Resolução do tipler-vol.1-Chap 15 SM

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1451 
Chapter 15 
Traveling Waves 
 
Conceptual Problems 
 
1 • [SSM] A rope hangs vertically from the ceiling. A pulse is sent up 
the rope. Does the pulse travel faster, slower, or at a constant speed as it moves 
toward the ceiling? Explain your answer. 
 
Determine the Concept The speed of a transverse wave on a uniform rope 
increases with increasing tension. The waves on the rope move faster as they 
move toward the ceiling because the tension increases due to the weight of the 
rope below the pulse. 
 
2 • A pulse on a horizontal taut string travels to the right. If the rope’s 
mass per unit length decreases to the right, what happens to the speed of the pulse 
as it travels to the right? (a) It slows down. (b) It speeds up. (c) Its speed is 
constant. (d) You cannot tell from the information given. 
 
Determine the Concept The speed v of a pulse on the string varies with the 
tension FT in the string and its mass per unit length μ according to μTFv = . 
Because the rope’s mass per unit length decreases to the right, the speed of the 
pulse increases. ( )b is correct. 
 
3 • As a sinusoidal wave travels past a point on a taut string, the arrival 
time between successive crests is measured to be 0.20 s. Which of the following is 
true? (a) The wavelength of the wave is 5.0 m. (b) The frequency of the wave is 
5.0 Hz. (c) The velocity of propagation of the wave is 5.0 m/s. (d) The 
wavelength of the wave is 0.20 m. (e) There is not enough information to justify 
any of these statements. 
 
Determine the Concept The distance between successive crests is one 
wavelength and the time between successive crests is the period of the wave 
motion. Thus, T = 0.20 s and f = 1/T = 5.0 Hz. )(b is correct. 
 
4 • Two harmonic waves on identical strings differ only in amplitude. 
Wave A has an amplitude that is twice that of wave B’s. How do the energies of 
these waves compare? (a) EA = EB. (b) EA = 2EB. (c) EA = 4EB. (d) There is 
not enough information to compare their energies. 
 
Picture the Problem The average energy transmitted by a wave on a string is 
proportional to the square of its amplitude and is given by ( ) xAE ΔΔ 2221av μω= 
where A is the amplitude of the wave, μ is the linear density (mass per unit length) 
Chapter 15 
 
 
1452 
of the string, ω is the angular frequency of the wave, and Δx is the length of the 
string. 
 
Because the waves on the strings 
differ only in amplitude, the energy 
of the wave on string A is given by: 
 
2
AA cAE = 
 
Express the energy of the wave on 
string B: 
 
2
BB cAE = 
 
Divide the first of these equations by 
the second and simplify to obtain: 
 
2
B
A
2
B
2
A
B
A ⎟⎟⎠
⎞
⎜⎜⎝
⎛==
A
A
cA
cA
E
E 
 
Because AA = 2AB: 
42
2
B
B
B
A =⎟⎟⎠
⎞
⎜⎜⎝
⎛=
A
A
E
E ⇒ ( )c is correct. 
 
5 • [SSM] To keep all of the lengths of the treble strings (unwrapped 
steel wires) in a piano all about the same order of magnitude, wires of different 
linear mass densities are employed. Explain how this allows a piano manufacturer 
to use wires with lengths that are the same order of magnitude. 
 
Determine the Concept The resonant (standing wave) frequencies on a string are 
inversely proportional to the square root of the linear density of the string ( )λμTTf = . Thus extremely high frequencies (which might otherwise require 
very long strings) can be accommodated on relatively short strings if the strings 
are linearly less dense than the high frequency strings. High frequencies are not a 
problem as they use short strings anyway. 
 
6 • Musical instruments produce sounds of widely varying frequencies. 
Which sounds waves have the longer wavelengths? (a) The lower frequencies. 
(b) The higher frequencies. (c) All frequencies have the same wavelength. 
(d) There is not enough information to compare the wavelengths of the different 
frequency sounds. 
 
Determine the Concept Once the sound has been produced by a vibrating string, 
membrane, or air column, the speed with which it propagates is the product of its 
frequency and wavelength ( ).λfv = For a given medium and temperature, v is 
constant. Hence the wavelength and frequency for a given sound are inversely 
proportional and ( )a is correct. 
 
Traveling Waves 
 
 
1453
7 • In Problem 6, which sound waves have the higher speeds? (a) The 
lower frequency sounds. (b) The higher frequency sounds. (c) All frequencies 
have the same wave speed. (d) There is not enough information to compare their 
speeds. 
 
Determine the Concept Once the sound has been produced by a vibrating string, 
membrane, or air column, the wave speed with which it propagates depends on 
the properties of the medium in which it is propagating and is independent of the 
frequency and wavelength of the sound. ( )c is correct. 
 
8 • Sound travels at 343 m/s in air and 1500 m/s in water. A sound of 
256 Hz is made under water, but you hear the sound while walking along the side 
of the pool. In the air, the frequency is (a) the same, but the wavelength of the 
sound is shorter, (b) higher, but the wavelength of the sound stays the same, 
(c) lower, but the wavelength of the sound is longer, (d) lower, and the 
wavelength of the sound is shorter, (e) the same, and the wavelength of the sound 
stays the same. 
 
Determine the Concept In any medium, the wavelength, frequency, and speed of 
a sound wave are related through λ = v/f. Because the frequency of a wave is 
determined by its source and is independent of the nature of the medium, if v is 
greater in water than in air, λ will be shorter in air than in water. )(a is correct. 
 
9 • While out on patrol, the battleship Rodger Young hits a mine, begins to 
burn, and ultimately explodes. Sailor Abel jumps into the water and begins 
swimming away from the doomed ship, while Sailor Baker gets into a life raft. 
Comparing their experiences later, Abel tells Baker, ″I was swimming 
underwater, and heard a big explosion from the ship. When I surfaced, I heard a 
second explosion. What do you think it could be?″ Baker says, ″I think it was 
your imagination—I only heard one explosion. ″ Explain why Baker only heard 
one explosion, while Abel heard two. 
 
Determine the Concept There was only one explosion. Sound travels faster in 
water than air. Abel heard the sound wave in the water first, then, surfacing, 
heard the sound wave traveling through the air, which took longer to reach him. 
 
10 • True or false: A 60-dB sound has twice the intensity of a 30-dB sound. 
 
Picture the Problem The intensity level β, measured in decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is defined to be the threshold of 
hearing. 
 
Chapter 15 
 
 
1454 
Express the intensity level of the 60-
dB sound: 
( )
0
60logdB10dB60
I
I= ⇒ 0660 10 II = 
 
Express the intensity level of the 30-
dB sound: 
( )
0
30logdB10dB30
I
I= ⇒ 0330 10 II = 
 
Because I60 = 103I30, the statement is false. 
 
11 • [SSM] At a given location, two harmonic sound waves have the 
same displacement amplitude, but the frequency of sound A is twice the 
frequency of sound B. How do their average energy densities compare? (a) The 
average energy density of A is twice the average energy density of B. (b) The 
average energy density of A is four times the average energy density of B. (c) The 
average energy density of A is 16 times the average energy density of B. (d) You 
cannot compare the average energy densities from the data given. 
 
Determine the Concept The average energy density of a sound wave is given by 
2
0
2
2
1
av sρωη = where ρ is the average density of the medium, s0 is the 
displacement amplitude of the molecules making up the medium, and ω is the 
angular frequency of the sound waves. 
 
Express
the average energy density 
of sound A: 
 
2
A,0
2
AA2
1
A av, sωρη = 
 
The average energy density of sound 
B is given by: 
 
2
B,0
2
BB2
1
B av, sωρη = 
 
Dividing the first of these equation 
by the second yields: 2
B,0
2
BB2
1
2
A,0
2
AA2
1
B av,
A av,
s
s
ωρ
ωρ
η
η = 
 
Because the sound waves are 
identical except for their frequencies: 
2
B
A
2
B
A
2
B
2
A
B av,
A av,
2
2
⎟⎟⎠
⎞
⎜⎜⎝
⎛=⎟⎟⎠
⎞
⎜⎜⎝
⎛==
f
f
f
f
π
π
ω
ω
η
η
 
 
Because fA = 2fB: 
 4
2
2
B
B
B av,
A av, =⎟⎟⎠
⎞
⎜⎜⎝
⎛=
f
f
η
η ⇒ ( )b is correct. 
 
12 • At a given location, two harmonic sound waves have the same 
frequency but the amplitude of sound A is twice the amplitude of sound B. How 
do their average energy densities compare? (a) The average energy density of A is 
twice the average energy density of B. (b) The average energy density of A is four 
times the average energy density of B (c) The average energy density of A is 16 
Traveling Waves 
 
 
1455
times the average energy density of B (d) You cannot compare the average energy 
densities from the data given. 
 
Determine the Concept The average energy density of a sound wave is given by 
2
0
2
2
1
av sρωη = where ρ is the average density of the medium, s0 is the 
displacement amplitude of the molecules making up the medium, and ω is the 
angular frequency of the sound waves. 
 
Express the average energy density 
of sound A: 
2
A,0
2
AA2
1
A av, sωρη = 
 
The average energy density of sound 
B is given by: 
 
2
B,0
2
BB2
1
B av, sωρη = 
 
Dividing the first of these equation 
by the second yields: 2
B,0
2
BB2
1
2
A,0
2
AA2
1
B av,
A av,
s
s
ωρ
ωρ
η
η = 
 
Because the sound waves are 
identical except for their 
displacement amplitudes: 
2
B,0
A,0
2
B,0
2
A,0
B av,
A av, ⎟⎟⎠
⎞
⎜⎜⎝
⎛==
s
s
s
s
η
η
 
 
Because B,0A,0 2ss = 
 4
2
2
B,0
B,0
B av,
A av, =⎟⎟⎠
⎞
⎜⎜⎝
⎛=
s
s
η
η ⇒ ( )b is correct. 
 
13 • What is the ratio of the intensity of normal conversation to the sound 
intensity of a soft whisper (at a distance of 5.0 m)? (a) 103, (b) 2, (c) 10−3, (d) 1/2. 
Hint: See Table 15-1. 
 
Determine the Concept The problem is asking us for the ratio of the intensities 
corresponding to normal conversation and a soft whisper. Table 15-1 includes the 
quantities we need in order to find this ratio. 
 
Use Table 15-1 to find the ratio of 
I/I0, where I0 is the threshold of 
hearing, for normal conversation: 
 
6
onconversati
 normal0
10=⎟⎟⎠
⎞
⎜⎜⎝
⎛
I
I 
Use Table 15-1 to find the ratio of 
I/I0 for a soft whisper: 
 
3
whisper
soft 0
10=⎟⎟⎠
⎞
⎜⎜⎝
⎛
I
I 
Chapter 15 
 
 
1456 
Dividing the first of these equation 
by the second yields: 
 33
6
whisper
soft 0
onconversati
 normal0
10
10
10 ==
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛
I
I
I
I
 
( )a is correct. 
 
14 • What is the ratio of the intensity level of normal conversation to the 
sound intensity level of a soft whisper (at a distance of 5.0 m)? (a) 103, (b) 2, 
(c) 10−3, (d) 1/2. Hint: See Table 15-1. 
 
Determine the Concept The problem is asking us for the ratio of the sound 
intensity levels (decibel levels) corresponding to normal conversation and a soft 
whisper. Table 15-1 includes the quantities we need in order to find this ratio. 
 
From Table 15-1, the sound intensity 
level corresponding to normal 
conversation is: 
 
dB 60
onconversati
 normal =β 
From Table 15-1, the sound intensity 
level corresponding to a soft whisper 
is: 
 
dB 30
whisper
soft =β 
 
Dividing the first of these equation 
by the second yields: 
 
2
dB 30
dB 60
whisper
soft 
onconversati
 normal
==β
β
 
( )b is correct. 
 
15 • To increase the sound intensity level by 20 dB requires the sound 
intensity to increase by what factor? (a) 10, (b) 100, (c) 1000, (d) 2. 
 
Determine the Concept The sound intensity level is given by ( ) ( )0log dB 10 II=β where I is the intensity of a given sound and I0 is the 
intensity of the threshold of hearing. 
 
The sound intensity level for a sound 
whose intensity is I1 is given by: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
1
1 log dB 10 I
Iβ 
 
The sound intensity level for a sound 
whose intensity is I2 is given by: 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
2
2 log dB 10 I
Iβ 
Traveling Waves 
 
 
1457
Subtract the second equation from 
the first and simplify to obtain: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛=−
2
1
21 log dB 10 I
Iββ 
 
Solving for I1/I2 yields: dB 10
2
1
21
10
ββ −
=
I
I 
 
For a 20-dB increase in the sound 
intensity level: 
2dB 10
dB 20
2
1 1010 ==
I
I ⇒ ( )b is correct. 
 
16 • You are using a hand-held sound level meter to measure the intensity 
level of the roars produced by a lion prowling in the high grass. To decrease the 
measured sound intensity level by 20 dB requires the lion move away from you 
until its distance from you has increased by what factor? (a) 10, (b) 100, (c) 1000, 
(d) You cannot tell the required distance from the data given. 
 
Determine the Concept The sound intensity level is given by ( ) ( )0log dB 10 II=β where I is the intensity of a given sound and I0 is the 
intensity of the threshold of hearing. The intensity of the a given sound varies 
with distance from its source according to ( )2av 4 rPI π= . 
 
The sound intensity level for a sound 
whose intensity is I1 is given by: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
1
1 log dB 10 I
Iβ 
 
The sound intensity level for a sound 
whose intensity is I2 is given by: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
2
2 log dB 10 I
Iβ 
Subtract the second equation from 
the first and simplify to obtain: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛=−
2
1
21 log dB 10 I
Iββ 
 
Solving for I1/I2 yields: dB 10
2
1
21
10
ββ −
=
I
I 
 
Letting r1 be the lion’s initial 
distance from you and r2 be the 
lion’s final distance from you, 
substitute for I1 and I2 and simplify 
to obtain: 
 
dB 10
2
1
2
2
2
2
av
2
1
av
21
10
4
4 ββ
π
π −==
r
r
r
P
r
P
⇒ dB 10
1
2
21
10
ββ −
=
r
r 
 
Chapter 15 
 
 
1458 
For a 20-dB decrease in the sound 
intensity level: 1010
dB 10
dB 20
1
2 ==
r
r ⇒ ( )a is correct. 
 
17 • One end of a very light (but strong) thread is attached to an end of a 
thicker and denser cord. The other end of the thread is fastened to a sturdy post 
and you pull the other end of the cord so the thread and cord are taut. A pulse is 
sent down the thicker, denser cord. True or false: 
 
(a) The pulse that is reflected back from the thread-cord attachment point is 
inverted compared to the initial incoming pulse. 
(b) The pulse that continues past the thread-cord attachment point is not inverted 
compared to the initial incoming pulse. 
(c) The pulse that continues past the thread-cord attachment point has an 
amplitude that is smaller than the pulse that is reflected. 
 
(a) False. Because the reflection medium (cord) in which the pulse travels initially 
has a greater linear density than the transmission medium (thread), there is no 
phase shift in the reflected pulse. 
 
(b) True. Because the thread-cord attachment point of the media is more like a 
loose end than a fixed end, the pulse transmitted into the light thread is in phase 
with the incoming pulse in the thicker, denser cord. 
 
(c) True. Because the string is attached to a thicker, denser cord, the reflected 
pulse behaves almost as
though it was reflected from a fixed end and is, therefore, 
inverted. The pulse in the light thread is in phase with the incoming pulse in the 
thicker, denser cord. Because energy is conserved and there is some energy 
transmitted and some reflected, the transmitted wave cannot have an amplitude as 
large as the amplitude of the pulse in the thicker, denser cord. 
 
18 • Light traveling in air strikes a glass surface at an incident angle of 45°. 
True or false: (a) The angle between the reflected light ray and the incident ray is 
90°. (b) The angle between the reflected light ray and the refracted light ray is less 
than 90°. 
 
Determine the Concept The ray diagram shows the relationship between the 
incident, reflected, and refracted rays. 
Traveling Waves 
 
 
1459
air glass
°45
θ1
rayIncident 
ray Reflected
ray Refracted
glassair vv >
 
(a) True. Because the angle of incidence is equal to the angle of reflection, the 
sum of these angles in this example is 90°. Note: this is a special case and is not 
generally true. 
 
(b) False. Because light travels slower in glass than in air, the refracted ray makes 
an angle of less than 45° with the normal. Hence the angle it makes with the 
reflected ray is greater than 90°. 
 
19 • [SSM] Sound waves in air encounter a 1.0-m wide door into a 
classroom. Due to the effects of diffraction, the sound of which frequency is least 
likely to be heard by all the students in the room, assuming the room is full? 
(a) 600 Hz, (b) 300 Hz, (c) 100 Hz, (d) All the sounds are equally likely to be 
heard in the room. (e) Diffraction depends on wavelength not frequency, so you 
cannot tell from the data given. 
 
Determine the Concept If the wavelength is large relative to the door, the 
diffraction effects are large and the waves spread out as they pass through the 
door. Because we’re interested in sounds that are least likely to be heard 
everywhere in the room, we want the wavelength to be short and the frequency to 
be high. Hence ( )a is correct. 
 
20 • Microwave radiation in modern microwave ovens has a wavelength on 
the order of centimeters. Would you expect significant diffraction if this radiation 
was aimed at a 1.0-m wide door? Explain. 
 
Determine the Concept No. Because the wavelength of the radiation is small 
relative to the door, the diffraction effects are small and the waves do not spread 
out significantly as they pass through the door. 
 
 
Chapter 15 
 
 
1460 
21 •• [SSM] Stars often occur in pairs revolving around their common 
center of mass. If one of the stars is a black hole, it is invisible. Explain how the 
existence of such a black hole might be inferred by measuring the Doppler 
frequency shift of the light observed from the other, visible star. 
 
Determine the Concept The light from the visible star will be shifted about its 
mean frequency periodically due to the relative approach toward and recession 
away from Earth as the star revolves about the common center of mass. 
 
22 •• Figure 15-30 shows a wave pulse at time t = 0 moving to the right. 
(a) At this particular time, which segments of the string are moving up? 
(b) Which segments are moving down? (c) Is there any segment of the string at 
the pulse that is instantaneously at rest? Answer these questions by sketching the 
pulse at a slightly later time and a slightly earlier time to see how the segments of 
the string are moving. 
 
Determine the Concept The graph shown below shows the pulse at an earlier 
time (t < 0) and later time (t > 0). (a) One can see that at t = 0, the portion of the 
string between 1 cm and 2 cm is moving down, (b) the portion between 2 cm and 
3 cm is moving up, and (c) the string at x = 2 cm is instantaneously at rest. 
0 1 2 3 4 5 6 7 8 9 10
x , cm 
y
t = 0
t > 0
t < 0
 
 
23 •• Make a sketch of the velocity of each string segment versus position 
for the pulse shown in Figure 15-30. 
 
Determine the Concept The velocity of the string at t = 0 is shown. Note that the 
velocity is negative for 1 cm < x < 2 cm and is positive for 2 cm < x < 3 cm. 
Traveling Waves 
 
 
1461
0 1 2 3 4 5 6 7 8 9 10
x , cm
v y
 
 
24 •• An object of mass m hangs on a very light rope that is connected to the 
ceiling. You pluck the rope just above the object, and a wave pulse travels up to 
the ceiling and back. Compare the round-trip time for such a wave pulse to the 
round-trip time of a wave pulse on the same rope if an object of mass 9m is hung 
on the rope instead. (Assume the rope does not stretch, that is, the mass-to-ceiling 
distance is the same in each case.) 
 
Determine the Concept If the mass providing the tension in the rope is increased 
by a factor of n, then the tension in the rope increases by the same factor and the 
speed of the pulse on the rope increases by a factor of n and the round trip time 
for the pulse is decreased by a factor of n1 . 
 
Let l represent the distance the pulse 
travels. Then the time for the round 
trip is given by: 
 
v
t l= (1) 
Express the speed of the wave when 
the object of mass m provides the 
tension in the rope: 
 
μ
m
m
F
v T,= 
Express the speed of the wave when 
the object of mass 9m provides the 
tension in the rope: 
 
μ
m
m
F
v T,99 = 
Substituting for mv9 in equation (1) 
yields: 
mm
m FF
t
T,9T,9
9
μ
μ
ll == (2) 
 
Similarly, the travel time when the 
object whose mass is m is providing 
the tension is given by: mm
m FF
t
T,T,
μ
μ
ll == (3) 
 
Chapter 15 
 
 
1462 
Divide equation (2) by equation (3) 
and simplify to obtain: 
m
m
m
m
F
F
t
t
T,9
T,9 = 
 
Because mm FF T,T,9 9= : 
3
1
T,
T,9
9
==
m
m
m
m
F
F
t
t ⇒ mm tt 319 = 
 
25 •• The explosion of a depth charge beneath the surface of the water is 
recorded by a helicopter hovering above its surface, as shown in Figure 15-31. 
Along which path⎯A, B, or C⎯will the sound wave take the least time to reach 
the helicopter? Explain why you chose the path you did. 
 
Determine the Concept Path C. Because the wave speed is highest in the water, 
and more of path C is underwater than are paths A or B, the sound wave will 
spend the least time on path C. 
 
26 •• Does a speed of Mach 2 at an altitude of 60,000 feet mean the same as 
a speed of Mach 2 near ground level? Explain clearly. 
 
Determine the Concept No. The term Mach 2 means that the speed is twice the 
speed of sound at a given altitude. Because the speed of sound is lower at high 
altitudes than at ground level (due to lower density and colder temperature), Mach 
2 means less than twice the speed of sound at sea level. 
 
Estimation and Approximation 
 
27 •• Many years ago, Olympic 100 m dashes were started by the sound 
from a starter’s pistol, with the starter positioned several meters down the track, 
just on the inside of the track. (Today, the pistol that is used is often only a 
trigger, which is used to electronically activate speakers behind each sprinter’s 
starting blocks. This method avoids the problem of one runner hearing sound 
before the other runners.) Estimate the time advantage the inside lane (relative to 
the runner at the outside lane of 8 runners) would have if all runners started when 
they heard the sound from the starter’s pistol. 
 
Picture the Problem Due to the differences in the distances from the starter to 
the inside- and outside-lane sprinters, the sound of the starting pistol took more 
time to reach the outside-lane sprinter than it did to reach the inside-lane sprinter. 
 
Express the difference in time for the 
two sprinters in terms of the 
distances the sound must travel and 
the speed of sound:
vv
ttt
lane
 inside
lane
 outside
lane
 inside
lane
 ousideΔ
ll
−=
−=
 
Traveling Waves 
 
 
1463
 
Assuming that the separation of the 
inside- and outside-lane sprinters is 
w: 
 
22
lane
 inside
lane
 outside w+= ll 
Substitute for 
lane 
outsidel to obtain: 
v
w
t lane
 inside
22
lane
 inside
Δ
ll −+
= 
 
Assuming that the separation of the 
inside- and outside-lane sprinters is 
7.0 m and that the starter is 5.0 m 
from the inside-lane sprinter yields: 
( ) ( )
ms 11
m/s 343
m 0.5m 0.7m 0.5
Δ
22
=
−+=t
 
 
Remarks: Because a 100-m sprint is often decided by one or two hundredths 
of a second, this difference could be significant. This difference led to the 
invention of the electronic starter’s sound. 
 
28 •• Estimate the speed of the bullet as it passes through the helium balloon 
in Figure 15-32. Hint: A protractor would be beneficial. 
 
Picture the Problem You can use a protractor to measure the angle of the shock 
cone and then estimate the speed of the bullet using .sin uv=θ The speed of 
sound in helium at room temperature (293 K) is 977 m/s. 
 
Relate the speed of the bullet u to the 
speed of sound v in helium and the 
angle of the shock cone θ : 
 
u
v=θsin ⇒ θsin
vu = 
Measuring θ yields: 
 
°≈ 70θ 
Substitute numerical values and 
evaluate u: km/s0.170sin
m/s977 =°=u 
 
29 •• The new student townhouses at a local college are in the form of a 
semicircle half-enclosing the track field. To estimate the speed of sound in air, an 
ambitious physics student stood at the center of the semicircle and clapped his 
hands rhythmically at a frequency at which he could not hear the echo of the clap, 
because it reached him at the same time as his next clap. This frequency was 
about 2.5 claps/s. Once he established this frequency, he paced off the distance to 
the townhouses, which was 30 double strides. Assuming that the length of each 
stride is equal to half his height (5 ft 11 in), estimate the speed of sound in air 
using these data. How far off is your estimation from the commonly accepted 
value of 343 m/s? 
 
Chapter 15 
 
 
1464 
Picture the Problem Let d be the distance to the townhouses. We can relate the 
speed of sound to the distance to the townhouses to the frequency of the clapping 
for which no echo is heard. 5 ft 11 in is equal to 1.8 m. 
 
Relate the speed of sound to the 
distance it travels to the townhouses 
and back to the elapsed time: 
 
t
dv Δ=
2 
Express d in terms of the number of 
strides and distance covered per 
stride: 
 
( )( ) m54m/stride8.1strides30 ==d 
Relate the elapsed time Δt to the 
frequency f of the clapping: 
 
s40.0
claps/s5.2
11
Δ ===
f
t 
 
Substitute numerical values and 
evaluate v: 
 
( ) km/s27.0m/s 270
s40.0
m542 ===v 
 
The percent difference between 
this result and 343 m/s is: %21m/s343
m/s270m/s343diff % =−= 
 
Speed of Waves 
 
30 • (a) The bulk modulus of water is 2.00 × 109 N/m2. Use this value to 
find the speed of sound in water. (b) The speed of sound in mercury is 1410 m/s. 
What is the bulk modulus of mercury (ρ = 13.6 × 103 kg/m3)? 
 
Picture the Problem The speed of sound in a fluid is given by ρBv = where 
B is the bulk modulus of the fluid and ρ is its density. 
 
(a) Express the speed of sound in 
water in terms of its bulk modulus: 
 
ρ
Bv = 
Substitute numerical values and 
evaluate v: 
km/s1.41
km/s 414.1
kg/m101.00
N/m102.00
33
29
=
=×
×=v
 
 
(b) Solving ρBv = for B yields: 2vB ρ= 
 
Substitute numerical values and 
evaluate B: 
( )( )
210
233
N/m102.70
m/s1410kg/m1013.6
×=
×=B
 
Traveling Waves 
 
 
1465
31 • Calculate the speed of sound waves in hydrogen gas (M = 2.00 g/mol 
and γ = 1.40) at T = 300 K. 
 
Picture the Problem The speed of sound in a gas is given by MRTv γ= where 
R is the gas constant, T is the absolute temperature, M is the molecular mass of the 
gas, and γ is a constant that is characteristic of the particular molecular structure 
of the gas. Because hydrogen gas is diatomic, γ = 1.4. 
 
Express the dependence of the speed 
of sound in hydrogen gas on the 
absolute temperature: 
 
M
RTv γ= 
Substitute numerical values and 
evaluate v: 
( )( )
km/s32.1
kg/mol1000.2
K300KJ/mol314.84.1
3
=
×
⋅= −v
 
 
32 • A 7.00-m-long string has a mass of 100 g and is under a tension of 
900 N. What is the speed of a transverse wave pulse on this string? 
 
Picture the Problem The speed of a transverse wave pulse on a string is given by 
μTFv = where FT is the tension in the string, m is its mass, L is its length, and 
μ is its mass per unit length. 
 
Express the dependence of the speed 
of the pulse on the tension in the 
string: 
μ
TFv = where μ is the mass per unit 
length of the string. 
 
Substitute numerical values and 
evaluate v: m/s251
m7.00
kg0.100
N900 ==v 
 
33 •• [SSM] (a) Compute the derivative of the speed of a wave on a string 
with respect to the tension dv/dFT, and show that the differentials dv and dFT obey 
TT2
1 FdFvdv = . (b) A wave moves with a speed of 300 m/s on a string that is 
under a tension of 500 N. Using the differential approximation, estimate how 
much the tension must be changed to increase the speed to 312 m/s. (c) Calculate 
ΔFT exactly and compare it to the differential approximation result in Part (b). 
Assume that the string does not stretch with the increase in tension. 
 
 
Chapter 15 
 
 
1466 
Picture the Problem (a) The speed of a transverse wave on a string is given by 
μTFv = where TF is the tension in the wire and μ is its linear density. We can 
differentiate this expression with respect to FT and then separate the variables to 
show that the differentials satisfy .TT2
1 FdFvdv = (b) We’ll approximate the 
differential quantities to determine by how much the tension must be changed to 
increase the speed of the wave to 312 m/s. (c) We can use μTFv = to obtain 
an exact expression for ΔFT, 
 
(a) Evaluate dv/dFT: 
TT
T
T 2
11
2
1
F
v
F
F
dF
d
dF
dv ⋅==⎥⎦
⎤⎢⎣
⎡= μμ 
 
Separate the variables to obtain: 
T
T
2
1
F
dF
v
dv = 
 
(b) Solve the equation derived in Part 
(a) for dFT: v
dvFdF TT 2= 
 
Approximate dFT with ΔFT and dv 
with Δv to obtain: 
 
v
vFF Δ2Δ TT = 
 
Substitute numerical values and 
evaluate ΔFT: ( )
N40
m/s300
m/s 300m/s312N5002Δ T
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=F
 
 
(c) The exact value for (ΔF)exact is 
given by: 
 
( ) T,1T,2exactΔ FFF −= (1) 
Express the wave speeds for the two 
tensions: μ
T,1
1
F
v = and μ
T,2
2
F
v = 
 
Dividing the second equation by the 
first and simplifying yields: 
 
T,1
T,2
T,1
T,2
1
2
F
F
F
F
v
v ==
μ
μ ⇒
2
1
2
T,1T,2 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
v
vFF 
 
Traveling Waves 
 
 
1467
Substituting for FT,2 in equation (1) 
yields: 
 
( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −⎟⎟⎠
⎞
⎜⎜⎝
⎛=
−⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
Δ
2
1
2
F,1
T,1
2
1
2
F,1exactT
v
vF
F
v
vFF
 
 
Substitute numerical values and 
evaluate (ΔFT)exact: 
 
( ) ( )
N 8.40
1
m/s 300
m/s 312N 500Δ
2
exactT
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −⎟⎠
⎞⎜⎝
⎛=F
 
 
The percent error between the exact 
and approximate values for ΔFT is: 
( )
( )
%2
N 8.40
N 40.0N 8.40
Δ
ΔΔ
exactT
TexactT
≈
−=−
F
FF
 
 
34 •• (a) Compute the derivative of the speed of sound in air with respect to 
the absolute temperature, and show that the differentials dv and dT obey 
TdTvdv 21= . (b) Use this result to estimate the percentage change
in the speed 
of sound when the temperature changes from 0 to 27°C. (c) If the speed of sound 
is 331 m/s at 0°C, estimate its value at 27°C using the differential approximation. 
(d) How does this approximation compare with the result of an exact calculation? 
 
Picture the Problem The speed of sound in a gas is given by MRTv γ= where 
R is the gas constant, T is the absolute temperature, M is the molecular mass of the 
gas, and γ is a constant that is characteristic of the particular molecular structure 
of the gas. We can differentiate this expression with respect to T and then separate 
the variables to show that the differentials satisfy .21 TdTvdv = We’ll 
approximate the differential quantities to estimate the percentage change in the 
speed of sound when the temperature increases from 0 to 27°C. Lacking 
information regarding the nature of the gas, we can express the ratio of the speeds 
of sound at 300 K and 273 K to obtain an expression that involves just the 
temperatures. 
 
(a) Evaluate dv/dT: 
T
v
M
R
RT
M
M
RT
dT
d
dT
dv
2
1
2
1
=
⎟⎠
⎞⎜⎝
⎛=⎥⎦
⎤⎢⎣
⎡= γγ
γ
 
 
Chapter 15 
 
 
1468 
Separate the variables to obtain: 
T
dT
v
dv
2
1= 
 
(b) Approximate dT with ΔT and dv 
with Δv and substitute numerical 
values to obtain: 
 
%0.5
K 273
K 273K 300
2
1Δ =⎟⎠
⎞⎜⎝
⎛ −=
v
v 
 
(c) Using a differential 
approximation, approximate the 
speed of sound at 300 K: ⎟⎠
⎞⎜⎝
⎛ Δ+=
Δ+≈
v
vv
v
vvvv
1K273
K273K273K300
 
 
Substitute numerical values and 
evaluate v300 K: 
 
( )( )
m/s347
0495.01m/s331K300
=
+=v
 
 
(d) Use MRTv γ= to express 
the speed of sound at 300 K: 
( )
M
Rv K300K300
γ= 
 
Use MRTv γ= to express the 
speed of sound at 273 K: 
( )
M
Rv K273K732
γ= 
 
Divide the first of these equations by 
the second and solve for and evaluate 
v300 K: 
( )
( ) 273
300
K273
K300
K273
K300 ==
M
R
M
R
v
v
γ
γ
 
and 
( ) m/s347
273
300m/s331K300 ==v 
 
Note that these two results agree to three significant figures. 
 
35 ••• Derive a convenient formula for the speed of sound in air at 
temperature t in Celsius degrees. Begin by writing the temperature as T = T0 + 
ΔT, where T0 = 273 K corresponds to 0°C and ΔT = t, the Celsius temperature. 
The speed of sound is a function of T, v(T). To a first-order approximation, you 
can write ( ) ( ) ( ) TdTdvTvTv T Δ00 +≈ , where dv / dT( )T 0 is the derivative 
evaluated at T = T0. Compute this derivative, and show that the result leads to 
 
v = 331 m / s( ) 1+ t
2T0
⎛ 
⎝ ⎜ 
⎞ 
⎠ ⎟ = 331 + 0.606t( ) m/s. 
Traveling Waves 
 
 
1469
Picture the Problem We can use the approximate expression for v(T) given in 
the problem statement and the result of Problem 34 to show that ( )m/s 606.0331 tv += . 
 
The speed of sound as a function of 
T is given by: 
( ) ( ) T
dT
dvTvTv
T
Δ
0
0 ⎟⎠
⎞⎜⎝
⎛+≈ (1) 
 
From Problem 44, dv/dT is given by: ( )
T
Tv
dt
dv
2
1= 
 
Evaluating 
0TT
dT
dv
=
⎥⎦
⎤ yields: ( )
0
0
2
1
0
T
Tv
dT
dv
TT
=⎥⎦
⎤
=
 (2) 
 
Substitute equation (2) in equation (1) 
and simplify to obtain: ( ) ( ) ( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
⎟⎟⎠
⎞
⎜⎜⎝
⎛+≈
0
0
0
0
0
2
Δ1
Δ
2
1
T
TTv
T
T
TvTvTv
 
 
For T0 = 273° K, ( ) m/s 3310 =Tv , 
and tT =Δ : ( ) ( )
( )m/s 606.0331
K 2732
1m/s 331
t
tv
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛ +=
 
 
The Wave Equation 
 
36 • Show explicitly that the following functions satisfy the wave equation 
2
2
22
2 1
t
y
vx
y
∂
∂=∂
∂ : (a) y(x,t) = k(x + vt)3, (b) y(x, t) = Aeik(x – vt) , where A and k are 
constants and 1−=i , and (c) y(x,t) = ln k(x – vt). 
 
Picture the Problem To show that each of the functions satisfies the wave 
equation, we’ll need to find their first and second derivatives with respect to x and 
t and then substitute these derivatives in the wave equation. 
 
(a) Find the first two spatial 
derivatives of ( ) ( ) :3vtxktx,y += ( )
23 vtxk
x
y +=∂
∂ 
 and 
( )vtxk
x
y +=∂
∂ 62
2
 (1) 
 
Chapter 15 
 
 
1470 
Find the first two temporal 
derivatives of ( ) ( ) :3vtxktx,y += ( )
23 vtxkv
t
y +=∂
∂ 
 and 
( )vtxkv
t
y +=∂
∂ 2
2
2
6 (2) 
 
Express the ratio of equation (1) to 
equation (2): 
 
( )
( ) 22
2
2
2
2
1
6
6
vvtxkv
vtxk
t
y
x
y
=+
+=
∂
∂
∂
∂
 
confirming that ( ) ( )3, vtxktxy += 
satisfies the general wave equation. 
 
(b) Find the first two spatial 
derivatives of ( ) ( )vtxikAetx,y −= : 
 
( )vtxikikAe
x
y −=∂
∂ , ( )vtxikAeki
x
y −=∂
∂ 22
2
2
 
or 
( )vtxikAek
x
y −−=∂
∂ 2
2
2
 (3) 
 
Find the first two temporal 
derivatives of ( ) ( )vtxikAetx,y −= : ( )vtxikikvAet
y −−=∂
∂ , 
( )vtxikAevki
t
y −=∂
∂ 222
2
2
 
or 
 ( )vtxikAevk
t
y −−=∂
∂ 22
2
2
 (4) 
 
Express the ratio of equation (3) to 
equation (4): 
 
( )
( ) 222
2
2
2
2
2
1
vAevk
Aek
t
y
x
y
vtxik
vtxik
=−
−=
∂
∂
∂
∂
−
−
 
confirming that ( ) ( )vtxikAetxy −=, 
satisfies the general wave equation. 
 
(c) Find the first two spatial 
derivatives of ( ) ( )vtxktx,y −= ln : vtx
k
x
y
−=∂
∂ 
and 
( )2
2
2
2
vtx
k
x
y
−−=∂
∂ (5) 
 
Traveling Waves 
 
 
1471
Find the first two temporal 
derivatives of ( ) ( )vtxktx,y −= ln : vtx
vk
t
y
−−=∂
∂ 
and 
( )2
22
2
2
vtx
kv
t
y
−−=∂
∂ (6) 
 
Express the ratio of equation (5) 
to equation (6): 
 
( )
( )
2
2
22
2
2
2
2
2
2
1
v
vtx
kv
vtx
k
t
y
x
y
=
−−
−−=
∂
∂
∂
∂
 
confirming that ( ) ( )vtxktx,y −= ln 
satisfies the general wave equation. 
 
37 • Show that the function y = A sin kx cos ωt satisfies the wave equation. 
 
Picture the Problem The wave equation is 2
2
22
2 1
t
y
vx
y
∂
∂=∂
∂ . To show that 
tkxAy ωcossin= satisfies this equation, we’ll need to find the first and second 
derivatives of y with respect to x and t and then substitute these derivatives in the 
wave equation. 
 
Find the first two spatial derivatives 
of tkxAy ωcossin= : tkxAkx
y ωcoscos=∂
∂ 
and 
tkxAk
x
y ωcossin22
2
−=∂
∂ (1) 
 
Find the first two temporal 
derivatives of tkxAy ωcossin= : tkxAt
y ωω sinsin−=∂
∂ 
and 
tkxA
t
y ωω cossin22
2
−=∂
∂ (2) 
 
Chapter 15 
 
 
1472 
 
Express the ratio of equation (1) to 
equation (2): 
 
2
2
2
2
2
2
2
2
2
1
cossin
cossin
v
k
tkxA
tkxAk
t
y
x
y
=
=−
−=
∂
∂
∂
∂
ωωω
ω
 
confirming that tkxAy ωcossin= 
satisfies the general wave equation. 
 
Harmonic Waves on a String 
 
38 • One end of a string 6.0 m long is moved up and down with simple 
harmonic motion at a frequency of 60 Hz. If the wave crests travel the length of 
the string in 0.50 s, find the wavelength of the waves on the string. 
 
Picture the Problem We can find the wavelength of the waves on the string 
using the relationship λfv = . 
 
Express the wavelength of the 
waves: 
vf =λ ⇒
f
v=λ 
 
The speed of the wave is equal to 
the distance divided by the time: 
m/s12
s0.50
m6.0
Δ
Δ ===
t
xv 
 
Substitute numerical values to 
obtain: 
cm20
s60
m/s12
1 == −λ 
 
39 • [SSM] A harmonic wave on a string with a mass per unit length of 
0.050 kg/m
and a tension of 80 N, has an amplitude of 5.0 cm. Each point on the 
string moves with simple harmonic motion at a frequency of 10 Hz. What is the 
power carried by the wave propagating along the string? 
 
Picture the Problem The average power propagated along the string by a 
harmonic wave is v,AP 2221av μω= where v is the speed of the wave, and μ, ω, and 
A are the linear density of the string, the angular frequency of the wave, and the 
amplitude of the wave, respectively. 
 
Express and evaluate the power 
propagated along the string: 
 
vAP 2221av μω= 
Traveling Waves 
 
 
1473
The speed of the wave on the string 
is given by: μ
TFv = 
 
Substitute for v and simplify to obtain: 
 ( )
( ) μπ
μπμμω
T
2
T
22
2
1T22
2
1
av
2
2
FfA
FAfFAP
=
==
 
 
Substitute numerical values and evaluate Pav: 
 ( )( )[ ] ( )( ) W9.9kg/m0.050N80m0.050s102 21av == −πP 
 
40 • A 2.00-m-long rope has a mass of 0.100 kg. The tension is 60.0 N. An 
oscillator at one end sends a harmonic wave with an amplitude of 1.00 cm down 
the rope. The other end of the rope is terminated so that all of the energy of the 
wave is absorbed and none is reflected. What is the frequency of the oscillator if 
the power transmitted is 100 W? 
 
Picture the Problem The average power propagated along the rope by a 
harmonic wave is v,AP 2221av μω= where v is the velocity of the wave, and μ, ω, 
and A are the linear density of the string, the angular frequency of the wave, and 
the amplitude of the wave, respectively. 
 
Rewrite the power equation in terms 
of the frequency of the wave: 
 
vAfvAP 2222221av 2 μπμω == 
 
Solve for the frequency and simplify 
to obtain: v
P
AvA
Pf μπμπ
av
22
av 2
2
1
2
== 
 
The wave speed is given by: 
μ
TFv = 
 
Substitute for v and simplify to 
obtain: 
T
av
T
av 2
2
12
2
1
F
P
AF
P
A
f μπ
μμ
π == 
 
Chapter 15 
 
 
1474 
 
Substitute numerical values and evaluate f: 
 
( )
( )
( )
Hz 171
N 0.60
m 2.00
kg 0.100
 W1002
m 0100.02
1 =
⎟⎠
⎞⎜⎝
⎛
= πf 
 
41 •• The wave function for a harmonic wave on a string is 
y(x, t) = (1.00 mm) sin(62.8 m–1 x + 314 s–1 t). (a) In what direction does this 
wave travel, and what is the wave’s speed? (b) Find the wavelength, frequency, 
and period of this wave. (c) What is the maximum speed of any point on the 
string? 
 
Picture the Problem ( ) ( )tkxAtxy ω−= sin, describes a wave traveling in the +x 
direction. For a wave traveling in the −x direction, we have ( ) ( )tkxAtxy ω+= sin, . 
We can determine A, k, and ω by examination of the wave function. The 
wavelength, frequency, and period of the wave can, in turn, be determined from k 
and ω. 
 
(a) Because the sign between the kx and ωt terms is positive, the wave is traveling 
in the −x direction. 
 
The speed of the wave is: 
m/s00.5
m8.62
s314
1
1
=== −
−
k
v ω 
 
(b) The coefficient of x is k and: 
λ
π2=k ⇒
k
πλ 2= 
 
Substitute numerical values and 
evaluate λ: cm0.10m8.62
2
1 == −πλ 
 
The coefficient of t is ω and: 
Hz0.50
2
s314
2
1
===
−
ππ
ωf 
 
The period of the wave motion is the 
reciprocal of its frequency: 
 
s0200.0
s0.50
11
1 === −fT 
 
(c) Express and evaluate the 
maximum speed of any string 
segment: 
( )( )
m/s0.314
rad/s314mm00.1max
=
== ωAv
 
 
Traveling Waves 
 
 
1475
42 •• A harmonic wave on a string with a frequency of 80 Hz and an 
amplitude of 0.025 m travels in the +x direction with a speed of 12 m/s. (a) Write 
a suitable wave function for this wave. (b) Find the maximum speed of a point on 
the string. (c) Find the maximum acceleration of a point on the string. 
 
Picture the Problem ( ) ( )tkxAx,ty ω−= sin describes a wave traveling in the +x 
direction. We can find ω and k from the data included in the problem statement 
and substitute in the general equation. The maximum speed of a point on the 
string can be found from ωAv =max and the maximum acceleration from 
.2max ωAa = 
 
(a) Express the general form of the 
equation of a harmonic wave 
traveling to the right: 
 
( ) ( )tkxAx,ty ω−= sin 
Evaluate ω: 
 
( )
12
11
s100.5
s503s8022
−
−−
×=
=== ππω f
 
 
Determine k: 11 m42
m/s12
s503 −− ===
v
k ω 
 
Substitute to obtain: ( ) ( ) ( ) ( )[ ]txx,ty 121 s100.5m42sinm025.0 −− ×−= 
 
(b) The maximum speed of a point 
on the string is: 
( )( )
m/s13
s503m0.025 1max
=
== −ωAv
 
 
(c) Express the maximum 
acceleration of a point on the string: 
 
2
max ωAa = 
 
Substitute numerical values and 
evaluate amax: 
( )( ) 221max km/s3.6s503m0.025 == −a
 
43 •• A 200-Hz harmonic wave with an amplitude equal to 1.2 cm moves 
along a 40-m-long string that has a mass of 0.120 kg and a tension of 50 N. 
(a) What is the average total energy of the waves on a 20-m-long segment of 
string? (b) What is the power transmitted past a given point on the string? 
 
Picture the Problem The average power propagated along the string is 
.2221av vAP μω= The average energy on a segment of length L is the average 
power multiplied by the time for the wave to travel the distance L 
Chapter 15 
 
 
1476 
 
(a) Express the power transmitted 
past a given point on the string. The 
average energy passing a point in 
time t is the average power 
multiplied by the time: 
 
vAP 2221av μω= 
and 
tPE avav = 
The time for the wave to travel the 
length L is L/v: v
Lt = 
 
Combine the previous steps and 
solve for the average energy on the 
string: 
 
( ) 2222221
22
2
122
2
1
av
22 ALfAfL
AL
v
LAvE
μππμ
ωμωμ
==
==
 
Substitute numerical values and evaluate Eav: 
 
( ) ( )( ) J 8.6m0.012m 20s200
m40
kg0.1202 2212av =⎟⎟⎠
⎞
⎜⎜⎝
⎛= −πE 
 
(b) Express Pav: 222
1
av AvP ωμ= 
 
To calculate Pav we need an 
expression for the wave speed v: 
 
μ
TFv = 
Substitute for v to obtain: 
 ( )
μμπ
μπμ
T222
T22
2
1
av
2
2
FAf
FAfP
=
=
 
 
Substitute numerical values and evaluate Pav: 
 
( ) ( ) W44
m 20
kg 060.0
N 50m 012.0s 200
m 20
kg 060.02 2212av =⎟⎠
⎞⎜⎝
⎛= −πP 
 
44 •• On a real string, some of the energy of the wave dissipates as the wave 
travels down the string. Such a situation can be described by a wave function 
whose amplitude A(x) depends on x: y = A(x) sin (kx – ωt) where 
A(x) = (A0e–bx). What is the power transported by the wave as a function of x, 
where x > 0? 
Traveling Waves 
 
 
1477
Picture the Problem The power propagated along the string by a harmonic wave 
is vAP 2221 μω= where v is the velocity of the wave, and μ, ω, and A are the linear 
density of the string, the angular frequency of the wave, and the amplitude of the 
wave, respectively. 
 
Express the power associated with 
the wave at the origin: 
 
vAP 2221 μω= (1) 
Express the amplitude of the 
wave at x: 
 
( ) bxeAxA −= 0 
Substitute in equation (1) to obtain: ( ) ( )
bx
bx
veA
veAxP
22
0
2
2
1
2
0
2
2
1
−
−
=
=
μω
μω
 
 
45 •• [SSM] Power is to be transmitted along a taut string by means of 
transverse harmonic waves. The wave speed is 10 m/s and the linear mass density 
of the string is 0.010 kg/m. The power source oscillates with an amplitude of 
0.50 mm. (a) What average power is transmitted along the string if the frequency 
is 400 Hz? (b) The power transmitted can be increased by increasing the tension 
in the string, the frequency of the source, or the amplitude of the waves. By how 
much would each of these quantities have to increase to cause an increase in 
power
by a factor of 100 if it is the only quantity changed? 
 
Picture the Problem The average power propagated along a string by a 
harmonic wave is vAP 2221av μω= where v is the speed of the wave, and μ, ω, and 
A are the linear density of the string, the angular frequency of the wave, and the 
amplitude of the wave, respectively. 
 
(a) Express the average power 
transmitted along the string: 
 
vAfvAP 2222221av 2 μπμω == 
 
Substitute numerical values and 
evaluate Pav: 
( )( )
( ) ( )
mW79
m/s10m100.50
s400kg/m0.0102
23
212
av
=
××
=
−
−πP
 
 
(b) Because 2av fP ∝ , increasing the frequency by a factor of 10 would increase 
the power by a factor of 100. 
Chapter 15 
 
 
1478 
 
Because 2av AP ∝ , increasing the amplitude by a factor of 10 would increase 
the power by a factor of 100. 
 
Because vP ∝av and Fv ∝ , increasing the tension by a factor of 104 would 
increase v by a factor of 100 and the power by a factor of 100. 
 
46 ••• Two very long strings are tied together at the point x = 0. In the region 
x < 0, the wave speed is v1, while in the region x > 0, the speed is v2. A sinusoidal 
wave is incident on the knot from the left (x < 0); part of the wave is reflected and 
part is transmitted. For x < 0, the displacement of the wave is describabed by 
y(x,t) = A sin (k1x – ωt) + B sin(k1x + ωt), while for x > 0, y(x,t) = C sin (k2x – ωt), 
where ω/k1 = v1 and ω/k2 = v2. (a) If we assume that both the wave function y and 
its first spatial derivative ∂y/∂x must be continuous at x = 0, show that 
C/A = 2v2/( v1 + v2), and that B/A = (v1 – v2)/( v1 + v2). (b) Show that 
B2 + (v1/v2)C2 = A2. 
 
Picture the Problem We can use the assumption that both the wave function and 
its first spatial derivative are continuous at x = 0 to establish equations relating A, 
B, C, k1, and k2. Then, we can solve these simultaneous equations to obtain 
expressions for B and C in terms of A, v1, and v2. 
 
(a) Let y1(x, t) represent the wave 
function in the region x < 0, and 
y2(x, t) represent the wave function 
in the region x > 0. Express the 
continuity of the two wave functions 
at x = 0: 
( ) ( )tyty ,0,0 21 = 
and ( )[ ] ( )[ ]
( )[ ]tkC
tkBtkA
ω
ωω
−=
++−
0sin
0sin0sin
2
11 
or ( ) ( )tCtBtA ωωω −=+− sinsinsin 
 
Because the sine function is odd; that 
is, ( ) θθ sinsin −=− : 
 
tCtBtA ωωω sinsinsin −=+− 
and 
CBA =− (1) 
 
Differentiate the wave functions with 
respect to x to obtain: ( )
( )txkBk
txkAk
x
y
ω
ω
++
−=∂
∂
11
11
1
cos
cos
 
and 
( )txkCk
x
y ω−=∂
∂
22
2 cos 
 
Traveling Waves 
 
 
1479
 
Express the continuity of the slopes 
of the two wave functions at x = 0: 
 
∂y1
∂x
x =0
= ∂y 2∂x
x =0
 
and ( )[ ] ( )[ ]
( )[ ]tkCk
tkBktkAk
ω
ωω
−=
++−
0cos
0cos0cos
22
1111 
or ( )
( )tCk
tBktAk
ω
ωω
−=
+−
cos
coscos
2
11 
 
Because the cosine function is even; 
that is, ( ) θθ coscos =− : tCktBktAk ωωω coscoscos 211 =+ and 
CkBkAk 211 =+ (2) 
 
Multiply equation (1) by k1 and add 
it to equation (2) to obtain: 
 
( )CkkAk 2112 += 
Solving for C yields: 
 Akk
A
kk
kC
1221
1
1
22
+=+= 
 
Solve for C/A and substitute ω/v1 
for k1 and ω/v2 for k2 to obtain: 
 21
2
2112
2
1
2
1
2
vv
v
vvkkA
C
+=+=+= 
 
Substitute in equation (1) to obtain: 
A
vv
vBA ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+=− 12
22 
 
Solving for B/A yields: 
21
21
vv
vv
A
B
+
−= 
 
 
Chapter 15 
 
 
1480 
(b) We wish to show that 
 
 B2 + (v1/v2)C2 = A2 
 
Use the results of (a) to obtain the 
expressions B = −[(1 − α)/(1 + α)] A 
and C = 2A/(1 + α), where α = v1/v2. 
 
Substitute these expressions into 
 
 B2 + (v1/v2)C2 = A2 
 
and check to see if the resulting 
equation is an identity: 
( )
( )
( )
( )
( )
( )
11
1
1
1
1
1
21
1
1
421
1
1
41
1
1
2
1
1
1
2
1
1
2
2
2
2
2
2
2
2
22
22
2
2
2
22
2
12
=
=+
+
=+
++
=+
++−
=+
+−
=⎟⎠
⎞⎜⎝
⎛
++⎟⎠
⎞⎜⎝
⎛
+
−
=⎟⎠
⎞⎜⎝
⎛
++⎟⎠
⎞⎜⎝
⎛
+
−
=+
α
α
α
αα
α
ααα
α
αα
ααα
α
ααα
α AAA
AC
v
vB
 
 
The equation is an identity: 
Therefore, 22
2
12 AC
v
vB =+ 
 
Remarks: Our result in (a) can be checked by considering the limit of B/A as 
v2/v1 → 0. This limit gives B/A = +1, telling us that the transmitted wave has 
zero amplitude and the incident and reflected waves superpose to give a 
standing wave with a node at x = 0. 
 
Harmonic Sound Waves 
 
47 • A sound wave in air produces a pressure variation given by 
( ) ( )txtxp 343
2
cos75.0, −= π , where p is in pascals, x is in meters, and t is in 
seconds. Find (a) the pressure amplitude, (b) the wavelength, (c) the frequency, 
and (d) the wave speed. 
 
Picture the Problem The pressure variation is of the form ( ) ( )vtxkpx,tp −= cos0 where k is the wave number, p0 is the pressure amplitude, 
and v is the wave speed. We can find λ from k and f from ω and k. 
 
(a) By inspection of the equation: Pa75.00 =p 
 
Traveling Waves 
 
 
1481
(b) Because :
2
2 π
λ
π ==k 
 
m00.4=λ 
(c) Solve 
k
f
k
v πω 2== for f to obtain: 
 
π2
kvf = 
 
Substitute numerical values and 
evaluate f: 
( )
Hz8.85
2
m/s343
2 == π
π
f 
 
(d) By inspection of the equation: m/s343=v 
 
48 • (a) Middle C on the musical scale has a frequency of 262 Hz. What is 
the wavelength of this note in air? (b) The frequency of the C an octave above 
middle C is twice that of middle C. What is the wavelength of this note in air? 
 
Picture the Problem The frequency, wavelength, and speed of the sound waves 
are related by v = fλ. 
 
(a) The wavelength of middle C is 
given by: 
m30.1
s262
m/s340
1 === −f
vλ 
 
(b) Evaluate λ for a frequency twice 
that of middle C: ( ) m649.0s2622 m/s340 1 === −fvλ 
 
49 • [SSM] The density of air is 1.29 kg/m3. (a) What is the displacement 
amplitude for a sound wave with a frequency of 100 Hz and a pressure amplitude 
of 1.00 × 10–4 atm? (b) The displacement amplitude of a sound wave of frequency 
300 Hz is 1.00 ×10–7 m. What is the pressure amplitude of this wave? 
 
Picture the Problem The pressure amplitude depends on the density of the 
medium ρ, the angular frequency of the sound wave μ, the speed of the wave v, 
and the displacement amplitude s0 according to .00 vsp ρω= 
 
(a) Solve 00 vsp ρω= for s0: 
v
ps ρω
0
0 = 
 
Chapter 15 
 
 
1482 
 
Substitute numerical values and evaluate s0: 
 ( )( )( )( )( ) m 4.36m 1064.3m/s 343s 100kg/m 1.292 Pa/atm 1001325.1atm 1000.1 513
54
0 μπ =×=
××= −−
−
s 
 
(b) Use 00 vsp ρω= to find p0: 
 ( )( )( )( ) mPa4.83m101.00m/s343s300kg/m29.12 7130 =×= −−πp 
 
50 • The density of air is 1.29 kg/m3. (a) What is the displacement 
amplitude of a sound wave that has a frequency of 500 Hz at the pain-threshold 
pressure amplitude of 29.0 Pa? (b) What is the displacement amplitude of a sound 
wave that has the same pressure amplitude as the wave in Part (a), but has a 
frequency of 1.00 kHz? 
 
Picture the Problem The pressure amplitude depends on the density of the 
medium ρ, the angular frequency of the sound wave μ, the wave speed v, and the 
displacement amplitude s0 according to 00 vsp ρω= . 
 
(a) Solve 00 vsp ρω= for s0: 
v
ps ρω
0
0 = 
 
Substitute numerical values and 
evaluate s0: ( )( )( )
m9.20
m/s343Hz500kg/m1.292
Pa0.29
30
μ=
=
π
s
 
 
(b) Proceed as in (a) with 
 f = 1.00 kHz: ( )( )(
)
m4.10
m/s343kHz 00.1kg/m1.292
Pa0.29
30
μ=
=
π
s
 
51 • A typical loud sound wave that has a frequency of 1.00 kHz has a 
pressure amplitude of about 1.00 × 10–4 atm. (a) At t = 0, the pressure is a 
maximum at some point x1. What is the displacement at that point at t = 0? 
(b) Assuming the density of air is 1.29 kg/m3, what is the maximum value of the 
displacement at any time and place? 
 
Picture the Problem The pressure or density wave is 90° out of phase with the 
displacement wave. When the displacement is zero, the pressure and density 
changes are either a maximum or a minimum. When the displacement is a 
Traveling Waves 
 
 
1483
maximum or minimum, the pressure and density changes are zero. We can use 
00 vsp ρω= to find the maximum value of the displacement at any time and place. 
 
(a) If the pressure is a maximum at x1 when t = 0, the displacement s is zero. 
 
(b) Solve 00 vsp ρω= for s0: 
v
ps ρω
0
0 = 
 
Substitute numerical values and evaluate s0: 
 ( )( )( )( )( ) m64.3m/s343kHz 1.00kg/m1.292 Pa/atm101.01325atm101.00 3
54
0 μ=××=
−
π
s 
 
52 • An octave represents a change in frequency by a factor of two. Over 
how many octaves can a typical person hear? 
 
Picture the Problem A human can hear sounds between roughly 20 Hz and 
20 kHz; a factor of 1000. An octave represents a change in frequency by a factor 
of 2. We can evaluate 2N = 1000 to find the number of octaves heard by a person 
who can hear this range of frequencies. 
 
Relate the number of octaves to the 
difference between 20 kHz and 
20 Hz: 
 
10002 =N 
Take the logarithm of both sides of 
the equation to obtain: 
 
310log2log =N ⇒ 32log =N 
 
Solving for N yields: 
1097.9
2log
3 ≈==N 
 
53 •• In the oceans, whales communicate by sound transmission through the 
water. A whale emits a sound of 50.0 Hz to tell a wayward calf to catch up to the 
pod. The speed of sound in water is about 1500 m/s. (a) How long does it take the 
sound to reach the calf if he is 1.20 km away? (b) What is the wavelength of this 
sound in the water? (c) If the whales are close to the surface, some of the sound 
energy might refract out into the air. What would be the frequency and 
wavelength of the sound in the air? 
 
 
 
 
 
Chapter 15 
 
 
1484 
Picture the Problem (a) We can use the definition of average speed to find the 
time required for the sound to travel to the calf. (b) We can use the relationship 
between wavelength, frequency, and speed to find the wavelength of the sound in 
water. (c) The frequency of the sound does not change as it travels from water to 
air, but its wavelength changes because of the difference in the speed of sound in 
water and in air. 
 
(a) Relate the time it takes the sound 
to reach the calf to the distance from 
the whale to the calf and the speed of 
sound in water: 
 
s 80.0
m/s1500
km 20.1
Δ ===
v
dt 
(b) The wavelength of this sound in 
water is the ratio of its speed in water 
to its frequency: 
 
m 30
Hz 50.0
m/s 1500water
water === f
vλ 
 
(c) Because the frequency does not 
change as it travels from water to air: 
 
Hz 0.50=f 
The wavelength of this sound in air 
is the ratio of its speed in air to its 
frequency: 
m .866
Hz 50.0
m/s 433air
air === f
vλ 
 
Waves in Three Dimensions: Intensity 
 
54 • A spherical sinusoidal source radiates sound uniformly in all 
directions. At a distance of 10.0 m, the sound intensity level is 1.00 × 10–4 W/m2. 
(a) At what distance from the source is the intensity 1.00 × 10–6 W/m2? (b) What 
power is radiated by this source? 
 
Picture the Problem The intensity of the sound from the spherical sinusoidal 
source varies inversely with the square of the distance from the source. The power 
radiated by the source is the product of the intensity of the radiation and the 
surface area over which it is distributed. 
 
(a) Relate the intensity I1 at a 
distance R1 from the source to the 
energy per unit time (power) arriving 
at the point of interest: 
1
2
11 av,2
1
1 av,
1 44
IRP
R
P
I ππ =⇒= 
 
Traveling Waves 
 
 
1485
 
At a distance R2 from the source: 
 2
2
22 av,2
2
2 av,
2 44
IRP
R
P
I ππ =⇒= 
 
 
Because 2 av,1 av, PP = : 
2
1
122
2
21
2
1 44 I
IRRIRIR =⇒= ππ 
 
Substituting numerical values and 
evaluating R2 gives: ( )
m100
 W/m1000.1
 W/m1000.1m 0.10 26
24
2
=
×
×= −
−
R
 
 
(b) Solve 2
av
4 r
PI π= for Pav: 
 
IrP 2av 4π= 
Substitute numerical values and 
evaluate Pav: 
( ) ( )
mW126
W/m1000.1m0.104 242av
=
×= −πP
 
 
55 • [SSM] A loudspeaker at a rock concert generates a sound that has an 
intensity level equal to 1.00 × 10–2 W/m2 at 20.0 m and has a frequency of 
1.00 kHz. Assume that the speaker spreads its energy uniformly in three 
dimensions. (a) What is the total acoustic power output of the speaker? (b) At 
what distance will the sound intensity be at the pain threshold of 1.00 W/m2? 
(c) What is the sound intensity at 30.0 m? 
 
Picture the Problem Because the power radiated by the loudspeaker is the 
product of the intensity of the sound and the area over which it is distributed, we 
can use this relationship to find the average power, the intensity of the radiation, 
or the distance to the speaker for a given intensity or average power. 
 
(a) Use IrP 2av 4π= to find the total 
acoustic power output of the speaker: 
( ) ( )
W3.50 W27.50
W/m1000.1m0.204 222av
==
×= −πP
 
 
(b) Relate the intensity of the sound 
at 20 m to the distance from the 
speaker: 
 
( )2av22 m 0.204 W/m1000.1 π
P=× −
 
 
Chapter 15 
 
 
1486 
 
Relate the threshold-of-pain 
intensity to the distance from the 
speaker: 
 
2
av2
4
W/m00.1
r
P
π= 
Divide the first of these equations by 
the second and solve for r: 
 
( )( ) m00.2m0.201000.1 22 =×= −r 
 
(c) Use 2
av
4 r
PI π= to find the intensity 
at 30.0 m: 
( ) ( )
23
2
W/m1045.4
m0.304
W3.50m0.30
−×=
= πI 
 
56 •• When a pin of mass 0.100 g is dropped from a height of 1.00 m, 0.050 
percent of its energy is converted into a sound pulse that has a duration of 0.100 s. 
(a) Estimate how far away the dropped pin can be heard if the minimum audible 
intensity is 1.00 × 10–11 W/m2. (b) Your result in Part (a) is much too large in 
practice because of background noise. If you assume that the intensity must be at 
least 1.00 × 10–8 W/m2 for the sound to be heard, estimate how far away the 
dropped pin can be heard. (In both parts, assume that the intensity is P/4πr2.) 
 
Picture the Problem We can use conservation of energy to find the acoustical 
energy resulting from the dropping of the pin. The power developed can then be 
found from the given time during which the energy was transformed from 
mechanical to acoustical form. We can find the range at which the dropped pin 
can be heard from I = P/4π r2. 
 
(a) Assuming that I = P/4π r2, 
express the distance at which one can 
hear the dropped pin: 
 
I
Pr π4= 
Use conservation of energy to 
determine the sound energy 
generated when the pin falls: 
 
( )( )
( )( )
J10905.4
m00.1m/s81.9
kg10100.000050.0
7
2
3
−
−
×=
×
×=
= mghE ε
 
 
The power of the sound pulse is 
given by: 
W104.905
s0.100
J104.905
Δ
6
7
−
−
×=
×==
t
EP
 
 
Traveling Waves 
 
 
1487
Substitute numerical values and 
evaluate r: ( )
km20.0
W/m1000.14
W10905.4
211
6
=
×
×= −
−
πr 
 
(b) Repeat the last step in (a) with 
I = 1.00 × 10–8 W/m2: ( ) m2.6W/m1000.14 W10905.4 28
6
=×
×= −
−
πr 
 
*Intensity Level 
 
57 • [SSM] What is
the intensity level in decibels of a sound wave that 
has an intensity of (a) 1.00 × 10–10 W/m2 and (b) 1.00 × 10–2 W/m2? 
 
Picture the Problem The intensity level β of a sound wave, measured in 
decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is defined to be 
the threshold of hearing. 
 
(a) Using its definition, calculate the 
intensity level of a sound wave whose 
intensity is 1.00 × 10–10 W/m2: 
 
( )
dB0.2010log10
W/m10
W/m101.00logdB10
2
212
201
==
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×= −
−
β
 
 
(b) Proceed as in (a) with 
I = 1.00 × 10–2 W/m2: ( )
dB10010log10
W/m10
W/m101.00logdB10
10
212
22
==
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×= −
−
β
 
 
58 • What is the intensity of a sound wave if, at a particular location, the 
intensity level is (a) β = 10 dB and (b) β = 3 dB? 
 
Picture the Problem (a) and (b) The intensity level β of a sound wave, measured 
in decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is defined to 
be the threshold of hearing. 
 
(a) Solve ( ) ( )0logdB10 II=β for I 
to obtain: 
 
( )
0
dB1010 II β= 
Evaluate I for β = 10 dB: ( ) ( )
( ) 211212 00
dB10dB10
W/m10W/m1010
1010
−− ==
== III
 
 
Chapter 15 
 
 
1488 
(b) Proceed as in (a) with β = 3 dB: ( ) ( )
( ) 212212 00
dB10dB3
W/m102W/m102
210
−− ×==
== III
 
 
 
59 • At a certain distance, the sound intensity level of a dog’s bark is 
50 dB. At that same distance, the sound intensity of a rock concert is 10,000 times 
that of the dog’s bark. What is the sound intensity level of the rock concert? 
 
Picture the Problem The intensity level β of a sound wave, measured in 
decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is defined to be 
the threshold of hearing. 
 
Express the sound intensity level of 
the rock concert: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
concert
concert logdB10 I
Iβ (1) 
 
Express the sound intensity level of 
the dog’s bark: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
doglogdB10dB50
I
I
 
 
Solve for the intensity of the dog’s 
bark: 
( )
27
2125
0
5
dog
W/m10
W/m101010
−
−
=
== II
 
 
Express the intensity of the rock 
concert in terms of the intensity 
of the dog’s bark: 
( )
23
274
dog
4
concert
W/m10
W/m101010
−
−
=
== II
 
 
Substitute in equation (1) and 
evaluate βconcert: ( )
( ) dB9010logdB10
W/m10
W/m10logdB10
9
212
23
concert
==
⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
−
β
 
 
60 • What fraction of the acoustic power of a noise would have to be 
eliminated to lower its sound intensity level from 90 to 70 dB? 
 
Picture the Problem We can express the intensity levels at both 90 dB and 70 dB 
in terms of the intensities of the sound at those levels. By subtracting the two 
expressions, we can solve for the ratio of the intensities at the two levels and then 
find the fractional change in the intensity that corresponds to a decrease in 
intensity level from 90 dB to 70 dB. 
 
Traveling Waves 
 
 
1489
Express the intensity level at 90 dB: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
90logdB10dB90
I
I 
 
Express the intensity level at 70 dB: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
70logdB10dB70
I
I 
Express Δβ = β90 − β70: 
( ) ( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=
=Δ
70
90
0
70
0
90
logdB10
logdB10logdB10
dB20
I
I
I
I
I
I
β
 
Solving for I90 yields: 7090 100II = 
 
Express the fractional change in the 
intensity from 90 dB to 70 dB: 
%99
100
100
70
7070
90
7090 =−=−
I
II
I
II 
 
Because IP ∝ , the fractional change in power is %99 . 
 
61 •• A spherical source radiates sound uniformly in all directions. At a 
distance of 10 m, the sound intensity level is 80 dB. (a) At what distance from the 
source is the intensity level 60 dB? (b) What power is radiated by this source? 
 
Picture the Problem The intensity at a distance r from a spherical source varies 
with distance from the source according to .4 2av rPI π= We can use this 
relationship to relate the intensities corresponding to an 80-dB intensity level (I80) 
and the intensity corresponding to a 60-dB intensity level (I60) to their distances 
from the source. We can relate the intensities to the intensity levels 
through ( ) ( )0logdB10 II=β . 
 
(a) Express the intensity of the sound 
where the intensity level is 80 dB: 280
av
80 4 r
PI π= 
 
Express the intensity of the sound 
where the intensity level is 60 dB: 
 
2
60
av
60 4 r
PI π= 
Chapter 15 
 
 
1490 
Divide the first of these equations by 
the second to obtain: 
 
( )
2
2
60
2
60
av
2
av
60
80
m100
4
m104 r
r
P
P
I
I ==
π
π 
 
Solving for r60 yields: ( )
60
80
60 m10 I
Ir = 
 
Find the intensity of the 80-dB 
sound level radiation: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
80logdB10dB80
I
I 
and 
24
0
8
80 W/m1010
−== II 
 
Find the intensity of the 60-dB 
sound level radiation: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
60logdB10dB60
I
I 
and 
26
0
6
60 W/m1010
−== II 
 
Substitute numerical values for I80 and 
I60 and evaluate r60: ( ) km10.0W/m10
W/m10m10 26
24
60 == −
−
r 
 
(b) Using the intensity corresponding 
to an intensity level of 80 dB, 
express and evaluate the power 
radiated by this source: 
( ) ( )[ ]
W13.0
m104πW/m10 22480
=
== −AIP
 
 
62 •• Harry and Sally are sitting on opposite sides of a circus tent when an 
elephant trumpets a loud blast. If Harry experiences a sound intensity level of 65 
dB and Sally experiences only 55 dB, what is the ratio of the distance between 
Sally and the elephant to the distance between Harry and the elephant? 
 
Picture the Problem The intensity of the sound heard by Harry and Sally 
depends inversely on the square of the distance between the elephant and each of 
them. We can use the definition of sound intensity (decibel) level 
( ( ) ( )0logdB 10 II=β ) and the definition of intensity ( API av= ) to find the ratio 
of these distances. 
 
Traveling Waves 
 
 
1491
Express the sound intensity level at 
Harry’s location: ( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
2
H
av
0
H
H
4
logdB 10
logdB 10
Ir
P
I
I
π
β
 
 
Similarly, the sound intensity level at 
Sally’s location is: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
2
S
av
S 4
logdB 10
Ir
P
πβ 
 
The difference in decibel level’s at 
the two locations is given by: 
 
SHΔ βββ −= 
 
Substituting for βH, βS, and Δβ yields: 
 
( ) ( ) dB 10dB 55dB 65
4
logdB 10
4
logdB 10Δ
0
2
S
av
0
2
H
av =−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=
Ir
P
Ir
P
ππβ 
 
Simplifying this expression yields: 
 1log 2
H
2
S =⎟⎟⎠
⎞
⎜⎜⎝
⎛
r
r ⇒ 12
H
2
S 10=
r
r 
 
Solving for the ratio HS rr yields: 
 
2.310
H
S ==
r
r 
 
63 •• Three noise sources produce intensity levels of 70, 73, and 80 dB 
when acting separately. When the sources act together, the resultant intensity is 
the sum of the individual intensities. (a) Find the sound intensity level in decibels 
when the three sources act at the same time. (b) Discuss the effectiveness of 
eliminating the two least intense sources in reducing the intensity level of the 
noise. 
 
Picture the Problem We can find the intensities of the three sources from their 
intensity levels and, because their intensities are additive, find the intensity level 
when all three sources are acting. 
 
(a) Express the sound intensity level 
when the three sources act at the 
same time: 
 
( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
807370
0
sources3
sources3
logdB10
logdB10
I
III
I
Iβ
 
 
Chapter 15 
 
 
1492 
Find the intensities of each of the 
three sources: ( ) 0770
0
70 10logdB10dB70 II
I
I =⇒⎟⎟⎠
⎞
⎜⎜⎝
⎛= 
( ) 03.773
0
73 10logdB10dB73 II
I
I =⇒⎟⎟⎠
⎞
⎜⎜⎝
⎛= 
and 
( ) 0880
0
80 10logdB10dB80 II
I
I =⇒⎟⎟⎠
⎞
⎜⎜⎝
⎛= 
 
Substituting in the expression for β3 sources yields: 
 
( ) ( ) ( )
dB81
101010logdB10101010logdB10 83.77
0
0
8
0
3.7
0
7
sources3
=
++=⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++=
I
IIIβ
 
 
(b) Find the intensity level with the 
two least intense sources eliminated: ( ) dB8010logdB10
0
0
8
80 =⎟⎟⎠
⎞
⎜⎜⎝
⎛=
I
Iβ 
 
Eliminating the 70-dB and 73-dB sources does not reduce the intensity level 
significantly. 
 
64 •• Show that if two people are different distances away from a sound 
source, the difference Δβ between the intensity levels reaching the people, in 
decibels, will always be the same, no matter the power radiated by the source. 
 
Picture the Problem Let the two people be identified by the numerals 1 and 2 
and use the definition of the intensity level, in decibels, to express Δβ as a 
function of the distances r1 and r2 of the two people from the source. 
 
Express the difference in the sound 
intensity level heard by the two 
people: 
 
12Δ βββ −= 
The sound level intensity (decibel 
level) heard by the second person is 
given by: 
( )
( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
2
2
av
02
av
0
2
2
4
logdB 10
logdB 10
logdB 10
Ir
P
IA
P
I
I
π
β
 
Traveling Waves 
 
 
1493
In like manner: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
2
1
av
1 4
logdB 10
Ir
P
πβ 
 
Substitute in the expression for Δβ to obtain: 
 
( ) ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
2
1
av
0
2
2
av
4
logdB 10
4
logdB 10Δ
Ir
P
Ir
P
ππβ 
 
Simplifying yields: 
 ( )
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
=
2
2
2
1
0
2
1
av
0
2
2
av
logdB 10
4
4logdB 10Δ
r
r
Ir
P
Ir
P
π
πβ
 
 
This result shows that the difference in sound level intensities depends only on 
the distances to the source and not on the source’s power output. 
 
65 ••• Everyone at a party is talking equally loudly. One person is talking to 
you and the sound intensity level at your location is 72 dB. Assuming that all 38 
people at the party are at the same distance from you as the person who you are 
talking to, find the sound intensity level at your location. 
 
Picture the Problem The sound intensity level can be found from the intensity of 
the sound due to the 38 people. When 38 people are talking, the intensities add. 
 
Express the sound level when all 
38 people are talking: ( )
( ) ( )
( )
dB88
dB7238logdB10
logdB1038logdB10
38logdB10
0
1
0
1
38
=
+=
⎟⎟⎠
⎞
⎜⎜⎝
⎛+=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
I
I
I
Iβ
 
 
An equivalent but longer solution: 
 
 
Express the sound intensity level 
when all 38 people are talking: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
1
38
38logdB10
I
Iβ 
Chapter 15 
 
 
1494 
Express the sound intensity level 
when only one person is talking: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛==
0
1
1 logdB10dB72 I
Iβ 
 
Solving for I1 yields: ( )
25
2122.7
0
2.7
1
W/m101.58
W/m101010
−
−
×=
== II
 
 
Express the sound intensity when all 
38 people are talking: 
 
138 38II = 
The sound intensity level is: ( ) ( )
dB88
W/m10
W/m101.5838logdB10 212
25
38
=
⎥⎦
⎤⎢⎣
⎡ ×= −
−
β
 
 
66 ••• When a violinist pulls the bow across a string, the force with which the 
bow is pulled is fairly small, about 0.60 N. Suppose the bow travels across the A 
string, which vibrates at 440 Hz, at 0.50 m/s. A listener 35 m from the performer 
hears a sound of 60-dB intensity. Assuming that the sound radiates uniformly in 
all directions, with what efficiency is the mechanical energy of bowing converted 
to sound energy? 
 
Picture the Problem Let η represent the efficiency with which mechanical 
energy is converted to sound energy. Because we’re given information regarding 
the rate at which mechanical energy is delivered to the string and the rate at which 
sound energy arrives at the location of the listener, we’ll take the efficiency to be 
the ratio of the sound power delivered to the listener divided by the power 
delivered to the string. We can calculate the power input directly from the given 
data. We’ll calculate the intensity of the sound at 35 m from its intensity level at 
that distance and use this result to find the power output. 
 
Express the efficiency of the 
conversion of mechanical energy to 
sound energy: 
 
in
out
P
P=η 
Find the power delivered by the bow 
to the string: 
 
( )( ) W0.30m/s0.50N0.60in === FvP 
 
Using ( ) ( )0logdB10 II=β , find the 
intensity of the sound at 35 m: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
m35logdB10dB60
I
I
 
and 
26
0
6
m35 W/m1000.110
−×== II 
Traveling Waves 
 
 
1495
The rate at which sound energy is 
emitted is given by: 
( )( )
W0.0154
m35W/m101.004π 226out
=
×== −IAP 
 
Substitute numerical values and 
evaluate η: %1.5W0.30
W0.0154 ==η 
 
67 ••• [SSM] The noise intensity level at some location in an empty 
examination hall is 40 dB. When 100 students are writing an exam, the noise level 
at that location increases to 60 dB. Assuming that the noise produced by each 
student contributes an equal amount of acoustic power, find the noise intensity 
level at that location when 50 students have left. 
 
Picture the Problem Because the sound intensities are additive, we’ll find the 
noise intensity level due to one student by subtracting the background noise 
intensity from the intensity due to the students and dividing by 100. Then, we’ll 
use this result to calculate the intensity level due to 50 students. 
 
Express the intensity level due to 50 
students: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
1
50
50logdB10
I
Iβ 
Find the sound intensity when 100 
students are writing the exam: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
100logdB10dB60
I
I 
and 
26
0
6
100 W/m1010
−== II 
 
Find the sound intensity due to the 
background noise: ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
0
backgroundlogdB10dB40
I
I
 
and 
28
0
4
background W/m1010
−== II 
 
Express the sound intensity due to 
the 100 students: 26
2826
background100
W/m10
W/m10W/m10
−
−−
≈
−=− II
 
Find the sound intensity due to 1 
student: 
 
28background100 W/m10
100
−=− II 
Substitute numerical values and 
evaluate the noise intensity level 
due to 50 students: 
( ) ( )
dB57
W/m10
W/m101.0050logdB10 212
28
50
=
×= −
−
β
 
Chapter 15 
 
 
1496 
String Waves Experiencing Speed Changes 
68 • A 3.00-m-long piece of string with a mass of 25.0 g is tied to 4.00 m 
of heavy twine with a mass of 75.0 g and the combination is put under a tension 
of 100 N. If a transverse pulse is sent down the less dense string, determine the 
reflection and transmission coefficients at the junction point. 
 
Picture the Problem Let the subscript ″s″ refer to the string and the subscript ″t″ 
to the heavy twine. We can use the definitions of the reflection and transmission 
coefficients and the expression for the speed of waves on a string (Equation 15-3) 
to find the speeds of the pulse on the string and the heavy twine. 
 
Use their definitions to express the 
reflection and transmission 
coefficients: 
 t
s
t
s
st
st
1
1
v
v
v
v
vv
vvr
+
−
=+
−= (1) 
and 
t
sst
t
1
22
v
vvv
v
+
=+=τ
(2) 
 
Use Equation 15-3 to express vt and 
vs: t
T
t μ
Fv = and 
s
T
s μ
Fv = 
 
Dividing the expression for vs by the 
expression for vt and simplifying 
yields: 
 s
t
t
T
s
T
t
s
μ
μ
μ
μ ==
F
F
v
v 
 
Substitute for ts vv in equation (1) to 
obtain: 
s
t
s
t
1
1
μ
μ
μ
μ
+
−
=r (3) 
 
Express the ratio of μt to μs: 
ts
st
s
s
t
t
s
t
l
l
l
l
m
m
m
m
==μ
μ 
 
Traveling Waves 
 
 
1497
Substituting in equation (3) yields: 
ts
st
ts
st
1
1
l
l
l
l
m
m
m
m
r
+
−
= 
 
Substitute numerical values and 
evaluate r: 
 
( )( )( )( )
( )( )( )( )
20.0
m 00.4kg 100.25
m 00.3kg 100.751
m 00.4kg 100.25
m 00.3kg 100.751
3
3
3
3
−=
×
×+
×
×−
=
−
−
−
−
r
 
 
Substitute for ts vv in equation (2) to 
obtain: 
ts
st
s
t 1
2
1
2
l
l
m
m+
=
+
=
μ
μτ 
 
Substitute numerical values and 
evaluateτ: 
 
( )( )( )( )
80.0
m 00.4kg 100.25
m 00.3kg 100.751
2
3
3
=
×
×+
=
−
−τ
 
 
Remarks: Because 1=+τr , we could have used this relationship to find τ. 
 
69 • [SSM] Consider a taut string with a mass per unit length μ1, carrying 
transverse wave pulses that are incident upon a point where the string connects to 
a second string with a mass per unit length μ2. (a) Show that if μ2 = μ1, then the 
reflection coefficient r equals zero and the transmission coefficient τ equals +1. 
(b) Show that if μ2 >> μ1, then r ≈ –1 and τ ≈ 0; and (c) if μ2 << μ1 then r ≈ +1 
and τ ≈ +2. 
 
Picture the Problem We can use the definitions of the reflection and 
transmission coefficients and the expression for the speed of waves on a string 
(Equation 15-3) to r and t in terms of the linear densities of the strings. 
 
 
 
Chapter 15 
 
 
1498 
(a) Use their definitions to express 
the reflection and transmission 
coefficients: 
 2
1
2
1
12
12
1
1
v
v
v
v
vv
vvr
+
−
=+
−= (1) 
and 
2
112
2
1
22
v
vvv
v
+
=+=τ (2) 
 
Use Equation 15-3 to express v2 and 
v1: 2
T
2 μ
Fv = and 
1
T
1 μ
Fv = 
 
Dividing the expression for v1 by the 
expression for v2 and simplifying 
yields: 
 1
2
2
T
1
T
2
1
μ
μ
μ
μ ==
F
F
v
v 
 
Substitute for 21 vv in equation (1) to 
obtain: 
1
2
1
2
1
1
μ
μ
μ
μ
+
−
=r (3) 
 
Substitute for 21 vv in equation (2) to 
obtain: 
1
21
2
μ
μτ +
= (4) 
 
If μ2 = μ1: 0
11
11 =+
−=r 
and 
1
11
2
1
2
1
2
=+=+
=
μ
μτ 
 
 
 
 
 
 
Traveling Waves 
 
 
1499
(b) From equations (1) and (2), if 
μ2 >> μ1 then v1 >> v2: 
 
1
1
1
12
12 −=−≈+
−=
v
v
vv
vvr 
and 
0
1
22
2
112
2 ≈
+
=+=
v
vvv
vτ 
 
(c) If μ2 << μ1, then v2 >> v1: 
 1
1
1
2
1
2
1
12
12 ≈
+
−
=+
−=
v
v
v
v
vv
vvr 
and 
2
1
22
2
112
2 ≈
+
=+=
v
vvv
vτ 
 
70 •• Verify the validity of ( ) 22121 τvvr += (Equation 15-36) by 
substituting the expressions for r and τ into it. 
 
Picture the Problem Making the indicated substitutions will lead us to the 
identity 1 = 1. 
 
The reflection and transmission 
coefficients are given by: 12
12
vv
vvr +
−= and
12
22
vv
v
+=τ 
 
Substituting into the equation given 
in the problem statement yields: 
 
2
12
2
2
1
2
12
12 21 ⎟⎟⎠
⎞
⎜⎜⎝
⎛
++⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−=
vv
v
v
v
vv
vv 
 
Simplify this expression algebraically to obtain: 
 
( )
( ) ( ) ( )
( )
( )
1
242
4
1 2
12
2
12
2
12
2
112
2
2
2
12
12
2
112
2
2
2
12
2
12
2
2
12
=
+
+=+
++=+
++−=+
⎟⎟⎠
⎞
⎜⎜⎝
⎛+−
=
vv
vv
vv
vvvv
vv
vvvvvv
vv
v
vvvv
 
 
71 ••• Consider a taut string that has a mass per unit length μ1 carrying 
transverse wave pulses of the form y = f(x – v1t) that are incident upon a point P 
where the string connects to a second string with mass per unit length μ2. Derive ( ) 22121 τvvr += by equating the power incident on point P to the power 
reflected at P plus the power transmitted at P. 
Chapter 15 
 
 
1500 
Picture the Problem Choose the direction of propagation of the incident pulse as 
the +x direction and let x = 0 at point P. Energy is conserved as the incident pulse 
is partially reflected and partially transmitted at point P. 
 
From the conservation of energy we 
have: 
 
trin PPP =+ (1) 
From Equation 15-20, the power 
transmitted in the direction of 
increasing x is given by: 
x
y
x
yFP ∂
∂
∂
∂−= T 
 
Substituting in equation (1) yields: 
 
⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂−=⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂−+⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂−
x
y
x
yF
x
y
x
yF
x
y
x
yF ttTrrTininT 
or, upon simplification, 
x
y
x
y
x
y
x
y
x
y
x
y
∂
∂
∂
∂=∂
∂
∂
∂+∂
∂
∂
∂ ttrrinin (2) 
 
Because the incident pulse is given 
by ( )tvxfy 1in −= , the reflected and 
transmitted pulses are given by: 
 
( )tvxrfy 1r −−= 
and 
[ ]⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= tvx
v
vfy 2
2
1
t τ 
 
Evaluating 
x
y
∂
∂ and
t
y
∂
∂ using the 
chain rule yields: 
 
xd
df
x
y
∂
∂=∂
∂ η
η and td
df
t
y
∂
∂=∂
∂ η
η 
where η is the argument of the wave 
function. 
 
For the transmitted pulse: [ ]
2
1
2
2
1t
v
v
d
dftvx
v
v
xd
df
x
y
ητητ =⎟⎟⎠
⎞
⎜⎜⎝
⎛ −∂
∂=∂
∂ 
and 
[ ] ( )
ητ
ητητ
d
dfv
v
d
dftvx
v
v
td
df
t
y
1
12
2
1t
−=
−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −∂
∂=∂
∂
 
 
Traveling Waves 
 
 
1501
 
For the reflected pulse: ( ) ηη d
dfrtvx
xd
dfr
x
y −=−−∂
∂=∂
∂
1
r 
and 
( ) ( )
η
ηη
d
dfrv
v
d
dfrtvx
td
dfr
t
y
1
11
r
−=
−=−−∂
∂=∂
∂
 
 
For the incident pulse: ( ) ηη d
dftvx
xd
df
x
y =−∂
∂=∂
∂
1
in 
and 
( ) ( )
η
ηη
d
dfv
v
d
dftvx
td
df
t
y
1
11
in
−=
−=−∂
∂=∂
∂
 
Substitute in equation (2) to obtain: 
 
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞⎜⎜⎝
⎛−+⎟⎟⎠
⎞⎜⎜⎝
⎛− ητητηηηη d
dfv
v
v
d
df
d
dfrv
d
dfr
d
dfv
d
df
1
2
1
11 
 
Simplifying and rearranging terms 
yields: 
2
2
121 τ
v
vr += 
 
The Doppler Effect 
 
72 • A sound source is moving at 80 m/s toward a stationary listener that is 
standing in still air. (a) Find the wavelength of the sound in the region between 
the source and the listener. (b) Find the frequency heard by the listener. 
 
Picture the Problem We can use Equation 15-38 (
s
s
f
uv ±=λ ) to find the 
wavelength of the sound between the source and the listener and Equation15-41a 
( s
s
r
r fuv
uvf ±
±= ) to find the frequency heard by the listener. 
 
(a) Apply Equation 15-38 to find λ: 
 
m32.1
s200
m/s80m/s343
1
s
s
s
s
=
−=−=±= −f
uv
f
uvλ
 
Chapter 15 
 
 
1502 
(b) Apply Equation 15-41a to obtain 
fr: 
( )
Hz261
s200
m/s80m/s343
m/s343
0
1
s
s
s
r
r
=
−=
−
±=±
±=
−
sfuv
vf
uv
uvf
 
 
73 • Consider the situation described in Problem 72 from the reference 
frame of the source. In this frame, the listener and the air are moving toward the 
source at 80 m/s
and the source is at rest. (a) At what speed, relative to the source, 
is the sound traveling in the region between the source and the listener? (b) Find 
the wavelength of the sound in the region between the source and the listener. 
(c) Find the frequency heard by the listener. 
 
Picture the Problem (a) In the reference frame of the source, the speed of sound 
from the source to the listener is reduced by the speed of the air. (b) We can find 
the wavelength of the sound in the region between the source and the listener 
from v = fλ. (c) Because the sound waves in the region between the source and the 
listener will be compressed by the motion of the listener, the frequency of the 
sound heard by the listener will be higher than the frequency emitted by the 
source and can be calculated using s
s
r
r fuv
uvf ±
±= (Equation 15-41a). 
 
(a) The speed of sound in the 
reference frame of the source is: 
 
m/s263
m/s80m/s433wind
=
−=−= uvv'
 
 
(b) Noting that the frequency is 
unchanged, express the wavelength 
of the sound: 
 
m32.1
s200
m/s263
1 === −f
v'λ 
 
(c) Apply Equation 15-41a to obtain: 
( )
Hz261
s200
m/s263
m/s80m/s263
0
1
s
r
s
s
r
r
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
⎟⎠
⎞⎜⎝
⎛
±
+=±
±=
−
f
v'
uv'f
uv'
uv'f
 
 
74 • A sound source is moving away from the stationary listener at 
80 m/s. (a) Find the wavelength of the sound waves in the region between the 
source and the listener. (b) Find the frequency heard by the listener. 
 
Traveling Waves 
 
 
1503
Picture the Problem We can use ( ) ss fuv ±=λ ( Equation 15-38) to find the 
wavelength of the sound in the region between the source and the listener 
and s
s
r
r fuv
uvf ±
±= (Equation 15-41a) to find the frequency heard by the listener. 
Because the sound waves in the region between the source and the listener will be 
spread out by the motion of the listener, the frequency of the sound heard by the 
listener will be lower than the frequency emitted by the source. 
 
(a) Because the source is moving 
away from the listener, use the 
positive sign in the numerator of 
Equation 15-38 to find the 
wavelength of the sound between the 
source and the listener: 
 
m12.2
s200
m/s80m/s343
1
s
s
s
s
=
+=
+=±=
−
f
uv
f
uvλ
 
(b) Because the listener is at rest and 
the source is receding, ur = 0 and the 
denominator of Equation 15-41a is 
the sum of the two speeds: ( )
Hz162
s200
m/s80m/s343
m/s343
0
1
s
s
s
s
r
r
=
+=
+
±=±
±=
−
f
uv
vf
uv
uvf
 
 
75 • The listener is moving 80 m/s from the stationary source that is in rest 
relative to the air. Find the frequency heard by the listener. 
 
Picture the Problem Because the listener is moving away from the source, we 
know that the frequency he/she will hear will be less than the frequency emitted 
by the source. We can use s
s
r
r fuv
uvf ±
±= (Equation 15-41a), with us = 0 and the 
minus sign in the numerator, to determine its value. 
 
Relate the frequency heard by the 
listener to that of the source: 
 ( )
Hz153
s200
m/s343
m/s80m/s343
0
1
s
r
s
s
r
r
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
⎟⎠
⎞⎜⎝
⎛
±
−=±
±=
−
f
v
uvf
uv
uvf
 
 
76 •• You have made the trek to observe a Space Shuttle landing. Near the 
end of its descent, the ship is traveling at Mach 2.50 at an altitude of 5000 m. 
(a) What is the angle that the shock wave makes with the line of flight of the 
shuttle? (b) How far are you from the shuttle by the time you hear its shock wave, 
Chapter 15 
 
 
1504 
assuming the shuttle maintains both a constant heading and a constant 5000-m 
altitude after flying directly over your head? 
 
Picture the Problem The diagram shows the position of the shuttle at time t after 
it was directly over your head (located at point P) Let u represent the speed of the 
shuttle and v the speed of sound. We can use trigonometry to determine the angle 
of the shock wave as well as the location of the shuttle x when you hear the shock 
wave. 
 
 
(a) Referring to the diagram, express 
θ in terms of v, u, and t: 5.2
11sin ===
vuut
vtθ 
 
Solving for θ yields: °=°=⎟⎠
⎞⎜⎝
⎛= − 6.2358.23
50.2
1sin 1θ 
 
(b) Using the diagram, relate θ to 
the altitude h of the shuttle and the 
distance x: 
 
x
h=θtan ⇒ θtan
hx = 
Substitute numerical values and 
evaluate x: 
km5.11
tan23.58
m5000 =°=x 
 
77 •• The SuperKamiokande neutrino detector in Japan is a water tank the 
size of a 14-story building. When neutrinos collide with electrons in water, most 
of their energy is transferred to the electrons. As a consequence, the electrons then 
fly off at speeds that approach c. The neutrino is counted by detecting the shock 
wave, called Cerenkov radiation, that is produced when the high-speed electrons 
travel through the water at speeds greater than the speed of light in water. If the 
maximum angle of the Cerenkov shock-wave cone is 48.75°, what is the speed of 
light in water? 
 
Picture the Problem The angle θ of the Cerenkov shock wave is related to the 
speed of light in water v and the speed of light in a vacuum c according to 
.sin cv=θ 
 
Traveling Waves 
 
 
1505
Relate the speed of light in water v to 
the angle of the Cerenkov cone: 
 
c
v=θsin ⇒ θsincv = 
Substitute numerical values and 
evaluate v: 
( )
m/s10254.2
75.48sinm/s10998.2
8
8
×=
°×=v
 
 
78 •• You are in charge of calibrating the radar guns for a local police 
department. One such device emits microwaves at a frequency of 2.00 GHz. 
During the trials, you have it arranged so that these waves are reflected from a car 
moving directly away from the stationary emitter. In this situation, you detect a 
frequency difference (between the received microwaves and the ones sent out) of 
293 Hz. Find the speed of the car. 
 
Picture the Problem Because the car is moving away from the stationary emitter 
at a speed ur, the frequency fr it receives will be less than the frequency emitted 
by the emitter. The microwaves reflected from the car, moving away from a 
stationary detector, will be of a still lower frequency fr′. We can use the Doppler 
shift equations to derive an expression for the speed of the car in terms of 
difference of these frequencies. 
 
Express the frequency fr received 
by the moving car in terms of fs, ur, 
and c: 
 
s
r
s
s
r
r 0
f
c
ucf
uc
ucf ⎟⎠
⎞⎜⎝
⎛
±
−=±
±= (1) 
The waves reflected by the car are 
like waves re-emitted by a source 
moving away from the radar gun: 
 
r
s
r
s
r
r fuc
cf
uc
uc'f ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+=±
±= (2) 
 
Substitute equation (1) in equation (2) 
to eliminate fr: 
 
s
1
rr
s
r
r
s
s
r
s
r
s
r
11 
1
1
f
c
u
c
uf
c
u
c
u
f
uc
ucf
c
uc
uc
c'f
−
⎟⎠
⎞⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛ −=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
+
−
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−=⎟⎠
⎞⎜⎝
⎛ −⎟⎟⎠
⎞
⎜⎜⎝
⎛
+=
 
 
Because cu r << , c
u
c
u r
1
r 11 −≈⎟⎠
⎞⎜⎝
⎛ +
−
. 
Substituting for 
1
r1
−
⎟⎠
⎞⎜⎝
⎛ +
c
u and 
simplifying yields: 
s
2
r
s
rr
r
1
11 
f
c
u
f
c
u
c
u'f
⎟⎠
⎞⎜⎝
⎛ −=
⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ −≈
 
Chapter 15 
 
 
1506 
 
Because cu r << , c
u
c
u r
2
r 211 −≈⎟⎠
⎞⎜⎝
⎛ − . 
Substituting for 
2
r1 ⎟⎠
⎞⎜⎝
⎛ −
c
u gives: 
 
s
r
r
21 f
c
u'f ⎟⎠
⎞⎜⎝
⎛ −≈ 
The frequency difference detected at 
the source is given by: 
 
s
r
s
r
srs
2
21Δ
f
c
u
f
c
uf'fff
=
⎟⎠
⎞⎜⎝
⎛ −−=−=
 
 
Solving for ur yields:
f
f
cu Δ
2 s
r = 
 
Substitute numerical values and 
evaluate ur: ( ) ( )
km/h1.79
h
s3600
m10
km1
s
m0.22
Hz293
GHz2.002
m/s10998.2
3
8
r
=
××=
×=u
 
 
79 •• [SSM] The Doppler effect is routinely used to measure the speed of 
winds in storm systems. As the manager of a weather monitoring station in the 
Midwest, you are using a Doppler radar system that has a frequency of 625 MHz 
to bounce a radar pulse off of the raindrops in a swirling thunderstorm system 
50 km away. You measure the reflected radar pulse to be up-shifted in frequency 
by 325 Hz. Assuming the wind is headed directly toward you, how fast are the 
winds in the storm system moving? Hint: The radar system can only measure the 
component of the wind velocity along its ″line of sight.″ 
 
Picture the Problem Because the wind is moving toward the weather station 
(radar device), the frequency fr the raindrops receive will be greater than the 
frequency emitted by the radar device. The radar waves reflected from the 
raindrops, moving toward the stationary detector at the weather station, will be of 
a still higher frequency fr′. We can use the Doppler shift equations to derive an 
expression for the radial speed u of the wind in terms of difference of these 
frequencies. 
 
Use Equation 15-41a to express the 
frequency fr received by the 
raindrops in terms of fs, ur, and c: 
s
r
s
s
r
r fc
ucf
uc
ucf ⎟⎠
⎞⎜⎝
⎛ +=±
±= (1) 
 
Traveling Waves 
 
 
1507
The waves reflected by the drops are 
like waves re-emitted by a source 
moving toward the source at the 
weather station: 
 
r
s
r
s
r
r fuc
cf
uc
uc'f ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−=±
±= (2) 
 
Substitute equation (1) in equation (2) 
to eliminate fr: 
 
s
1
rr
s
r
r
s
s
r
s
r
s
r
11 
1
1
f
c
u
c
uf
c
u
c
u
f
uc
ucf
c
uc
uc
c'f
−
⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ +=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
−
+
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
+=⎟⎠
⎞⎜⎝
⎛ +⎟⎟⎠
⎞
⎜⎜⎝
⎛
−=
 
 
Because cu r << , c
u
c
u r
1
r 11 +≈⎟⎠
⎞⎜⎝
⎛ −
−
. 
Substituting for 
1
r1
−
⎟⎠
⎞⎜⎝
⎛ −
c
u and 
simplifying yields: 
 
s
2
r
s
rr
r
1
11 
f
c
u
f
c
u
c
u'f
⎟⎠
⎞⎜⎝
⎛ +=
⎟⎠
⎞⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛ +≈
 
Because cu r << , c
u
c
u r
2
r 211 +≈⎟⎠
⎞⎜⎝
⎛ + . 
Substituting for 
2
r1 ⎟⎠
⎞⎜⎝
⎛ +
c
u gives: 
 
s
r
r
21 f
c
u'f ⎟⎠
⎞⎜⎝
⎛ +≈ 
The frequency difference detected at 
the source is: 
 
s
r
ss
r
sr
2
21Δ
f
c
u
ff
c
uf'ff
=
−⎟⎠
⎞⎜⎝
⎛ +=−=
 
 
Solving for ur yields: f
f
cu Δ
2 s
r = 
 
Substitute numerical values and 
evaluate ur: ( ) ( )
mi/h174
m/s4470.0
mi/h1m/s 95.77
Hz253
MHz2562
m/s10998.2 8
r
=
×=
×=u
 
 
Chapter 15 
 
 
1508 
80 •• A stationary destroyer is equipped with sonar that sends out 40-MHz-
pulses of sound. The destroyer receives pulses back from a submarine directly 
below with a time delay of 80 ms at a frequency of 39.958 MHz. If the speed of 
sound in seawater is 1.54 km/s, (a) what is the depth of the submarine? (b) What 
is its vertical speed? 
 
Picture the Problem Let the depth of the submarine be represented by D and its 
vertical speed by u. The submarine acts as both a receiver and source. We can 
apply the definition of average speed to determine the depth of the submarine and 
use the Doppler shift equations to derive an expression for the vertical speed of 
the submarine in terms of the frequency difference. The frequency received is 
lower than the frequency of the source, so the sub is descending. 
 
(a) Using the definition of average 
speed, relate the depth of the 
submarine to the time delay between 
the transmitted and reflected pulses: 
 
tvD Δ=2 ⇒ tvD Δ= 21 
 
Substitute numerical values and 
evaluate D: 
 
( )( ) m62ms80km/s1.5421 ==D 
 
(b) Use s
s
r
r fuv
uvf ⎟⎟⎠
⎞
⎜⎜⎝
⎛
±
±= to express 
the frequency fsub received by the 
submarine: 
 
0sub fv
uvf ⎟⎠
⎞⎜⎝
⎛ −= (1) 
where u is the vertical speed of the 
submarine. 
Use s
s
r
r fuv
uvf ⎟⎟⎠
⎞
⎜⎜⎝
⎛
±
±= to express 
the frequency fr′ received by the 
destroyer: 
subr fuv
v'f ⎟⎠
⎞⎜⎝
⎛
+= (2) 
 
 
Substitute equation (1) in equation 
(2) to eliminate fsub: 
 
0r ' fuv
uvf ⎟⎠
⎞⎜⎝
⎛
+
−= ⇒ v
ff
ffu '
'
r0
r0
+
−= 
 
Substitute numerical values and evaluate u: 
 
( ) m/s81.0km/s 54.1
MHz 958.39MHz 000.40
MHz 958.39MHz 000.40 =⎟⎠
⎞⎜⎝
⎛
+
−=u 
and so, because u is positive, the sub is descending. 
 
Traveling Waves 
 
 
1509
81 •• A police radar unit transmits microwaves of frequency 3.00 × 1010 Hz, 
and their speed in air is 3.00 × 108 m/s. Suppose a car is receding from the 
stationary police car at a speed of 140 km/h. (a) What is the frequency difference 
between the transmitted signal and the signal received from the receding car? 
(b) Suppose the police car is, instead, moving at a speed of 60 km/h in the same 
direction as the other vehicle. What is the difference in frequency between the 
emitted and the reflected signals? 
 
Picture the Problem The radar wave strikes the speeding car at frequency fr . 
This frequency is less than fs because the car is moving away from the source. The 
frequency shift is given by Equation 15-42 (the low-speed, relative to light, 
approximation). The car then acts as a moving source emitting waves of 
frequency fr. The police car detects waves of frequency fr′ < fr because the source 
(the speeding car) is moving away from the police car. The total frequency shift 
is the sum of the two frequency shifts. 
 
(a) Express the frequency difference 
Δf as the sum of the frequency 
difference sr1Δ fff −= and the 
frequency difference rr2Δ f'ff −= : 
 
21 ΔΔΔ fff += (1) 
 
Using Equation 15-42, substitute for 
the frequency differences in equation 
(1): 
 
( )rsrsΔ ffc
uf
c
uf
c
uf +−=−−= (2) 
where rs uuu ±= is the speed of the 
source relative to the receiver. 
 
Apply Equation 15-42 to 1Δf to 
obtain: 
 
c
u
f
ff
f
f −=−=
s
sr
s
1Δ 
where we’ve used the minus sign 
because we know the frequency 
difference is a downshift. 
 
Solving for fr yields: 
 sr
1 f
c
uf ⎟⎠
⎞⎜⎝
⎛ −= 
 
Substitute for fr in equation (2) and 
simplify to obtain: 
s
ss
2
1Δ
f
c
u
c
u
f
c
uf
c
uf
⎟⎠
⎞⎜⎝
⎛ −−=
⎟⎟⎠
⎞⎜⎜⎝
⎛ ⎟⎠
⎞⎜⎝
⎛ −+−=
 
 
Chapter 15 
 
 
1510 
Because u/c is negligible compared 
to 2: c
uff s2Δ −≈ 
 
Substitute numerical values and evaluate Δf: 
 
( ) kHz 78.7
m/s1000.3
s3600
h1
h
km140
Hz103.002Δ 8
10 −=×
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×
×−≈f 
 
(b) Use the result derived in (a) with u equal to the difference in the speeds of the 
car and the police cruiser to obtain: 
 
( ) kHz .44
m/s1000.3
s3600
h1
h
km 60
h
km140
Hz103.002Δ 8
10 −=×
⎟⎟⎠
⎞
⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ −
×−≈f 
 
82 •• In modern medicine, the Doppler effect is routinely used to measure 
the rate and direction of blood flow in arteries and veins. High frequency 
″ultrasound″ (sound at frequencies above the human hearing range) is typically 
employed. Suppose you are in charge of measuring the blood flow in a vein 
(located in the lower leg of an older patient) that returns blood upward to the 
heart. Her varicose veins indicate that perhaps the one-way valves in the vein are 
not working properly and that the blood is ″pooling″ in the veins and perhaps 
even that the blood
flow is backward toward her feet. Employing sound with a 
frequency of 50.0 kHz, you point the sound source from above her thigh region 
down towards her feet and measure the sound reflected from that vein area to be 
lower than 50.0 kHz. (a) Was your diagnosis of the valve condition correct? If so, 
explain. (b) Estimate the instrument’s frequency difference capability to enable 
you to measure speeds down to 1.00 mm/s. Take the speed of sound in flesh to be 
the same as that in water, 1500 m/s. 
 
Picture the Problem (a) Whether your diagnosis was correct depends on whether 
the signal reflected from the blood is upshifted or downshifted. (b) Applying the 
Doppler shift equations with the source stationary and the receiver moving 
initially and then a second time with the source moving and the receiver 
stationary will allow us to estimate the instrument’s frequency difference 
capability. 
 
(a) Because the received sound frequency is less than the frequency that was sent 
out, the blood it reflected from must have been moving away from the source, or 
toward the feet of the patient. The blood is flowing the wrong way and your 
diagnosis is correct. 
 
Traveling Waves 
 
 
1511
(b) The blood receives a frequency 
that is downshifted according to: 
 
s
r
s
r
s
s
r
r
1
0
f
v
u
f
v
uvf
uv
uvf
⎟⎠
⎞⎜⎝
⎛ −=
⎟⎠
⎞⎜⎝
⎛
±
−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
±
±=
 (1) 
 
The frequency given by equation (1) 
is the frequency emitted by the 
moving blood back toward the 
receiver. This results in a second 
downshift: 
 
''
'''
f
v
uf
v
u
f
uv
vf
uv
uvf
s
s
s
1
s
s
s
s
s
r
r
11
0
⎟⎠
⎞⎜⎝
⎛ −≈⎟⎠
⎞⎜⎝
⎛ +=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
±=⎟⎟⎠
⎞
⎜⎜⎝
⎛
±
±=
− (2) 
The source frequency in equation (2) 
is the received frequency given by 
equation (1). Note that the receiver 
and emitter speeds are the same … 
the speed of the blood. Let this speed 
be u to obtain: 
 
ssr 2111 fv
uf
v
u
v
uf ' ⎟⎠
⎞⎜⎝
⎛ −≈⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ −≈ 
Express the magnitude of the 
difference in frequencies: 
 
s
ssrs
2
21Δ
f
v
u
f
v
uffff '
=
⎟⎠
⎞⎜⎝
⎛ −−≈−=
 
 
Substitute numerical values and 
evaluate Δf: ( )
Hz .0330
kHz 0.50
m/s 1500
m/s .0012Δ
=
⎟⎠
⎞⎜⎝
⎛≈f
 
 
83 •• [SSM] A sound source of frequency fs moves with speed us relative 
to still air toward a receiver who is moving away with speed ur relative to still air 
away from the source. (a) Write an expression for the received frequency f′r. (b) 
Use the result that (1 – x)–1 ≈ 1 + x to show that if both us and ur are small 
compared to v, then the received frequency is approximately 
 
s
rel1' f
v
ufr ⎟⎠
⎞⎜⎝
⎛ += 
 
 where urel = us – ur is the velocity of the source relative to the receiver. 
 
 
 
 
Chapter 15 
 
 
1512 
Picture the Problem The received and transmitted frequencies are related 
through s
s
r
r fuv
uvf ±
±= (Equation 15-41a), where the variables have the meanings 
given in the problem statement. Because the source and receiver are moving in the 
same direction, we use the minus signs in both the numerator and denominator. 
 
(a) Relate the received frequency fr 
to the frequency fs of the source: 
 
s
1
sr
s
s
r
s
s
r
r
11
1
1
f
v
u
v
u
f
v
u
v
u
f
uv
uvf
−
⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ −=
−
−
=±
±=
 
 
(b) Expand ( ) 1s1 −− vu binomially 
and discard the higher-order terms: 
 
v
u
v
u s
1
s 11 +≈⎟⎠
⎞⎜⎝
⎛ −
−
 
Substitute to obtain: 
 
s
rs
s
srrs
s
sr
r
1
1
11
f
v
uu
f
v
u
v
u
v
u
v
u
f
v
u
v
uf
⎟⎠
⎞⎜⎝
⎛ −+≈
⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛−−+=
⎟⎠
⎞⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛ −=
 
because both us and ur are small 
compared to v. 
 
Because urel = us − ur 
s
rel
r 1 fv
uf ' ⎟⎠
⎞⎜⎝
⎛ +≈ 
 
84 ••• To study the Doppler shift on your own, you take an electronic tone 
generating device that is set to a frequency of middle C (262Hz) to a campus 
wishing well known as ″The Abyss. ″ When you hold the device at arm’s length 
(1.00 m), you measure its intensity level to be 80.0 dB. You then drop the tuner 
down the hole, listening to its sound as it falls. After the tuner has fallen for 
5.50 s, what frequency do you hear? 
 
 
 
 
 
 
Traveling Waves 
 
 
1513
Picture the Problem As the tuner falls, its speed increases and so the frequency 
you hear is Doppler shifted. Our concern, however, is with the frequency you hear 
5.50 s after you’ve dropped the tuner. The sound that you hear at this time was 
emitted sometime before the tuner had fallen for 5.50 s. Hence we must answer 
the question ″At what time was the sound emitted that you heard 5.50 s after 
dropping the tuner?″ 
 
Apply Equation 15-41a to express 
the frequency you’ll hear after 5.50 
s: 
 
s
fall
s
s
r
r fvv
vf
uv
uvf +=±
±= (1) 
Relate the elapsed time of 5.50 s to 
the time required for the tuner to fall 
and the time for the sound to return 
to you: 
 
returnfalltot ΔΔΔ ttt += (2) 
Use a constant-acceleration equation 
to relate the time-of-fall to the fall 
distance: 
 
( )2fall21 ΔΔ tgy = ⇒ g
yt 2ΔΔ fall = 
The return time depends on how far 
the tuner has fallen and on the speed 
of sound in air: 
 
v
yt ΔΔ return = ⇒ returnΔΔ tvy = 
Substituting for Δy in the expression 
for Δtfall yields: g
tvt returnfall
Δ2
Δ = 
 
Square both sides of the equation, 
use equation (2) to substitute for 
Δtreturn, and write the result explicitly 
as a quadratic equation to obtain: 
 
( ) ( ) 0Δ2Δ2Δ totfall2fall =−+ tg
vt
g
vt 
Substitute numerical values to obtain: 
 
( ) ( ) ( ) ( )( ) 0
m/s81.9
s 50.5m/s 3432
Δ
m/s81.9
m/s 3432
Δ 2fall2
2
fall =−+ tt 
or 
( ) ( )( ) 0s 6.384Δs 93.69Δ 2fall2fall =−+ tt 
 
 
 
Chapter 15 
 
 
1514 
Use your graphing calculator or the 
quadratic formula to solve this 
equation for its positive root: 
 
s 124.5Δ fall =t 
Because fallfall Δtgv = , equation (1) 
becomes: 
s
fall
r Δ
f
tgv
vf += 
 
Substitute numerical values and evaluate fr: 
 
( )( ) ( ) Hz 229Hz 262s 124.5m/s 81.9m/s 343 m/s 343 2r =+=f 
 
85 •• You are in a hot-air balloon carried along by a 36-km/h wind and have 
a sound source with you that emits a sound of 800 Hz as it approaches a tall 
building. (a) What is the frequency of the sound heard by an observer at the 
window of this building? (b) What is the frequency of the reflected sound heard 
by you? 
 
Picture the Problem The simplest way to approach this problem is to transform 
to a reference frame in which the balloon is at rest. In that reference frame, the 
speed of sound is v = 343 m/s, and ur = 36 km/h = 10 m/s. Then, we can use the 
equations for a moving receiver and a moving source to find the frequencies heard 
at the window and on the balloon. 
 
(a) Use Equation 15-41a to express 
the received frequency in terms of 
the frequency of the source: 
 
s
r
s
r
s
s
r
s
s
r
r
1
01
1
1
1
f
v
u
f
v
v
u
f
v
u
v
u
f
uv
uvf
⎟⎠
⎞⎜⎝
⎛ +=
±
+
=
±
±
=±
±=
 
 
Substitute numerical values and 
evaluate fr: ( )
kHz82.0
Hz823Hz800
m/s343
m/s101r
=
=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=f
 
 
Traveling Waves 
 
 
1515
 
(b) Treating the tall building as a moving source, express the frequency of the 
reflected sound heard by a person riding in the balloon: 
 
r
s
r
s
r
s
r
r
s
r
r
1
1
1
01
1
1
f
v
uf
v
u
vf
v
u
v
u
f
uv
uv'f
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
−
=
−
±
=
±
±
=±
±= 
 
Substitute numerical values and 
evaluate fr′: ( )
kHz0.85
Hz823
m/s343
m/s101
1
r
=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
−
='f
 
 
86 •• A car is approaching a reflecting wall. A stationary observer behind 
the car hears a sound of frequency 745 Hz from the car horn and a sound of 
frequency 863 Hz from the wall. (a) How fast is the car traveling? (b) What is the 
frequency of the car horn? (c) What frequency does the car driver hear reflected 
from the wall? 
 
Picture the Problem We can relate the frequencies fr and fr′ heard by the 
stationary observer behind the car to the speed of the car ur and the frequency of 
the car’s horn fs. Dividing these equations will eliminate the frequency of the car’s 
horn and allow us to solve for the speed of the car. We can then substitute to find 
the frequency of the car’s horn. We can find the frequency heard by the driver as 
a moving receiver by relating this frequency to the frequency reflected from the 
wall. 
 
(a) Use Equation 15-41a to relate the 
frequency heard by the observer 
directly from the car’s horn to the 
speed of the car: 
 
s
s
s
s
s
s
r
s
s
r
r
1
1
1
01
1
1
f
v
u
f
v
u
vf
v
u
v
u
f
uv
uvf
+
=
+
±
=
±
±
=±
±=
(1) 
 
Chapter 15 
 
 
1516 
Relate the frequency reflected from 
the wall to the speed of the car: 
 
s
s
s
s
s
s
r
r
1
1
1
01
f
v
u
f
v
u
vf
uv
uv'f
−
=
−
±
=±
±=
 (2) 
 
Divide equation (2) by equation (1) 
to obtain: 
v
u
v
u
f
'f
s
s
r
r
1
1
−
+
= ⇒ v
'f'f
f'fu
rr
rr
+
−= 
 
Substitute numerical values and 
evaluate us: 
( )
km/h6.90
h
s3600
m10
km1
s
m17.25
m/s343
Hz745Hz863
Hz745Hz863
3
s
=
××=
+
−=u
 
 
(b) Solve equation (1) for fs to 
obtain: r
r
s 1 fv
uf ⎟⎠
⎞⎜⎝
⎛ += 
 
Substitute numerical values and 
evaluate fs: ( )
Hz800
Hz745
m/s343
m/s25.171s
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=f
 
 
(c) The driver is a moving receiver 
and so we can relate the frequency 
heard by the driver to the frequency 
reflected by the wall (the frequency 
heard by the stationary observer): 
'f
v
uf rrdriver 1 ⎟⎠
⎞⎜⎝
⎛ += 
 
Substitute numerical values and 
evaluate fdriver: ( )
Hz926
H863
m/s343
m/s25.171driver
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ += zf
 
 
 
 
Traveling Waves 
 
 
1517
87 •• The driver of a car traveling at 100 km/h toward a vertical wall briefly 
sounds the horn. Exactly 1.00 s later she hears the echo and notes that its 
frequency is 840 Hz. How far from the wall was the car when the driver sounded 
the horn and what is the frequency of the horn? 
 
Picture the Problem Let t = 0 when the driver sounds her horn and let the 
distance to the wall at that instant be d. The received and transmitted frequencies 
are related through s
s
r
r fuv
uvf ±
±= (Equation 15-41a). Solving this equation for fs 
will allow us to determine the frequency of the car horn. We can use the total 
distance the sound travels (car-to-wall plus wall-back to-car … now closer to the 
wall) to determine the distance to the wall when the horn was briefly sounded. 
 
Use Equation 15-41a to relate the 
frequency heard by the driver to 
her speed and to the frequency of 
her horn: 
 
s
s
r
s
s
r
s
s
r
r
1
1
1
1
f
v
u
v
u
f
v
u
v
u
f
uv
uvf
−
+
=
±
±
=±
±= 
 
Solving for fs yields: 
r
r
s
s
1
1
f
v
u
v
u
f
+
−
= 
 
Substitute numerical values and evaluate fs: 
 
( ) ( ) Hz 714Hz840
m/s343
m/s78.271
m/s343
m/s78.271
Hz840
m/s343
s3600
h1
h
km100
1
m/s343
s3600
h1
h
km100
1
s =
+
−
=
×
+
×
−
=f 
 
Relate the distance d to the wall at 
t = 0 to the distance she travels in 
time Δt = 1 s, her speed us, and the 
speed of sound v: 
 
( ) tvtudd ΔΔs =−+ ⇒ ( ) tvud Δs21 += 
Substitute numerical values and 
evaluate d: 
( )( )
m185
s1.00m/s343m/s27.782
1
=
+=d
 
 
Chapter 15 
 
 
1518 
88 •• You are on a transatlantic flight traveling due west at 800 km/h. An 
experimental plane flying at Mach 1.6 and 3.0 km to the north of your plane is 
also on an east-to-west course. What is the distance between the two planes when 
you hear the sonic boom from the experimental plane? 
 
Picture the Problem You’ll hear the sonic boom when the surface of its cone 
reaches your plane. In the following diagram, the experimental plane is at C and 
your plane is at P. The distance h = 3.0 km. The distance between the planes 
when you hear the sonic boom is d. We can use trigonometry to determine the 
angle of the shock wave as well as the separation of the planes when you hear the 
sonic boom. 
 
 
Using the Pythagorean theorem, 
relate the separation of the planes d, 
to the distance h and the angle θ : 
 
θ2222 cosdhd += ⇒ θ2cos1
1
−= hd 
Express θ in terms of v, u, and t: 
6.1
11sin ===
vuut
vtθ 
so 
°=⎟⎠
⎞⎜⎝
⎛= − 7.38
6.1
1sin 1θ 
 
Substitute numerical values and 
evaluate d: ( ) km8.47.38cos1
1km0.3 2 =°−=d 
 
89 ••• The Hubble space telescope has been used to determine the existence 
of planets orbiting distant stars. The planet orbiting the star will cause the star to 
″wobble″ with the same period as the planet’s orbit. Because of wobble, light 
from the star will be Doppler-shifted up and down periodically. Estimate the 
maximum and minimum wavelengths of light of nominal wavelength 500 nm 
emitted by the Sun that is Doppler-shifted by the motion of the Sun due to the 
planet Jupiter. 
 
 
 
Traveling Waves 
 
 
1519
Picture the Problem The Sun and Jupiter orbit about their effective mass located 
at their common center of mass. We can apply Newton’s second law to the Sun to 
obtain an expression for its orbital speed about the Sun-Jupiter center of mass and 
then use this speed in the Doppler shift equation to estimate the maximum and 
minimum wavelengths resulting from the Jupiter-induced motion of the Sun. 
 
Letting v be the orbital speed of the 
sun about the center of mass of the 
Sun-Jupiter system, express the 
Doppler shift of the light due to this 
motion when the Sun is approaching 
Earth: 
 
c
v
c
v
c
c
v
c
v
f
'
cf'
−
+
=
−
+
==
1
1
1
1
λλ 
 
Solving for λ′ and simplifying 
yields: 
2121
1
11
11
1
1
−
−
⎟⎠
⎞⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛ −=
⎟⎠
⎞⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛ −=
+
−
=
c
v
c
v
c
v
c
v
c
v
c
v
'
λ
λλλ
 
 
Because v << c, we can expand 
21
1 ⎟⎠
⎞⎜⎝
⎛ −
c
v and 
21
1
−
⎟⎠
⎞⎜⎝
⎛ +
c
v 
binomially to obtain: 
 
c
v
c
v
2
11
21
−≈⎟⎠
⎞⎜⎝
⎛ − 
and 
c
v
c
v
2
11
21
−≈⎟⎠
⎞⎜⎝
⎛ +
−
 
 
Substitute for 
21
1 ⎟⎠
⎞⎜⎝
⎛ −
c
v and 
21
1
−
⎟⎠
⎞⎜⎝
⎛ +
c
v to obtain: 
 
c
v
c
v
c
v
c
v
−≈⎟⎠
⎞⎜⎝
⎛ −=
+
−
1
2
1
1
1 2
 
 
When the Sun is receding from 
Earth: 
 c
v
c
v
c
v
c
v
+≈⎟⎠
⎞⎜⎝
⎛ +=
−
+
1
2
1
1
1 2
 
 
Hence the motion of the Sun will 
give an observed Doppler shift of: ⎟⎠
⎞⎜⎝
⎛ ±≈
c
v' 1λλ (1) 
 
Chapter 15 
 
 
1520 
Apply Newton’s second law to the 
Sun: 
 
cm
2
S2
cm
effS
r
vM
r
MGM = ⇒
cm
eff
r
GMv = 
 
Measured from the center of the Sun, 
the distance to the center of mass of 
the Sun-Jupiter system is: 
 
( )
Js
JJ-s
Js
JJ-sS
cm
0
MM
Mr
MM
MrMr +=+
+= 
 
The effective mass is related to the 
masses of the Sun and Jupiter 
according to: 
 
Jseff
111
MMM
+= ⇒
Js
Js
eff MM
MMM += 
 
Substitute for Meff and rcm to obtain: 
 
J-s
s
Js
JJ-s
Js
Js
r
GM
MM
Mr
MM
MMG
v =
+
+= 
 
Using m1078.7 11J-s ×=r as the mean orbital radius of Jupiter, substitute 
numerical values and evaluate v: 
 ( )( ) m/s10306.1
m1078.7
kg1099.1kg/mN10673.6 4
11
302211
×=×
×⋅×=
−
v 
 
Substitute numerical values in 
equation (1) to obtain: ( )
( )( )5
8
4
1036.41nm500
m/s10998.2
m/s10306.11nm500'
−×±=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
×
×±≈λ
 
 
The maximum and minimum 
wavelengths are: 
nm 02.500max =λ 
and 
nm 98.499min =λ 
 
General Problems 
 
90 • At time t = 0, the shape of a wave pulse on a string is given by the 
function ( ) ( )( )223 m 2.00/m 120.00, xxy += , where x is in meters. (a) Sketch 
y(x, 0) versus x. (b) Give the wave function y(x,t) at a general time t if the pulse is 
moving in the +x direction with a speed of 10.0 m/s and if the pulse is moving in 
the −x direction with a speed of 10.0 m/s. 
 
 
Traveling Waves 
 
 
1521
Picture the Problem The equation of a wave traveling in the +x direction is of 
the form ( ) )( vtxfx,ty −= and that of a wave traveling in the −x direction 
is ( ) )( vtxfx,ty += . 
 
(a) The pulse at t = 0 shown below was plotted using a spreadsheet program: 
0.000
0.005
0.010
0.015
0.020
0.025
0.030
0.035
-6 -4 -2 0 2 4 6
x (m)
y(
x
,0
) (
m
)
 
 
(b) If the pulse is moving in the +x 
direction, the wave function must be 
of the form: 
 
( ) ( )[ ]txfvtxfx,ty m/s0.10)( −=−= 
because v = 10.0 m/s 
 
Replace x with ( )tx m/s0.10− to 
obtain: ( ) ( ) ( )[ ]22
3
m/s0.10m00.2
m120.0
tx
x,ty −+=
 
 
If the pulse is moving in the −x 
direction, the wave function must 
be of the form: 
 
( ) ( )[ ]txfvtxfx,ty m/s0.10)( +=+= 
because v = 10.0 m/s 
 
Replace x with ( )tx m/s0.10+ to 
obtain: ( ) ( ) ( )[ ]22
3
m/s0.10m00.2
m120.0,
tx
txy ++= 
 
 
 
Chapter 15 
 
 
1522 
91 • [SSM] A whistle that has a frequency of 500 Hz moves in a circle of 
radius 1.00 m at 3.00 rev/s. What are the maximum and minimum frequencies 
heard by a stationary listener in the plane of the circle and 5.00 m away from its 
center? 
 
Picture the Problem The pictorial 
representation depicts the whistle 
traveling in a circular path of radius 
r = 1.00 m. The stationary listener will 
hear the maximum frequency when the 
whistle is at point 1 and the minimum 
frequency when it is at point 2. These 
maximum and minimum frequencies 
are determined by f0 and the tangential 
speed us = 2π r/T. We can relate the 
frequencies heard at point P to the 
speed of the approaching whistle at 
point 1 and the speed of the receding 
whistle at point 2. 
su
m 00.1=r
m 00.5
1 2
P
su
 
 
Relate the frequency heard at point P 
to the speed of the approaching 
whistle at point 1: 
 
s
s
max
1
1 f
v
uf −
= 
Because ωru =s : 
smax
1
1 f
v
rf ω−
= 
 
Substitute numerical values and evaluate fmax: 
 
( )
( ) Hz529Hz500
m/s343
rev
rad2
s
rev3.00m00.1
1
1
max =
⎟⎠
⎞⎜⎝
⎛ ×
−
=
π
f 
 
Relate the frequency heard at point P 
to the speed of the receding whistle 
at point 2: 
 
s
s
min
1
1 f
v
uf +
= 
Traveling Waves 
 
 
1523
 
Substitute numerical values and evaluate fmin: 
 
( )
( ) Hz474Hz500
m/s343
rev
rad2
s
rev3.00m00.1
1
1
min =
⎟⎠
⎞⎜⎝
⎛ ×
+
=
π
f 
 
92 • Ocean waves move toward the beach with a speed of 8.90 m/s and a 
crest-to-crest separation of 15.0 m. You are in a small boat anchored off shore. 
(a) At what frequency do the wave crests reach you boat? (b) You now lift anchor 
and head out to sea at a speed of 15.0 m/s. At what frequency do the wave crests 
reach your boat now? 
 
Picture the Problem (a) The crest-to-crest separation of the waves is their 
wavelength. We can find the frequency of the waves from v = fλ. (b) When you 
lift anchor and head out to sea you’ll become a moving receiver and we can 
apply s
s
r
r fuv
uvf ±
±= to calculate the frequency you’ll observe. 
 
(a) The frequency of the ocean 
waves is the ratio of their speed to 
their wavelength: 
 
Hz0.593
m15.0
m/s8.90
0 === λ
vf 
 
(b) Express the frequency of the 
waves in terms of their speed and the 
speed of a moving receiver: 
 
s
r
s
r
s
s
r
r 1 fv
uf
v
uvf
uv
uvf ⎟⎠
⎞⎜⎝
⎛ +=+=±
±= 
 
Substitute numerical values and 
evaluate fr: ( )
Hz1.59
Hz0.593
m/s8.90
m/s15.01r
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=f
 
 
93 •• A 12.0-m-long wire that has an 85.0-g mass is stretched under a 
tension of 180 N. A pulse is generated at the left end of the wire, and 25.0 ms 
later a second pulse is generated at the right end of the wire. Where do the pulses 
first meet? 
 
 
 
 
 
Chapter 15 
 
 
1524 
Picture the Problem Let t be the time of travel of the left-hand pulse and the 
subscripts L and R refer to the pulse coming from the left and right, respectively. 
Because the pulse traveling from the right starts later than the pulse from the left, 
its travel time is t − Δt, where Δt = 25.0 ms. Both pulses travel at the same speed 
and the sum of the distances they travel is 12.0 m. 
 
Express the total distance the two 
pulses travel: 
 
( )ttvvtddd Δ−+=+= RL 
Solve for vt to obtain: 
 
( )tvdvt Δ+= 21 
The speed of the pulse is given by: 
Lm
FFv TT == μ 
 
Substituting for v yields: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ += t
Lm
Fdvt ΔT21 
Substitute numerical values and evaluate vt: 
 
( )( ) ( ) m99.7s1025.0
kg0.085
m12.0N180m12.0 321 =⎥⎦
⎤⎢⎣
⎡ ×+= −vt 
from the left end of the wire. 
 
94 •• You are parked on the shoulder of a highway. Find the speed of a car 
in which the tone of the car’s horn drops by 10 percent as it passes you. (In other 
words, the total drop in frequency between the ″approach″ value and the 
″recession″ value is 10%.) 
 
Picture the Problem Let the frequency of the car’s horn be fs, the frequency you 
hear as the car approaches be fr, and the frequency you hear as the car recedes be 
fr′. We can use s
s
r
r fuv
uvf ±
±= to express the frequencies heard as the car 
approaches and recedes and then use these frequencies to express the fractional 
change in frequency as the car passes you. 
 
Express the fractional change in 
frequency as the car passes you: 
 
10.0Δ
r
=
f
f
 
 
Relate the frequency heard as the car 
approaches to the speed of the car: 
 
s
s
s
s
s
s
r
r
1
10 f
v
ufuv
vf
uv
uvf
−
=−
±=±
±= 
Traveling Waves 
 
 
1525
Express the frequency heard as the 
car recedes in terms of the speed of 
the car: 
s
s
r
1
1 f
v
u'f +
= 
 
Divide the second of these frequency 
equations by the first to obtain: 
v
u
v
u
f
'f
s
s
r
r
1
1
+
−
= 
and 
10.0
1
1
1Δ
s
s
rr
r
r
r =
+
−
−==−
v
u
v
u
f
f
f
'f
f
f 
 
Solving for us yields: vu
9.1
10.0
s = 
 
Substitute numerical values and 
evaluate us: 
( )
km/h65
h
s3600
10
km1
s
m05.18
m/s343
9.1
10.0
3
s
=
××=
=u
 
 
95 •• [SSM] A loudspeaker driver 20.0 cm in diameter is vibrating at 
800 Hz with an amplitude of 0.0250 mm. Assuming that the air molecules in the 
vicinity have the same amplitude of vibration, find (a) the pressure amplitude 
immediately in front of the driver, (b) the sound intensity,
and (c) the acoustic 
power being radiated by the front surface of the driver. 
 
Picture the Problem (a) and (b) The pressure amplitude can be calculated 
directly from ,00 vsp ρω= and the intensity from .20221 vsI ρω= (c) The power 
radiated is the intensity times the area of the driver. 
 
(a) Relate the pressure amplitude to 
the displacement amplitude, angular 
frequency, wave velocity, and air 
density: 
 
00 vsp ρω= 
 
Substitute numerical values and 
evaluate p0: 
( ) ( )[ ]
( )( )
2
3
13
0
N/m6.55
m100.0250m/s343
s8002kg/m1.29
=
××
=
−
−πp
 
Chapter 15 
 
 
1526 
(b) Relate the intensity to these same 
quantities: 
 
vsI 20
2
2
1 ρω= 
 
Substitute numerical values and 
evaluate I: 
( ) ( )[ ]
( ) ( )
22
23
213
2
1
W/m49.3W/m494.3
m/s343m100.0250
s8002kg/m1.29
==
××
=
−
−πI
 
 
(c) Express the power in terms of the 
intensity and the area of the driver: 
 
IrIAP 2π== 
Substitute numerical values and 
evaluate P: 
( ) ( )
W110.0
W/m3.494m100.0 22
=
= πP
 
 
96 •• A plane, harmonic, sound wave in air has an amplitude of 1.00 μm and 
an intensity of 10.0 mW/m2. What is the frequency of the wave? 
 
Picture the Problem The frequency of the wave is related to the density of the 
air, displacement amplitude, and velocity by .20
2
2
1 vsI ρω= 
 
Relate the intensity of the wave to 
the density of the air, displacement 
amplitude, velocity, and angular 
frequency: 
 
vsI 20
2
2
1 ρω= ⇒
v
I
s ρω
21
0
= 
Because fπω 2= : 
v
I
s
f ρπ
2
2
1
0
= 
 
Substitute numerical values and evaluate f: 
 
( ) ( )( )( ) kHz07.1m/s343kg/m1.29 W/m1000.12m1000.12 1 3
22
6 =××=
−
−πf 
 
97 •• Water flows at 7.0 m/s in a pipe of radius 5.0 cm. A plate with area 
equal to the cross-sectional area of the pipe is suddenly inserted to stop the flow. 
Find the force exerted on the plate. Take the speed of sound in water to be 
1.4 km/s. Hint: When the plate is inserted, a pressure wave propagates through 
the water at the speed of sound vs. The mass of water brought to a stop in time Δt 
is the water in a length of pipe equal to vsΔt. 
 
Traveling Waves 
 
 
1527
Picture the Problem The force exerted on the plate is due to the change in 
momentum of the water. We can use Newton’s second law in the form F = Δp/Δt 
to relate F to the mass of water in a length of pipe equal to vsΔt and to the speed 
of the water. This mass of water, in turn, is given by the product of its density and 
the volume of water in a length of pipe equal to vsΔt. 
 
Relate the force exerted on the plate 
to the change in momentum of the 
water: 
t
mv
t
pF Δ
Δ=Δ
Δ= w 
 
 
Express Δm in terms of the mass of 
water in a length of pipe equal to 
vsΔt: 
 
tAvVm Δ=Δ=Δ sρρ 
Substitute for Δm to obtain: 
 
ws AvvF ρ= 
Substitute numerical values and evaluate F: 
 ( )( ) ( )[ ]( ) kN77m/s7.0m0.050km/s1.4kg/m011.00 233 =×= πF 
 
98 •• A high-speed flash photography setup to capture a picture of a bullet 
exploding a soap bubble is shown in Figure 15-33. The shock wave from the 
bullet is to be detected by a microphone that will trigger the flash. The 
microphone is placed on a track that is parallel to, and 0.350 m below, the path of 
the bullet. The track is used to adjust the position of the microphone. If the bullet 
is traveling at 1.25 times the speed of sound, how far back from the soap bubble 
must the microphone be set to trigger the flash? (Assume that the flash itself is 
instantaneous once the microphone is triggered.) 
 
Picture the Problem Let d be the horizontal distance from the soap bubble to the 
position of the microphone. The angle θ of the shock wave is related to the speed 
of sound in air u and the speed of the bullet v according to .sin vu=θ We can 
determine θ from the given information and then use this angle to find d. 
 
Express d in terms of the angle of the 
shock wave and the distance from 
the soap bubble to the laboratory 
bench: 
 
θtan
m350.0=d (1) 
Relate the speed of the bullet to the 
angle of the shock-wave cone: 
 
v
u=θsin ⇒ ⎟⎠
⎞⎜⎝
⎛= −
v
u1sinθ 
Chapter 15 
 
 
1528 
Substitute for θ in equation (1) to 
obtain: ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛= −
v
u
d
1sintan
m350.0 
 
Substitute numerical values and 
evaluate d: 
cm3.26
25.1
sintan
m350.0
1
=
⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛= −
u
u
d 
 
99 •• A column of precision marchers keeps in step by listening to the band 
positioned at the head of the column. The beat of the music is for 100 paces/min. 
A television camera shows that only the marchers at the front and the rear of the 
column are actually in step. The marchers in the middle section are striding 
forward with the left foot when those at the front and rear are striding forward 
with the right foot. The marchers are so well trained, however, that they are all 
certain that they are in proper step with the music. How long is the column? 
 
Picture the Problem The source of the problem is that it takes a finite time for 
the sound to travel from the front of the line of marchers to the back. We can use 
the given data to determine the time required for the beat to reach the marchers in 
the back of the column and then use this time and the speed of sound to find the 
length of the column. 
 
Express the length of the column in 
terms of the speed of sound and the 
time required for the beat to travel 
the length of the column: 
 
tvL Δ= 
Calculate the time for the sound to 
travel the length of the column: 
 
s600.0min
100
1
Δ ==t 
 
Substitute numerical values and 
evaluate L: 
( )( ) m206s0.600m/s343 ==L 
 
100 •• A bat flying toward a stationary obstacle at 12.0 m/s emits brief, high-
frequency sound pulses at a repetition frequency of 80.0 Hz. What is the interval 
between the arrival times of the reflected pulses heard by the bat? 
 
Picture the Problem The interval between the arrival times of the reflected 
pulses heard by the bat is the reciprocal of the frequency of the reflected pulses. 
We can use s
s
r
r fuv
uvf ±
±= to relate the frequency of the reflected pulses to the 
speed of the bat and the frequency it emits. 
Traveling Waves 
 
 
1529
Relate the interval between the 
arrival times of the echo pulses heard 
by the bat to frequency of the 
reflected pulses: 
 
r
1
f
t =Δ 
Relate the frequency of the pulses 
received by the bat to its speed and 
the frequency it emits: 
 
s
s
r
s
s
r
s
s
r
r
1
1
f
v
u
v
u
f
uv
uvf
uv
uvf
−
+
=−
+=±
±= 
Substitute for fr to obtain: 
s
r
s
1
1
Δ
f
v
u
v
u
t
⎟⎠
⎞⎜⎝
⎛ +
−
= 
 
Substitute numerical values and 
evaluate Δt: 
( ) ms7.11s80.0
m/s343
m/s12.01
m/s343
m/s12.01
Δ
1
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−
=
−
t
 
 
101 •• Laser ranging to the moon is done routinely to accurately determine 
the Earth–moon distance. However, to determine the distance accurately, 
corrections must be made for the average speed of light in Earth’s atmosphere, 
which is 99.997 percent of the speed of light in vacuum. Assuming that Earth’s 
atmosphere is 8.00 km high, estimate the length of the correction. 
 
Picture the Problem Let d be the distance to the moon, h be the height of Earth's 
atmosphere, and v be the average speed of light in Earth’s atmosphere. We can 
express d ′, the distance measured when Earth’s atmosphere is ignored, in terms 
of the time for a pulse of light to make a round-trip from Earth to the moon and 
solve this equation for the length of correction d ′ − d. 
 
Express the roundtrip time for a 
pulse of light
to reach the moon and 
return: c
hd
v
h
ttt
−+=
+=
22
atmosphere sEarth' ofout atmosphere sEarth'
 
 
Express the "measured" distance d ′ 
when we do not account for the 
atmosphere: 
 hdhv
c
c
hd
v
hcctd'
−+=
⎟⎠
⎞⎜⎝
⎛ −+== 22
2
1
2
1
 
 
Chapter 15 
 
 
1530 
Solve for the length of correction 
d ′ − d: ⎟⎠
⎞⎜⎝
⎛ −=− 1
v
chdd' 
 
Substitute numerical values and 
evaluate d ′ − d: ( )
m2.0
1
99997.0
km00.8
≈
⎟⎠
⎞⎜⎝
⎛ −=−
c
cdd'
 
 
Remarks: This is larger than the accuracy of the measurements, which is 
about 3 to 4 cm. 
 
102 •• A tuning fork attached to a taut string generates transverse waves. The 
vibration of the fork is perpendicular to the string. Its frequency is 400 Hz and the 
amplitude of its oscillation is 0.50 mm. The string has a linear mass density of 
0.010 kg/m and is under a tension of 1.0 kN. Assume that there are no waves 
reflected at the far end of the string. (a) What are the period and frequency of 
waves on the string? (b) What is the speed of the waves? (c) What are the 
wavelength and wave number? (d) What is a suitable wave function for the waves 
on the string? (e) What is the maximum speed and acceleration of a point on the 
string? (f) At what minimum average rate must energy be supplied to the fork to 
keep it oscillating at a steady amplitude? 
 
Picture the Problem (a) The frequency of the waves on the string is the same as 
the frequency of the tuning fork and their period is the reciprocal of the 
frequency. (b) We can find the speed of the waves from the tension in the string 
and its linear density. (c) The wavelength can be determined from the frequency 
and the speed of the waves and the wave number from its definition. (d) The 
general form of the wave function for waves on a string is ( ) ( )tkxAx,ty ω±= sin , 
so, once we know k and ω, because A is given, we can write a suitable wave 
function for the waves on this string. (e) The maximum speed and acceleration of 
a point on the string can be found from the angular frequency and amplitude of 
the waves. (f) Finally, we can use vAP 2221av μω= to find the minimum average 
rate at which energy must be supplied to the tuning fork to keep it oscillating with 
a steady amplitude. 
 
(a) The frequency of the waves on 
the string is the same as the 
frequency of the tuning fork: 
 
Hz400=f 
The period of the waves on the wire 
is the reciprocal of their frequency: 
 
ms2.50
s400
11
1 === −fT 
 
Traveling Waves 
 
 
1531
(b) Relate the speed of the waves to 
the tension in the string and its linear 
density: 
 km/s0.32
m/s316.2
kg/m0.010
kN1.0T
=
=== μ
Fv
 
 
(c) Use the relationship between the 
wavelength, speed and frequency of 
a wave to find λ: 
 
cm79
cm79.05
s400
m/s316.2
1
=
=== −f
vλ
 
 
Using its definition, express and 
evaluate the wave number: 
1
1
2
m9.7
m95.7
m1005.79
22
−
−
−
=
=×==
π
λ
πk
 
 
(d) Determine the angular frequency 
of the waves: 
 
( ) 131 s102.51s40022 −− ×=== ππω f 
 
Substitute for A, k, and ω in the general form of the wave function to obtain: 
 
( ) ( ) ( )[ ( ) ]txx,ty 131 s1051.2m9.7sinmm50.0 −− ×−= 
 
(e) Relate the maximum speed of a 
point on the string to the amplitude 
of the waves and the angular 
frequency of the tuning fork: 
 
( )( )
m/s1.3
s102.51m100.50 133
max
=
××=
=
−−
ωAv
 
Express the maximum acceleration 
of a point on the string in terms of 
the amplitude of the waves and the 
angular frequency of the tuning fork: 
 
( )( )
2
2133
2
max
km/s2.3
s102.51m100.50
=
××=
=
−−
ωAa
 
(f) Express the minimum average 
power required to keep the tuning 
fork oscillating at a steady amplitude 
in terms of the linear density of the 
string, the amplitude of its 
vibrations, and the speed of the 
waves on the string: 
 
vAP 2221av μω= 
Chapter 15 
 
 
1532 
 
Substitute numerical values and evaluate Pav: 
 
( )( ) ( ) ( ) W5.2m/s316m100.50s102.51kg/m0.010 2321321av =××= −−P 
 
103 ••• A long rope with a mass per unit length of 0.100 kg/m is under a 
constant tension of 10.0 N. A motor drives one end of the rope with transverse 
simple harmonic motion at 5.00 cycles per second and an amplitude of 
40.0 mm. (a) What is the wave speed? (b) What is the wavelength? (c) What is 
the maximum transverse linear momentum of a 1.00-mm segment of the rope? 
(d) What is the maximum net force on a 1.00-mm segment of the rope? 
 
Picture the Problem Let Δm represent the mass of the segment of length 
Δx = 1.00 mm. We can find the wave speed from the given data for the tension in 
the rope and its linear density. The wavelength can be found from v = fλ. We’ll 
use the definition of linear momentum to find the maximum transverse linear 
momentum of the 1-mm segment and apply Newton’s second law to the segment 
to find the maximum net force on it. 
 
(a) Find the wave speed from the 
tension and linear density: 
 
m/s10.0
kg/m0.100
N10.0T === μ
Fv 
 
(b) Express the wavelength in terms 
of the speed and frequency of the 
wave: 
 
m2.00
s5.00
m/s10.0
1 === −f
vλ 
(c) Relate the maximum transverse 
linear momentum of the 1.00-mm 
segment to the maximum transverse 
speed of the wave: 
 
xAfxAmvp Δ=Δ=Δ= μπωμ 2maxmax 
 
Substitute numerical values and 
evaluate pmax: 
( )( )
( )( )
m/skg1026.1
m/skg10257.1
m0.0400m101.00
kg/m0.100s5.002
4
4
3
1
max
⋅×=
⋅×=
××
=
−
−
−
−πp
 
 
(d) The maximum net force acting on 
the segment is the product of the mass 
of the segment and its maximum 
acceleration: 
max
max
2
maxmax
2
ΔΔ
fp
pxAmaF
π
ωωμ
=
===
 
Traveling Waves 
 
 
1533
Substitute numerical values and 
evaluate Fmax: 
( )( )
mN95.3
m/skg101.257s5.002 41max
=
⋅×= −−πF
 
 
104 ••• In this problem, you will derive an expression for the potential energy 
of a segment of a string carrying a traveling wave (Figure 15-34). The potential 
energy of a segment equals the work done by the tension in stretching the string, 
which is ΔU = FT Δl − Δx( ), where FT is the tension, Δl is the length of the 
stretched segment, and Δx is its original length. (a) Use the binomial expansion to 
show that ( ) xxyx ΔΔΔΔΔ 221≈−l , and therefore xx
yFU Δ
Δ
Δ
Δ
2
T2
1 ⎟⎠
⎞⎜⎝
⎛≈ . 
(b) Compute ∂y/∂x from the wave function ( ) ( )tkxAtxy ω−= sin, (Equation 15-
15) and show that ( ) xtkxkAFU Δ−≈ ω222T21 cosΔ . 
 
Picture the Problem We can follow the step-by-step instructions outlined above 
to obtain the given expressions for ΔU. 
 
(a) Express the potential energy of a 
segment of the string: 
 
( )xFU ΔΔΔ T −= l 
For Δy/Δx << 1: 
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛+=
2
2
1
Δ
Δ1ΔΔ
x
yxl 
and 
x
x
y
x
x
yxx
Δ
Δ
Δ
Δ
Δ
Δ1ΔΔΔ
2
2
1
2
2
1
⎟⎠
⎞⎜⎝
⎛=
−
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛+=−l
 
 
Substitute to obtain: 
x
x
yF
x
yFU Δ
Δ
Δ
Δ
Δ
Δ
2
T2
1
2
2
1
T ⎟⎠
⎞⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛= 
 
(b) Differentiate ( ) ( )tkxAx,ty ω−= sin to obtain: 
 
( )tkxkA
x
y ω−=∂
∂ cos 
Approximate Δy/Δx by ∂y/∂x and 
substitute in our result from Part (a): 
( )( )
( )tkxxkAF
xtkxkAFU
ω
ω
−Δ=
Δ−=
222
T2
1
2
T2
1
cos
cosΔ
 
 
Chapter 15 
 
 
1534 
105 ••• One end of a heavy, 3.00-m-long rope is attached to a high ceiling and 
the rest of the rope is allowed to hang freely. Show that transverse waves on the 
rope must follow the equation ⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂
2
21
t
y
gx
yx
x
 instead of 2
2
22
2 1
t
y
vx
y
∂
∂=∂
∂ . 
 
Picture the Problem Let the +x 
direction be straight upward and choose 
the origin 3.00 m below the ceiling. Let 
y be the transverse direction in which 
the rope is displaced. Applying 
Newton’s second law to an element of 
length of the rope will lead to the given 
equation, which relates the spatial 
derivatives of y(x,t) to its time 
derivatives. We’ll consider only small 
values of the angles θ1 and θ2 (see the 
diagram to the right) between the 
tangent to the rope and the x axis. 
θ
θ
x
1
2
x
0
Δ
 1x
 2x
F
F
T
T
mΔ
 gmF )(g Δ=
 
 
Apply xx maF =∑ to the segment 
whose mass is Δm and whose length 
is Δx to obtain: 
 
( ) ( ) ( ) 0coscos 11T22T =Δ−− gmxFxF θθ
Note that the segment is not accelerated 
vertically. 
 
 
The mass of the segment is the 
product of its linear density and 
length: 
 
xm Δ=Δ μ 
 
Substituting for Δm yields: 
 ( ) ( ) 0coscos 11T22T =Δ−− xgxFxF μθθ (1) 
 
Apply the small-angle 
approximation ( 1cos =θ for 1<<θ ) 
and let x1 = x and x2 = x1 + Δx to 
obtain: 
 
( ) ( ) 01T2T =Δ−− xgxFxF μ 
or 
 ( ) ( ) xgxFxxF Δ=−Δ+ μTT 
 
Traveling Waves 
 
 
1535
 
Noting that the tension in the rope is 
zero at x = 0, solve for the tension at 
x2: 
 
( ) xgxF Δ=Δ μT ⇒ ( ) 22T xgxF μ= 
Using the differential approximation we have: 
 
( ) ( ) ( ) x
x
FxFx
x
FxFxF
xx
Δ∂
∂+=Δ⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂+=
= 1
T
1T
T
1T2T
1
 (2) 
where we’ve written 
1
T
x
F
∂
∂ in place of 
1
T
xxx
F
=∂
∂ in order to be more concise. 
 
Substituting equation (2) in equation 
(1) yields: 
 
( ) ( ) 01T
1
T
1T =Δ−−Δ∂
∂+ xgxFx
x
FxF μ 
Simplify this equation to obtain: 
0
1
T =−∂
∂ μg
x
F (3) 
 
The segment moves horizontally, and 
the net force acting in this direction is: 
 
( ) ( ) 11T22T sinsin θθ xFxFFy −=∑ 
Using the small-angle approximation θθ tansin ≈ yields: 
 
( ) ( ) ( ) ( )
1
1T
2
2T11T22T tantan x
yxF
x
yxFxFxFFy ∂
∂−∂
∂=−≈∑ θθ (4) 
where we’ve also used 
x
yslopetan ∂
∂==θ . 
 
Applying the differential 
approximation gives: 
 
x
x
y
x
yx
x
y
xx
y
x
y Δ∂
∂+∂
∂=Δ⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
∂
∂+∂
∂=∂
∂
2
1
2
1112
 
 
Substituting for ( )2T xF and 
2x
y
∂
∂ in equation (4) yields: 
 
( ) ( )
1
1T2
1
2
11
T
1T x
yxFx
x
y
x
yx
x
FxFFy ∂
∂−⎟⎟⎠
⎞
⎜⎜⎝
⎛ Δ∂
∂+∂
∂
⎟⎟⎠
⎞
⎜⎜⎝
⎛ Δ∂
∂+=∑ 
 
Chapter 15 
 
 
1536 
 
Expand and simplify this expression to obtain: 
 
( ) ( ) ( ) ( )
( ) x
x
y
x
Fx
x
yxF
x
yxFx
x
y
x
Fx
x
y
x
Fx
x
yxF
x
yxFFy
Δ∂
∂
∂
∂+Δ∂
∂≈
∂
∂−Δ∂
∂
∂
∂+Δ∂
∂
∂
∂+Δ∂
∂+∂
∂=∑
11
T
2
1
2
1T
1
1T
2
2
1
2
1
T
11
T
2
1
2
1T
1
1T
 
where we have neglected the term with ( )2xΔ . 
 
Applying yy maF =∑ to our length 
element yields: 
 
( ) 2
2
11
T
2
1
2
1T t
yxx
x
y
x
Fx
x
yxF ∂
∂Δ=Δ∂
∂
∂
∂+Δ∂
∂ μ 
Divide both sides of this equation by 
Δx to obtain: 
 
( ) 2
2
11
T
2
1
2
1T t
y
x
y
x
F
x
yxF ∂
∂=∂
∂
∂
∂+∂
∂ μ (5) 
Substituting gxμ for FT yields: 
 
( )
2
2
2
2
t
y
x
y
x
gx
x
ygx ∂
∂=∂
∂
∂
∂+∂
∂ μμμ 
where we have dropped the subscript 1 
by replacing x1 with x. This can be done 
without loss of generality because all 
the x′s in equation (5) are x1′s. 
 
Divide both sides of the equation by 
gμ to obtain: 
 
2
2
2
2 1
t
y
gx
y
x
yx ∂
∂=∂
∂+∂
∂ (6) 
Note that the expression on the left-
hand side of the equation can be 
written as: 
 
⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂=∂
∂+∂
∂
x
yx
xx
y
x
yx 2
2
, a result you 
can verify by using the chain rule to 
differentiate the right-hand side of this 
equation. 
 
Substituting in equation (6) yields: 
2
21
t
y
gx
yx
x ∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂