
161 
Relativity 
117-122 
Einstein's theory of special 
119 
Residual induction 
306 
Resistance 
272-274 
determination of 
276 
internal 
274-275 
of semiconductors 
273 
superconductors 
273 
temperature coefficient of 
273, 273 i 
thermometer 
5 

356 
Index 
Resistivity 
272, 273ί; Units 
272 
temperature coefficient 
273, 273 ί 
Resistor 
parallel and series connection 
273-274 
Resonance 
AC electrical 
315 
in forced oscillations 
197 
Restoring force 
192 
Retina 
56 
Reversible-heat engines 
229-230 
Right-hand rule 
for magnetic field of straight conductor 
284 
for magnetic moment of coil 
288 
for torque vector 
288 
R-L series circuit 
299-300 
Rockets 
159-160 
Rods (visual receptors) 
56 
Rolling motion 
185-186 
Rö mers observation 
63 
Rotation 
175-186 
dynamics 
180, 183 
equilibrium 
180 
kinematics 
180 
kinetic energy 
179-180 
Rotational 
analogues of Newton's laws 
180 
analogues of translational motion 
180 
equilibrium 
180 
inertia 
177 
motion 
179-186 
dynamics of 
180, 183 
energy in 
179-180 
quantities 
176 
simple harmonic work 
192-193 
work 
180-181 
Rupture point 
200 
Saturated vapor 
30 
pressure of water, 30i 
Saturation magnetization 
306 
Scalar 
field 
253 
product 
163, 328 
quantity 
99 
Second definition 
126 
Second law of motion (see also Newton's) 
136, 
159 
Second law of thermodynamics 
229 
Seebeck effect 
306-307 
Seebeck emf 
307 
Self-inductance (see also Inductance) 
296 
Semiconductors 
273 
Series connections 
capacitors 
263 
resistors 
273 
Series resonance 
315 
Shear 
200-201 
angle 
201 
modulus 
201, 204i 
strain 
201 
stress 
200 
Simple 
harmonic motion 
192 
amplitude 
191 
energy of 
193-194 
frequency 
192 
period 
191 
rotational 
193 
microscope 
57 
pendulum 
193 
Sines 
336Í 
Single slit diffraction 
88-89 
Sinusoidal wave motion 
68 
Slug definition and conversion factor 
126 
Snell, Willebrord 
39 
Snell's law 
39 
Solar system 
150, 150i 
Solenoid 
energy 
297 
magnetic field of 
297 
magnetic flux in 
297 
Solid angle 
260 
Solids 
amorphous 
241 
covalent bonded 
242 
ionic bonded 
242 
kinetic theory of 
241-242 
metallic bonded 
241 
thermal properties of 
13-15 
Sound 
245-249 
adiabatic character of 
245, 246 
frequencies 
246 
intensity 
246-247 
loudness 
246-247 
musical 
247-248 
pitch 
247 
production 
245 
psychological effects of 
247 
quality 
247 
speed 
109, 245-256, 246i 

Index 
357 
Sound (contd.) 
waves 
67, 245 
waveform 
245 
Source of emf (see also emf) 
274 
Specific heat 
7-10, 9t 
measurement methods 
8, 9 
of gases 
224, 224i 
of liquids 
8 
of solids 
8 
Spectra 
92 
Spectrum 
electromagnetic 
323, 323i 
infrared 
324 
ultraviolet 
324 
Specular reflection 
38 
Speed 
of light (see also Light) 
63 
of longitudinal waves 
68, 206 
of sound (see also Sound) 
109, 245-246 
of transverse waves 
68, 206 
Spherical aberration 
44 
Spherical mirrors 
45 
Springs 
192 
Standing waves 
70 
State, change of 
29 
Static equilibrium 
133 
Static friction 
143 
Statics 
of fluids 209-212 
Stefan-Boltzmann law 
26 
Steradian 
260 
Stoke's theorem 
334 
Straight conductor, magnetic field of 
293, 294 
Strain 
200-205 
bulk 
201 
longitudinal 
205 
shear 
201 
tensile 
201 
Streamlines 
213 
Stress 
bulk 
200 
longitudinal 
206 
shear 
200 
tensile 
200 
Sublimation 
31 
curve 
31 
Superconductors 
273, 305 
Superposition principle 
69 
Susceptibility 
electric 
265-266 
magnetic 
304 
Tangential acceleration 
149 
Tangents 
336i 
Telescope 
magnifying power 
58 
objective 
57-58 
terrestrial 
58 
Temperature 
1 
absolute scales 
3 
Centigrade scale 
2 
coefficient of resistance 
273 
critical 
30, 30i 
Fahrenheit scale 
2 
gradient 
22 
Kelvin scale 
3, 232-233 
Rankine scale 
3 
scales 
2-4 
temperature measurement 
2 
thermodynamic scale 
3, 232-233 
Tensile 
strain 
201 
stress 
200 
Thermal 
conductivity 
21, 22i 
emf 
307 
energy 
1 
equilibrium 
7 
expansion 
coefficient of 
13-16, 14i, 15i 
of water 
15 
properties 
of gases 
16-17, 22, 31-34 
of liquids 
15-16, 22, 29-31 
of solids 
13-15, 21-22, 29-31 
radiation 
24-26, 324 
units of energy 
7 
Thermocouple 
5, 306 
Thermodynamics 
219-235 
first law 
221 
second law 
229 
Thermodynamic 
processes 
222-223 
temperature scale 
3, 232-233 
Thermoelectric 
effects 
306-307 
Thermometers 
2-5 
gas 
4 
mercury 
4 
resistance 
5 
Thermometric properties 
2 
Thermopile 
324 
Thermostat 
15 

358 
Index 
Thin films, interference in 79-80, 82 
Third law of motion (see also Newton's) 
136-137 
Thomson effect 
307 
Thomson heat 
307 
Thomson, William (Lord Kelvin) 
307 
Threshold of hearing 
247 
Throttling process 
223 
Time-constants of circuits 
277, 299 
Time dilation 
120 
Torque 
176 
on loop, magnetic 
288 
vector 
176 
work done by 
180, 181 
Torsion 
constant 
130, 289 
Total 
energy of particle 
171 
internal reflection 
40 
Tourmaline 
104 
Trajectory of projectile 
145 
Transformation 
Galilean 
112-114 
Lorentz 
120 
Transformer 
315, 316 
Transverse waves 
67, 97-99 
amplitude 
68, 97-98 
sinusoidal 
68, 97-98 
speed of 
97 
Trigonometric functions, 336i 
Triple point 
4, 31 
Triple point diagram 
31 
Ultrasonic waves 
245 
Ultraviolet spectra 
324-325 
Units, system of 
British engineering 
126 
FPS 
see Introduction 
rationalized MKS (see Introduction) 
126 
Unit vector 
100 
Universal 
gas constant 
17 
gravitation 
128 
Van der Waals equation 
32 
Vaporization 
curve 
31 
latent heat of 
10, Hi 
Vapor pressure 
30 
Vector 
99-101,327 
addition 
101, 327 
component 
99-100, 327 
cross product 
176, 328 
diagrams for AC 
310-314 
direction 
99, 321 
electric fields 253 
magnitude 
99-100, 328 
position 
108 
Poynting 
322 
product 
176, 328 
quantities 
327 
rectangular components 
327-328 
representation 
99, 327 
sum 
327 
unit 
100, 328 
Vector analysis 
327-334 
Velocity 
108 
angular 
176 
average 
109 
instantaneous 
109 
magnitude 
109 
relative 
113 
relativistic addition of 
121-122 
root-mean-squared 
238 
tangential 
180 
Venturi meter 
215 
Virtual 
image 
47 
object 
47 
Visible radiation, wavelengths 
79 
Vitreous humor 
56 
Volt 
255 
Voltmeter 
289 
Volume expansion 
14 
Watt 
165 
Wave 
67-74 
amplitude 
68 
diffraction 
73-74 
elastic 
205-206 
electromagnetic 
322-325 
equation 
321 
interference 
69 
light 
67,79 
linear polarized 
101 
longitudinal 
67 
motion 
67 
sinusoidal 
68 
reflection 
71 
refraction 
71 
sinusoidal 
68 
sound 
67, 245 
speed 
68, 206 

Index 
359 
Wave (contd,) 
standing 
70 
transverse 
67 
sinusoidal 
68 
Wavelength 
68 
Weber 
293 
Weight 
127 
Wet and dry bulb hygrometer 
34 
Wheatstone bridge 
276 
Work 
163-165; Definition 
163; Units and con-
version factors 
164-165, 219 
in friction 
167-168 
in rotational motion 
180-181 
internal and external 
219 
Work energy balance 
172 
Xrays 
324-325 
X-ray tube 
324 
X-ray unit (XU) 
323 
Young's experiment 
78 
Young's modulus 
201, 204i 
Young, Thomas 
78 
Zero, absolute 
2, 233