CHAPTER 16 WAVES AND SOUND
CONCEPTUAL QUESTIONS
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6.
REASONING AND SOLUTION A rope of mass m is hanging down from the ceiling.
Nothing is attached to the loose end of the rope. A transverse wave is traveling up the rope.
The tension in the rope is not constant. The lower portion of the rope pulls down on the
higher portions of the rope. If we imagine that the rope is divided into small segments, we
see that the segments near the top of the rope are being pulled down by more weight than the
segments near the bottom. Therefore, the tension in the rope increases as we move up the
rope. The speed of a transverse wave on the rope is given by Equation 16.2:
vwave  F /(m / L) . From Equation 16.2 we see that, as the tension F in the rope increases,
the speed of the wave increases. Therefore, as the transverse wave travels up the rope, the
speed of the wave increases.
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9.
REASONING AND SOLUTION As the disturbance moves outward when a sound wave is
produced, it compresses the air directly in front of it. This compression causes the air
pressure to rise slightly, resulting in a condensation that travels outward. The condensation
is followed by a region of decreased pressure, called a rarefaction. Both the condensation
and rarefaction travel away from the speaker at the speed of sound. As the condensations
and rarefactions of the sound wave move away from the disturbance, the individual air
molecules are not carried with the wave. Each molecule executes simple harmonic motion
about a fixed equilibrium position. As the wave passes by, all the particles in the region of
the disturbance participate in this motion. There are no particles that are always at rest.
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10. REASONING AND SOLUTION Assuming that we can treat air as an ideal gas, then the
speed of sound in air is given by Equation 16.5, v   kT / m , where  is the ratio of the
specific heats c P / c V , k is Boltzmann's constant, T is the Kelvin temperature, and m is the
mass of a molecule of the gas.
We can see from Equation 16.5 that the speed of sound in air is proportional to the square
root of the Kelvin temperature of the gas. Therefore, on a hot day, the speed of sound in air
is greater than it is on a cold day. Hence, we would expect an echo to return to us more
quickly on a hot day as compared to a cold day, other things being equal.
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14. REASONING AND SOLUTION A source is emitting sound uniformly in all directions.
According to Equation 16.9, I  P /(4  r 2 ) , the intensity of such a source varies as 1/r2.
Thus, the intensity I at a point in space depends on the distance of that point from the source.
A flat surface faces the source. As suggested in the following figure, the distance between
the source and the flat sheet varies, in general, from point to point on the sheet. The figure
indicates that, as we move up the screen, the distance between the source and the screen
increases. Therefore, the sound intensity is not the same at all points on the screen.
uniformly emitting
source
flat screen
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15. REASONING AND SOLUTION Two people talk simultaneously and each creates an
intensity level of 65 dB at a certain point. The total intensity level at this point does not
equal 130 dB. Intensity levels are defined in terms of logarithms. According to
Equation 16.10, the sound intensity level in decibels is given by   10 dB log( I / I0 ) ,
where I0 is the threshold of human hearing. We can show, using Equation 16.10, that an
intensity level of 65 dB corresponds to a sound intensity of 3.2 10 –6 W/m 2 . If two people
simultaneously talk and each creates an intensity of 3.2 10 –6 W/m 2 at a certain point,
then the total intensity at that point is 6.4 10 –6 W/m 2 . However, since the intensity level
is defined in terms of logarithms, the intensity level at that point is (from Equation 16.10)
68 dB. While the second person doubles the intensity at the point in question, he only
increases the intensity level (or the loudness) by 3 dB.
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17. REASONING AND SOLUTION According to Table 16.1, the speed of sound in air at
20 C is v = 343 m/s, while its value in water at the same temperature is v = 1482 m/s. These
values influence the Doppler effect, because of which an observer hears a frequency fo that
is different from the frequency fs that is emitted by the source of sound. For purposes of this
question, we assume that fs = 1000 Hz and that the speed at which the source moves is
vs = 25 m/s. Our conclusions, however, will be valid for any values of fs and vs .
a. The Doppler-shifted frequency when the source approaches the observer is given by
Equation 16.11 as f o  fs 1  (vs / v)  . Applying this equation for air and water, we find
Air
fo 
Water
fo 
1000 Hz
 1079 Hz
1  (25 m s )/(343 m s )
1000 Hz
 1017 Hz
1  (25 m s )/(1482 m s )
The change in frequency due to the Doppler effect in air is greater.
b. The Doppler-shifted frequency when the source moves away from the observer is given
by Equation 16.12 as f o  fs 1  (vs / v)  . Applying this equation for air and water, we
find
Air
fo 
Water
fo 
1000 Hz
 932 Hz
1 + (25 m s )/(343 m s )
1000 Hz
 983 Hz
1 + (25 m s )/(1482 m s )
The change in frequency due to the Doppler effect in air is greater.
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Problems
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6.
REASONING The speed of a Tsunamis is equal to the distance x it travels divided by
the time t it takes for the wave to travel that distance. The frequency f of the wave is equal to its
speed divided by the wavelength , f = v/ (Equation 16.1). The period T of the wave is related
to its frequency by Equation 10.5, T = 1/f.
SOLUTION
a.
The speed of the wave is (in m/s)
v
b.
The frequency of the wave is
f 
c.
x 3700  103 m  1 h 


  190 m/s
t
5.3 h
 3600 s 
v


190 m/s
 2.5  104 Hz
3
750  10 m
(16.1)
The period of any wave is the reciprocal of its frequency:
1
1

 4.0  103 s
(10.5)
4
f 2.5  10 Hz
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T
9.
REASONING During each cycle of the wave, a particle of the string moves through a
total distance that equals 4A, where A is the amplitude of the wave. The number of wave cycles
per second is the frequency f of the wave. Therefore, the distance moved per second by a string
particle is 4Af. The time to move through a total distance L, then, is t = L/(4Af). According to
Equation 16.1, however, we have that f = v/, so that
t
L
L

4 Af 4 A v /  
SOLUTION Using the result obtained above, we find that




1.0  10 3 m 0.18 m 
L
t

 5.0  10 1 s
4 Av 4 2.0  10 –3 m  450 m/s 
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15.
SSM WWW
REASONING According to Equation 16.2, the linear density of the
2
string is given by (m / L)  F / v , where the speed v of waves on the middle C string is given by
Equation 16.1, v  f   / T .
SOLUTION Combining Equations 16.2 and 16.1 and using the given data, we obtain
F FT 2 (944 N)(3.82  10 –3 s)2
m/L  2  2 
 8.68  10–3 kg/m
2
v

(1.26 m)
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21.
REASONING AND SOLUTION If the string has length L, the time required for a wave
on the string to travel from the center of the circle to the ball is
t
L
(1)
vwave
The speed of the wave is given by text Equation 16.2
v wave 
F
m string / L
(2)
The tension F in the string acts as a centripetal force on the ball, so that
F  mball 2 r  mball 2 L
Eliminating the tension F from Equations (2) and (3) above yields
v wave 
m ball  2L

m string / L
m ball 2 L2
m ball  2
L
mstring
m string
Substituting this expression for vwave into Equation (1) gives
(3)
L
t
L
m ball 2
m string

m ball  2
0.0230 kg
 3.26  10 –3 s
2
(15.0 kg)(12.0 rad/s)

m string
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29.
SSM REASONING AND SOLUTION The speed of sound in an ideal gas is given by
Equation 16.5, v   kT/ m . The ratio of the speed of sound v2 in the container after the
temperature is raised to the speed v1 before the temperature change is
v2

v1
T2
T1
Thus, the new speed is
T2
405 K
 1730 m/s
T1
201 K
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v 2  v1
 (1220 m/s)
36.
REASONING AND SOLUTION The wheel must rotate at a frequency of
f = (2200 Hz)/20 = 110 Hz. The angular speed  of the wheel is
 = 2 f = 2 (110 Hz) = 690 rad/s 
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44.
REASONING AND SOLUTION Since the frequency (f = v/) is the same in each gas,
we have
vKr/Kr = vNe/Ne
then
Ne = Kr(vNe/vKr)
We know that
v
kT
m
The ratio of the speeds is
v Ne
v Kr

m Kr
m Ne
We need to calculate the mass of each type of atom in the gas:
mKr = (83.8  10–3 kg/mol)/(6.022  1023/mol) = 1.39  10–25 kg
mNe = (20.2  10–3 kg/mol)/(6.022  1023/mol) = 3.35  10–26 kg
So we have,
1.39  10 –25 kg
 2.55 m
m Ne
3.35  10 –26 kg
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m Kr
 Ne   Kr
50.
 1.25 m 
REASONING AND SOLUTION The energy carried by the sound into the ear is
Energy = IAt = (3.2  10–5 W/m2)(2.1  10–3 m2)(3600 s) = 2.4  10 –4 J
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55.
s) is
REASONING AND SOLUTION The energy delivered to the lasagna in 7.0 minutes (420
Q = cmT = [3400 J/(kg.C°)](0.25 kg)(80.0 C°) = 6.8  104 J
(12.4)
The power delivered to the lasagna is
P
Energy 6.8  10 4 J

 1.6  10 2 W
time
420 s
(6.10b)
Thus, the intensity of the microwaves in the oven is
P
1.6  10 2 W

 1.0  10 4 W/m 2
(16.8)
A 1.6  10 –2 m 2
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I
63.
SSM REASONING According to Equation 16.10, the sound intensity level  in
decibels (dB) is related to the sound intensity I according to   10 dB log ( I / I 0 ) , where the
quantity I0 is the reference intensity. Since the sound is emitted uniformly in all directions, the
2
intensity, or power per unit area, is given by I  P /(4r ) . Thus, the sound intensity at position
2
1 can be written as I1  P /(4r1 ) , while the sound intensity at position 2 can be written as
I2  P /(4r22 ) . Therefore, the difference in the sound intensity level 21 between the two
positions is
 I2 
 I1
  10 dB  log 
 I0 
 I0
 21   2  1  10 dB  log 

 I2 / I0
  10 dB  log 

 I1 / I 0

 I2 
  10 dB  log  
 I1 

 P /(4 r22 ) 
 r12
 21  10 dB  log 
 10 dB  log  2
2 

 P /(4 r1 ) 
 r2
r
  20 dB  log  1
r
 2

r
  10 dB  log  1



 r2




2

r 
   20 dB  log  1    20 dB  log 1/ 2   –6.0 dB

 2r 

 1
The negative sign indicates that the sound intensity level decreases.
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66.
REASONING If I1 is the sound intensity produced by a single person, then N I1 is the
sound intensity generated by N people. The sound intensity level generated by N people is given
by Equation 16.10 as
 NI 
 N  10 dB  log  1 
 I0 
where I0 is the threshold of hearing. Solving this equation for N yields

N
 I  10 dB
N   0 10
 I1 
(1)
We also know that the sound intensity level for one person is
1
I 
1  10 dB  log  1 
 I0 
or
I1  I 0 10
10 dB
(2)
Equations (1) and (2) are all that we need in order to find the number of people at the
football game.
SOLUTION Substituting the expression for I1 from Equation (2) into Equation (1) gives
the desired result.
N
N
I 0 10
10 dB
1

10
109 dB
10 dB
60.0 dB
10 dB
 79 400
I 0 1010 dB 10
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73.
SSM REASONING Since you detect a frequency that is smaller than that emitted by
the car when the car is stationary, the car must be moving away from you. Therefore, according
to Equation 16.12, the frequency fo heard by a stationary observer from a source moving away
from the observer is given by
 1 
f o  fs 
v 
1 s 
v 

where fs is the frequency emitted from the source when it is stationary with respect to the
observer, v is the speed of sound, and vs is the speed of the moving source. This expression can
be solved for vs .
SOLUTION We proceed to solve for vs and substitute the data given in the problem
statement. Rearrangement gives
vs
f
 s –1
v
fo
Solving for vs and noting that f o / f s  0.86 yields
 f

 1

vs  v  s –1  (343 m/s) 
–1  56 m/s
 0.86 
 fo

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77.
REASONING AND SOLUTION The speed of the Bungee jumper, vs, after she has
fallen a distance y is given by
v 2s  v 20  2ay
Since she falls from rest, v0 = 0 m/s, and
vs 
2ay 
2(–9.80 m/s 2 )(–11.0 m)  14.7 m/s
Then, from Equation 16.11
 1 
1


f o  fs 
  (589 Hz) 
  615 Hz .
 1  vs / v 
1  14.7 m/s  /  343 m/s  
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