1. 307
Chapter 15
Traveling Waves
Conceptual Problems
1 • 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.
5 • 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 denser that the high frequency strings. High frequencies are not a
problem as they utilize short strings anyway.
11 • At a given location, two harmonic sound waves have the same
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
Aav, sωρη =
The average energy density of sound
B is given by:
2
B,0
2
BB2
1
Bav, sωρη =
2. Chapter 15308
Dividing the first of these equation
by the second yields: 2
B,0
2
BB2
1
2
A,0
2
AA2
1
Bav,
Aav,
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
Bav,
Aav,
2
2
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
==
f
f
f
f
π
π
ω
ω
η
η
Because fA = 2fB:
4
2
2
B
B
Bav,
Aav,
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
f
f
η
η
⇒ ( )b is correct.
19 • Sound waves in air encounter a 1.0-m wide door into a classroom. Due
to the effects of refraction, 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.
21 •• 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.
Speed of Waves
33 •• (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 to 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.
3. Traveling Waves 309
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
dv
FdF TT 2=
Approximate dFT with ΔFT and dv
with Δv to obtain: v
v
FF
Δ
2Δ TT =
Substitute numerical values and
evaluate ΔFT:
( )
N40
m/s300
m/s300m/s312
N5002Δ 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
v F
F
F
F
v
==
μ
μ
⇒
2
1
2
F,1T,2 ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
v
v
FF
4. Chapter 15310
Substituting for FT,2 in equation (1)
yields: ( )
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
1
Δ
2
1
2
F,1
T,1
2
1
2
F,1exactT
v
v
F
F
v
v
FF
Substitute numerical values and
evaluate (ΔFT)exact: ( ) ( )
N8.40
1
m/s300
m/s312
N500Δ
2
exactT
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
=F
The percent error between the exact
and approximate values for ΔFT is:
( )
( )
%2
N8.40
N40.0N8.40
Δ
ΔΔ
exactT
TexactT
≈
−
=
−
F
FF
Harmonic Waves on a String
39 • 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 22
2
1
av 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 22
2
1
av μω=
The speed of the wave on the string
is given by: μ
TF
v =
Substitute for v to obtain:
μ
μω T22
2
1
av
F
AP =
5. Traveling Waves 311
Substitute numerical values and evaluate Pav:
( )( )( ) ( ) W9.9
kg/m0.05
N80
m0.050s10kg/m0.0504
2212
2
1
av == −
πP
45 •• 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 22
2
1
av 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 22222
2
1
av 2 μπμω ==
Substitute numerical values and
evaluate Pav:
( )( )
( ) ( )
mW79
m/s10m100.50
s400kg/m0.0102
23
212
av
=
××
=
−
−
πP
(b) Because 2
av fP ∝ , increasing the frequency by a factor of 10 would increase
the power by a factor of 100.
Because 2
av 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.
6. Chapter 15312
Harmonic Sound Waves
49 • (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.
Assuming the density of air is 1.29 kg/m3
, 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
p
s
ρω
0
0 =
Substitute numerical values and evaluate s0:
( )( )
( )( )( )
m4.36m1064.3
m/s343s100kg/m1.292
Pa/atm1001325.1atm1000.1 5
13
54
0 μ
π
=×=
××
= −
−
−
s
(b) Use 00 vsp ρω= to find p0:
( )( )( )( ) mPa4.83m101.00m/s343s300kg/m29.12 713
0 =×= −−
πp
Waves in Three Dimensions: Intensity
55 • 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 2
av 4π= to find the total
acoustic power output of the speaker:
( ) ( )
W3.50W27.50
W/m1000.1m0.204 222
av
==
×= −
πP
7. Traveling Waves 313
(b) Relate the intensity of the sound
at 20 m to the distance from the
speaker:
( )2
av22
m0.204
W/m1000.1
π
P
=× −
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
P
I
π
= to find the intensity
at 30.0 m:
( )
( )
23
2
W/m1045.4
m0.304
W3.50
m0.30
−
×=
=
π
I
*Intensity Level
57 • What is the intensity level in decibels of a sound wave of intensity
(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/m101.00
logdB10
2
212
201
==
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
= −
−
β
(b) Proceed as in (a) with
I = 1.00 × 10–2
W/m2
:
( )
dB10010log10
W/m10
W/m101.00
logdB10
10
212
22
==
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
= −
−
β
67 ••• 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.
8. Chapter 15314
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
50
logdB10
I
I
β
Find the sound intensity when 100
students are writing the exam:
( ) ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
0
100
logdB10dB60
I
I
and
26
0
6
100 W/m1010 −
== II
Find the sound intensity due to the
background noise:
( ) ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
0
background
logdB10dB40
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/m101.0050
logdB10 212
28
50
=
×
= −
−
β
String Waves Experiencing Speed Changes
69 • 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.
9. Traveling Waves 315
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.
(a) Use their definitions to express
the reflection and transmission
coefficients:
2
1
2
1
12
12
1
1
v
v
v
v
vv
vv
r
+
−
=
+
−
= (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
F
v = and
1
T
1
μ
F
v =
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
2
1
2
μ
μ
τ
+
= (4)
If μ2 = μ1:
0
11
11
=
+
−
=r
and
1
11
2
1
2
1
2
=
+
=
+
=
μ
μ
τ
10. Chapter 15316
(b) From equations (1) and (2), if
μ2 >> μ1 then v1 >> v2:
1
1
1
12
12
−=
−
≈
+
−
=
v
v
vv
vv
r
and
0
1
22
2
112
2
≈
+
=
+
=
v
vvv
v
τ
(c) If μ2 << μ1, then v2 >> v1:
1
1
1
2
1
2
1
12
12
≈
+
−
=
+
−
=
v
v
v
v
vv
vv
r
and
2
1
22
2
112
2
≈
+
=
+
=
v
vvv
v
τ
The Doppler Effect
79 •• 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 f
c
uc
f
uc
uc
f ⎟
⎠
⎞
⎜
⎝
⎛ +
=
±
±
= (1)
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 f
uc
c
f
uc
uc
'f ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=
±
±
= (2)
11. Traveling Waves 317
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
u
f
c
u
c
u
f
uc
uc
f
c
uc
uc
c
'f
−
⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
+=
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
−
+
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
+
=⎟
⎠
⎞
⎜
⎝
⎛ +
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=
Because cur << ,
c
u
c
u r
1
r
11 +≈⎟
⎠
⎞
⎜
⎝
⎛
−
−
.
Substituting for
1
r
1
−
⎟
⎠
⎞
⎜
⎝
⎛
−
c
u
and
simplifying yields:
s
2
r
s
rr
r
1
11
f
c
u
f
c
u
c
u
'f
⎟
⎠
⎞
⎜
⎝
⎛
+=
⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+≈
Because cur << ,
c
u
c
u r
2
r 2
11 +≈⎟
⎠
⎞
⎜
⎝
⎛
+ .
Substituting for
2
r
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
c
u
gives:
s
r
r
2
1 f
c
u
'f ⎟
⎠
⎞
⎜
⎝
⎛
+≈
The frequency difference detected at
the source is:
s
r
ss
r
sr
2
2
1Δ
f
c
u
ff
c
u
f'ff
=
−⎟
⎠
⎞
⎜
⎝
⎛
+=−=
Solving for ur yields:
f
f
c
u Δ
2 s
r =
Substitute numerical values and
evaluate ur: ( )
( )
mi/h174
m/s4470.0
mi/h1
m/s95.77
Hz253
MHz2562
m/s10998.2 8
r
=
×=
×
=u
83 •• 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
12. Chapter 15318
s
rel
1' f
v
u
fr ⎟
⎠
⎞
⎜
⎝
⎛
+=
where urel = us – ur is the velocity of the source relative to the receiver.
Picture the Problem The received and transmitted frequencies are related
through s
s
r
r f
uv
uv
f
±
±
= (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
uv
f
−
⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
−=
−
−
=
±
±
=
(b) Expand ( ) 1
s1
−
− 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
u
f
⎟
⎠
⎞
⎜
⎝
⎛ −
+≈
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
−−+=
⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
−=
because both us and ur are small
compared to v.
Because urel = us − ur
s
rel
r 1 f
v
u
f '
⎟
⎠
⎞
⎜
⎝
⎛
+≈
General Problems
91 • 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?
13. Traveling Waves 319
Picture the Problem The diagram
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
m00.1=r
m00.5
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
u
f
−
=
Because ωru =s :
smax
1
1
f
v
r
f
ω
−
=
Substitute numerical values and evaluate fmax:
( )
( ) Hz529Hz500
m/s343
rev
rad2
s
rev
3.00m00.1
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
u
f
+
=
Substitute numerical values and evaluate fmin:
( )
( ) Hz474Hz500
m/s343
rev
rad2
s
rev
3.00m00.1
1
1
min =
⎟
⎠
⎞
⎜
⎝
⎛
×
+
=
π
f
14. Chapter 15320
95 •• 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 .2
0
2
2
1
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
(b) Relate the intensity to these same
quantities:
vsI 2
0
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 2
π==
Substitute numerical values and
evaluate P:
( ) ( )
W110.0
W/m3.494m100.0 22
=
= πP
