Chapter 14
Sound
Quick Quizzes
1.
(c). The speed of sound in air is given by v = ( 331 m s) T 273 K . Thus, increasing the
absolute temperature, T, will increase the speed of sound. Changes in frequency,
amplitude or air pressure have no affect on the speed of sound.
2.
(c). The distance between you and the buzzer is increasing. Therefore, the intensity at your
location is decreasing. As the buzzer falls, it moves away from you with increasing speed.
This causes the detected frequency to decrease.
3.
(b). The speed of sound increases in the warmer air, while the speed of the sound source
(the plane) remains constant. Therefore, the ratio of the speed of the source to that of
sound (that is, the Mach number) decreases.
4.
(b) and (e). A string fastened at both ends can resonate at any integer multiple of the
fundamental frequency. Of the choices listed, only 300 Hz and 600 Hz are integer
multiples of the 150 Hz fundamental frequency.
5.
(d). In the fundamental mode, an open pipe has a node at the center and antinodes at each
end. The fundamental wavelength of the open pipe is then twice the length of the pipe
and the fundamental frequency is fopen = v 2L . When one end of the pipe is closed, the
fundamental mode has a node at the closed end and an antinode at the open end. In this
case, the fundamental wavelength is four times the length of the pipe and the
fundamental frequency is f close = v 4 L .
6.
(a). The change in the length of the pipe, and hence the fundamental wavelength, is
negligible. As the temperature increases, the speed of sound in air increases and this
causes an increase in the fundamental frequency, f 0 = v λ 0 .
7.
(b). Since the beat frequency is steadily increasing, you are increasing the difference
between the frequency of the string and the frequency of the tuning fork. Thus, your
action is counterproductive and you should reverse course by loosening the string.
501
502
CHAPTER 14
Answers to Even Numbered Conceptual Questions
2.
The resonant frequency depends on the length of the pipe. Thus, changing the length of
the pipe will cause different frequencies to be emphasized in the resulting sound.
4.
Air flowing fast by a rim of the pipe creates a "shshshsh" sound called edgetone noise, a
mixture of all frequencies, as the air turbulently switches between flowing on one side of
the edge and the other. The air column inside the pipe finds one or more of its resonance
frequencies in the noise. The air column starts vibrating with large amplitude in a
standing vibration mode and radiates this sound into the surrounding air.
6.
The distance around the opening of the bell must be an integer multiple of the
wavelength. Actually, the circumference being equal to the wavelength would describe
the bell moving from side to side without bending, which it can do without producing any
sound. A tuned bell is cast and shaped so that some of these vibrations will have their
frequencies constitute higher harmonics of a musical note, the strike tone. This tuning is
lost if a crack develops in the bell. The sides of the crack vibrate as antinodes. Energy of
vibration may be rapidly lost into thermal energy at the end of the crack, so the bell may
not ring for so long a time.
8.
The speed of light is so high that the arrival of the flash is practically simultaneous with
the lightning discharge. Thus, the delay between the flash and the arrival of the sound of
thunder is the time sound takes to travel the distance separating the lightning from you.
By counting the seconds between the flash and thunder and knowing the approximate
speed of sound in air, you have a rough measure of the distance to the lightning bolt.
10.
Refer to Table 14.2 to see that a rock concert has an intensity level of about 120 dB, the
turning of a page in a textbook about 30 dB, a normal conversation is about 50 dB,
background noise at a church is about 30 dB. This leaves a cheering crowd at a football
game to be about 60 dB.
12.
No. Adding two sounds of equal loudness will produce an intensity double that
associated with either individual sound. However, the decibel scale is a logarithmic
function of intensity, so doubling the intensity only increases the decibel level by 10 log 2 .
Thus, the decibel level with both sounds present will be 53 dB.
14.
A beam of radio waves of known frequency is sent toward a speeding car, which reflects
the beam back to a detector in the police car. The amount the returning frequency has been
shifted depends on the velocity of the oncoming car.
16.
Consider the level of fluid in the bottle to be adjusted so that the air column above it
v
resonates at the first harmonic. This is given by f =
. This equation indicates that as the
4L
length L of the column increases (fluid level decreases), the resonant frequency decreases.
Sound
503
18.
Walking makes the person's hand vibrate a little. If the frequency of this motion equals the
natural frequency of coffee sloshing from side to side in a cup, then a large-amplitude
vibration of the coffee will build up in resonance. To get off resonance and back to the
normal case of a small-amplitude disturbance producing a small-amplitude result, the
person can walk faster, walk slower, vary his speed, or get a larger or smaller cup.
Alternatively, even at resonance, he can reduce the amplitude by adding damping, as by
stirring high-fiber quick cooking oat meal into the hot coffee.
20.
A vibrating string is not able to set very much air into motion when vibrated alone. Thus it
will not be very loud. If it is placed on the instrument, however, the string's vibration sets
the sounding board of the guitar into vibration. A vibrating piece of wood is able to move
a lot of air, and the note is louder.
504
CHAPTER 14
Answers to Even Numbered Problems
2.
0.196 s
4.
1.7 cm to 17 m
6.
18.5 m
8.
316 K
5.0 × 10 -17 W
(b)
5.0 × 10 -5 W
1.32 × 10 -4 W m 2
(b)
81.2 dB
(a)
7.96 × 10 -2 W m 2
(b)
109 dB
18.
(a)
IA
=2
IB
(b)
IA
=5
IC
20.
(a)
10.0 kHz
(b)
3.33 kHz
22.
32.1 m s , behind the car
24.
41 kHz
26.
(a)
0.0217 m s
(b)
2 000 029 Hz
28.
(a)
57.1 s
(b)
56.6 km
30.
(a)
0.431 m
(b)
0.863 m
32.
(a)
0.345 m
(b)
constructive interference
34.
824.0 N
36.
2.55 kHz
38.
120 Hz
40.
(a)
(b)
2
42.
9.00 kHz
44.
(a)
(b)
4.25 cm
10.
(a)
12.
3.01 dB
14.
(a)
16.
4.9 × 10 −3 kg m
531 Hz
(c)
2.82 m
(c)
2 000 058 Hz
(c)
no standing wave will form
Sound
46.
n ( 206 Hz ) for n = 1 to 9 and
n ( 84.5 Hz ) for n = 2 to 23
48.
3.3 × 10 2 m s
50.
29.7 cm
52.
3.98 Hz
54.
2.94 cm
56.
(a)
58.
7.8 m
60.
200 m s
62.
(a)
0.655 m
(b)
13.0°C
64.
(a)
0.233 m
(b)
14 mm
66.
(a)
a path difference of λ 2 produces destructive interference
(b)
Along a hyperbola given by
362 Hz
68.
1.93 m s
70.
(b)
531 Hz
72.
(a)
55.8 m s
(b)
287 Hz
(c)
x2 y2
−
= 1.00 m 2
16 9
(b)
2.50 kHz
0.953 m and 1.20 m
505
506
CHAPTER 14
Problem Solutions
14.1
Since vlight >> vsound , we ignore the time required for the lightning flash to reach the
observer in comparison to the transit time for the sound. Then,
d ≈ ( 343 m s ) (16.2 s ) = 5.56 × 10 3 m = 5.56 km
14.2
The speed of sound in seawater at 25°C is 1 530 m s . Therefore, the time for the sound
to reach the sea floor and return is
t=
14.3
2 d 2 ( 150 m )
=
= 0.196 s
v 1 530 m s
The speed of the sound wave is v = λ f = ( 0.50 m )( 700 Hz ) = 3.5 × 10 2 m s . Thus, from
v = ( 331 m s) T 273 K , the temperature of the air is
2
2


 3.5 × 10 2 m s 
v
T = ( 273 K ) 
273
K
=
(
)
 331 m s  = 305 K = 32° C
 331 m s 
14.4
At T = 27°C = 300 K , the speed of sound in air is
v = ( 331 m s )
T
300 K
= ( 331 m s)
= 347 m s
273 K
273 K
The wavelength of the 20 Hz sound is λ =
the 20 000 Hz is λ =
v 347 m s
=
= 17 m , and that of
f
20 Hz
347 m s
= 1.7 × 10 −2 m=1.7 cm . Thus, range of wavelengths of
20 000 Hz
audible sounds at 27°C is 1.7 cm to 17 m .
Sound
14.5
507
Since the sound had to travel the distance between the hikers and the mountain twice,
the time required for a one-way trip was 1.50 s. The speed of sound in air at
T = 22.0°C = 295 K is
v = ( 331 m s )
T
295 K
= ( 331 m s)
= 344 m s
273 K
273 K
and the distance the sound traveled to the mountain was
d = ( 344 m s ) (1.50 s ) = 516 m
14.6
At a temperature of T = 10.0° C = 283 K , the speed of sound in air is
v = ( 331 m s )
T
283 K
= ( 331 m s)
= 337 m s
273 K
273 K
The elapsed time between when the stone was released and when the sound is heard is
the sum of the time t1 required for the stone to fall distance h and the time t2 required
for sound to travel distance h in air on the return up the well. That is, t1 + t2 = 2.00 s . The
gt12
h=
distance the stone falls, starting from rest, in time t1 is
2
Also, the time for the sound to travel back up the well is
Combining these two equations yields
t2 =
h
= 2.00 s − t1
v
 g 2
  t1 = 2.00 s − t1
2v
With v = 337 m s and g = 9.80 m s 2 , this becomes
(1.45 × 10
−2
)
s -1 t12 + t1 − 2.00 s = 0
Applying the quadratic equation yields one positive solution of t1 = 1.945 s , so the depth
of the well is
(
)
9.80 m s 2 (1.945 s )
gt12
h=
=
= 18.5 m
2
2
2
508
CHAPTER 14
14.7
From Table 14.1, the speed of sound in the saltwater is vw = 1 530 m s . At
T = 20°C = 293 K , the speed of the sound in air is
va = ( 331 m s )
T
293 K
= ( 331 m s )
= 343 m s
273 K
273 K
If d is the width of the inlet, the transit time for the sound in the water is tw =
that for the sound in the air is ta = tw + 4.50 s=
Thus,
d
, and
vw
d
.
va
 v v 
d
d
=
+ 4.50 s , or d = ( 4.50 s )  w a 
v a vw
 vw − v a 
 ( 1 530 m s )( 343 m s ) 
3
d = ( 4.50 s ) 
 = 1.99 × 10 m = 1.99 km
 ( 1 530 − 343 ) m s 
14.8
At absolute temperature T, the speed of sound in air is
v = ( 331 m s)
T
273 K
Thus, if the speed of sound in the air column is measured to be v = 356 m s , the
temperature of the air in the column must be
2
2


 356 m s 
v
T = ( 273 K ) 
= ( 273 K ) 
= 316 K

 331 m s 
 331 m s 
14.9
The decibel level β = 10 log ( I I 0 ) , where I 0 = 1.00 × 10 −12 W m 2 .
(a) If β = 100 , then log ( I I 0 ) = 10 giving I = 1010 I 0 = 1.00 × 10 −2 W m 2
(b) If all three toadfish sound at the same time, the total intensity of the sound
produced is I ′ = 3I = 3.00 × 10 −2 W m 2 , and the decibel level is
 3.00 × 10 −2 W m 2 
−12
W m 2 
 1.00 × 10
β ′ = 10 log 
(
)
= 10 log ( 3.00 ) 1010  = 10  log ( 3.00 ) + 10  = 105
Sound
14.10
509
The sound power incident on the eardrum is √ = IA where I is the intensity of the sound
and A = 5.0 × 10 −5 m 2 is the area of the eardrum.
(a) At the threshold of hearing, I = 1.0 × 10 −12 W m 2 , and
(
)(
)
√ = 1.0 × 10 −12 W m 2 5.0 × 10 −5 m 2 = 5.0 × 10 −17 W
(b) At the threshold of pain, I = 1.0 W m 2 , and
(
)(
)
√ = 1.0 W m 2 5.0 × 10 −5 m 2 = 5.0 × 10 −5 W
14.11
The intensity of a spherical sound wave at distance r from a point source is
I = √av 4π r 2 , where √av is the average power radiated by the source. Thus, at distances
r1 = 5.0 m and r2 = 10 km = 10 4 m , the intensities of the sound wave radiating out from
the elephant are
I1 =
√av
4π r12
2
and I 2 =
r 
√av
giving I 2 =  1  I1
2
4π r2
 r2 
From the defining equation, β = 10 log ( I I o ) , the intensity level of the sound at distance
r2 from the elephant is seen to be
2
 r  2 I 
 I2 
I 
I 
 r1 
r 
1
1
 = 10 log   + 10 log  1  = 20 log  1  + 10 log  1 
β2 = 10 log   = 10 log   
 r2 
 r2 
 I0 
 I0 
 I0 
  r2  I 0 
or
14.12
 5.0 m 
+ β1 = −66 dB + 103 dB = 37 dB
 10 4 m 
β2 = 20 log 
Observe that
I 2 200 W m 2
=
= 2.00
I1 100 W m 2
Thus, the difference in the intensity levels of these two sounds is
I I 
 I2 
I 
I 
− 10 log  1  = 10 log  2 0  = 10 log  2 

 I1 
 I0 
 I0 
 I1 I 0 
β2 − β1 = 10 log 
or
β 2 − β1 = 10 log ( 2.00 ) = 10 ( 0.301) = 3.01 dB
510
CHAPTER 14
14.13
From β = 10 log ( I I 0 ) , the intensity for sound level β is I = I 0 10 β 10
The intensity of sound produced by one machine (β = 80 dB) is
(
I1 = I 0 108.0
)
(
)
The intensity needed to reach β = 90 dB is I = I 0 109.0 . Thus, the total number of
machines the factory can accommodate without exceeding 90 dB is
(
(
)
)
9.0
I I 0 10
N= =
= 10
I1 I 0 108.0
Since the factory already contains one of these machines, you can add
9 additional machines without going over the limit.
14.14
(a) The intensity of the sound generated by the orchestra (β = 80 dB) is
IOrch = I 0 10β 10 = I 0 108.0 , and that produced by the crying baby (β = 75 dB) is
(
(
)
)
I b = I 0 107.5 . Thus, the total intensity of the sound engulfing you is
(
I = IOrch + I b = I 0 108.0 + 107.5
(
)(
)
)
= 1.0 × 10 −12 W m 2 1.32 × 108 = 1.32 × 10 −4 W m 2
(b) The combined sound level is
β = 10 log ( I I 0 ) = 10 log (1.32 × 10 8 ) = 81.2 dB
14.15
If the intensity of a sound is I = 4.00 µ W m 2 = 4.00 × 10 −6 W m 2 , the sound level in
decibels is
 4.00 × 10 −6 W m 2 
 I
6
10
log
=
 1.00 × 10 −12 W m 2  = 10 log 4.00 × 10 = 66.0 dB
 I 0 
β = 10 log 
(
)
Sound
14.16
(a)
I=
√
100 W
−2
W m2
=
2 = 7.96 × 10
2
4π r
4π (10.0 m )
 7.96 × 10 −2 W m 2 
 I
(b) β = 10 log   = 10 log 
−12
W m 2 
 I0 
 1.00 × 10
(
)
= 10 log 7.96 × 1010 = 109 dB
(c) At the threshold of pain (β = 120 dB), the intensity is I = 1.00 W m 2 . Thus, from
I = √ 4π r 2 , the distance from the speaker is
r=
14.17
√
=
4π I
100 W
= 2.82 m
4π 1.00 W m 2
(
)
(a) The intensity of sound at 10 km from the horn (where β = 50 dB) is
(
)
I = I 0 10β 10 = 1.0 × 10 −12 W m 2 10 5.0 = 1.0 × 10 −7 W m 2
Thus, from I = √ 4π r 2 , the power emitted by the source is
(
√ = 4π r 2 I = 4π 10 × 10 3 m
) (1.0 × 10
2
−7
)
W m 2 = 1.3 × 10 2 W
(b) At r = 50 m , the intensity of the sound will be
I=
√
1.3 × 10 2 W
−3
=
W m2
2 = 4.0 × 10
2
4π r
4π ( 50 m )
and the sound level is
 4.0 × 10 −3 W m 2 
 I
=
= 10 log 4.0 × 10 9 = 96 dB
10
log
−12
2


W m 
 I0 
 1.0 × 10
β = 10 log 
(
)
511
512
CHAPTER 14
14.18
The intensity at distance r from the source is
(b)
I A rC2 (100 m ) + ( 200 m )
=
=
= 5
I C rA2
(100 m ) 2
2
14.19
Source
2
2
( √ 4π )
√
=
2
4π r
r2
2
rA = 100 m
(a)
I A rB2 (100 m ) + (100 m )
=
=
= 2
I B rA2
(100 m ) 2
I=
rB
A
100 m
rC
B
100 m
C
The sound level for intensity I is β = 10 log ( I I 0 ) . Therefore,
I I 
 I2 
I 
I 
− 10 log  1  = 10 log  2 0  = 10 log  2 

 I1 
 I0 
 I0 
 I1 I 0 
β2 − β1 = 10 log 
Since I =
( √ 4π ) , the ratio of intensities is
√
=
2
4π r
r2
I 2  √ 4π   r12  r12
=
=
I1  r22   √ 4π  r22
2
 r2 
r 
r 
Thus, β2 − β1 = 10 log  12  = 10 log  1  = 20 log  1 
r 
r 
r 
2
14.20
2
2
The general expression for the observed frequency of a sound when the source and/or
the observer are in motion is
 v + vO 
fO = f S 

 v − vS 
Here, v is the velocity of sound in air, vO is the velocity of the observer, vS is the velocity
of the source, and fS is the frequency that would be detected if both the source and
observer were stationary.
(a) If fS = 5.00 kHz and the observer is stationary ( vO = 0 ) , the frequency detected
when the source moves toward the observer at half the speed of sound ( vS = + v 2 )
is
 v+0 
fO = ( 5.00 kHz ) 
 = ( 5.00 kHz )( 2 ) = 10.0 kHz
v−v 2
Sound
513
(b) When fS = 5.00 kHz and the source moves away from a stationary observer at half
the speed of sound ( vS = − v 2 ) , the observed frequency is
 v+0 
2
fO = ( 5.00 kHz ) 
 = ( 5.00 kHz )   = 3.33 kHz
3
v+v 2
14.21
When a stationary observer ( vO = 0 ) hears a moving source, the observed frequency is
 v + vO 
 v 
.
fO = f S 
= fS 

 v − vS 
 v − vS 
(a) When the train is approaching, vS = + 40.0 m s and
(f )
O approach


345 m s
= ( 320 Hz ) 
= 362 Hz
 345 m s − 40.0 m s 
After the train passes and is receding, vS = − 40.0 m s and
(f )
O recede


345 m s
= ( 320 Hz ) 
 = 287 Hz .
 345 m s − ( − 40.0 m s ) 
Thus, the frequency shift that occurs as the train passes is
∆fO = ( fO ) recede − ( fO ) approach = −75.2 Hz , or it is a 75.2 Hz drop
(b) As the train approaches, the observed wavelength is
λ=
(f )
v
O approach
=
345 m s
= 0.953 m
362 Hz
514
CHAPTER 14
14.22
Since the observer hears a reduced frequency, the source and observer are getting
farther apart. Hence, the bicyclist is behind the car .
With the bicyclist (observer) behind the car (source) and both moving in the same
direction, the observer moves toward the source ( vO > 0 ) while the source moves away
from the observer ( vS < 0 ) . Thus, vO = + vbicyclist = + vcar 3 and vS = − vcar where vcar is
the speed of the car.
 v + vcar 3
 v + vO 
= fS 
The observed frequency is fO = fS 

 v − vS 
 v − − vcar
(
)

 v + vcar 3 
 = fS 
,
 v + vcar 

 345 m s + vcar 3 
giving 415 Hz = ( 440 Hz ) 
 and vcar = 32.1 m s
 345 m s + vcar 
14.23
 v + vO 
Both source and observer are in motion, so fO = fS 
. Since each train moves
 v − vS 
toward the other, vO > 0 and vS > 0 . The speed of the source (train 2) is
vS = 90.0
km  1000 m   1 h 


 = 25.0 m s
h  1 km   3600 s 
and that of the observer (train 1) is vO = 130 km h = 36.1 m s . Thus, the observed
frequency is
 345 m s + 36.1 m s 
fO = ( 500 Hz ) 
= 595 Hz
 345 m s − 25.0 m s 
Sound
14.24
515
It is useful to consider this echo in two steps. First, consider the wall to be a stationary
observer ( vO = 0 ) with the source (the bat) approaching at vS = vbat . The frequency of the
 v 
wave reflecting from the wall is f reflect = fS 
. Next, consider the wall as a
 v − vbat 
stationary source ( vS = 0 ) of the reflected wave and the bat to be the approaching
observer ( vO = + vbat ) of this wave. The frequency detected by the bat will be
 v + vbat 
.
fO = f reflect 
 v 
Combining gives the frequency of the echo detected by the bat as
  v    v + vbat 
 v + vbat 
 345 + 5.0 
= ( 40 kHz ) 
= 41 kHz
fO =  f S 
 
 = fS 


 345 − 5.0 
v
 v − vbat 
  v − vbat  
14.25
With the train approaching the stationary observer ( vO = 0 ) at speed vt , the source
velocity is vS = + vt and the observed frequency is
 345 m s 
442 Hz = fS 

 345 m s − vt 
(1)
As the train recedes, the source velocity is vS = − vt and the observed frequency is
 345 m s 
441 Hz = fS 

 345 m s + vt 
Dividing equation (1) by (2) gives
(2)
442 345 m s + vt
=
,
441 345 m s − vt
and solving for the speed of the train yields vt = 0.391 m s
14.26
 115 min 
= 12.0 rad s
(a) ω = 2π f = 2π 
 60.0 s min 
and for harmonic motion,
(
)
vmax = ω A = (12.0 rad s ) 1.80 × 10 −3 m = 0.0217 m s
516
CHAPTER 14
(b) The heart wall is a moving observer ( vO = + vmax ) and the detector a stationary
source, so the maximum frequency reflected by the heart wall is
(f )
wall max
 v + vmax 
 1500 + 0.0217 
= fS 
= ( 2 000 000 Hz ) 
 = 2 000 029 Hz


v
1500


(c) Now, the heart wall is a moving source ( vS = + vmax ) and the detector a stationary
observer. The observed frequency of the returning echo is


v
1500


= ( 2 000 029 Hz ) 
f echo = ( f wall ) max 
 = 2 000 058 Hz


1500 − 0.0217 
 v − vmax 
14.27


 v 
v
For a source receding from a stationary observer, fO = fS 
 = fS 
 . Thus,
 v + vS 
 v − − vS 
the speed the falling tuning fork must reach is
(
)
 f

 512 Hz 
vS = v  S − 1 = ( 340 m s ) 
− 1 = 18.9 m s
 485 Hz 
 fO

The distance it has fallen from rest before reaching this speed is
v 2 − 0 (18.9 m s ) − 0
∆y1 = S
=
= 18.3 m
2 ay
2 9.80 m s 2
2
(
)
The time required for the 485 Hz sound to reach the observer is
t=
∆y1
18.3 m
=
= 0.0538 s
v
340 m s
During this time the fork falls an additional distance
∆y2 = vS t +
(
)
1 2
1
2
ay t = (18.9 m s ) ( 0.0538 s ) + 9.80 m s 2 ( 0.0538 s ) = 1.03 m
2
2
The total distance fallen before the 485 Hz sound reaches the observer is
∆y = ∆y1 + ∆y2 = 18.3 m+1.03 m= 19.3 m
Sound
14.28
517
(b) The half angle of the shock wave is given by
sin θ = vsound vsource = 1 3
Thus, θ = 19.47° . In Figure P14.28(b), we have
x=
20 000 m
h
=
= 56.6 × 10 3 m = 56.6 km
tan θ tan19.47 °
(a) The time required for the plane to travel the distance found above is
t=
14.29
x
vplane
=
56.6 × 10 3 m
= 57.1 s
3 ( 330 m s )
The half-angle of the cone of the shock wave is θ where
 vsound 
 1 
= sin −1 
= 42°

 1.5 
 vsource 
θ = sin −1 
As shown in the sketch, the angle between the direction of
propagation of the shock wave and the direction of the plane’s
velocity is
q
ur
vplane
q
fur
vshock
φ = 90° − θ = 90° − 42° = 48°
14.30
The wavelength of the sound produced by the speaker is
λ=
v 345 m s
=
= 0.863 m
f
400 Hz
(a) If destructive interference is now occurring, one can increase the path length by
λ 2 = 0.431 m to produce constructive interference. This is done by sliding the Utube out a distance of 0.431 m 2 = 0.216 m .
(b) With destructive interference currently taking place, one can increase the path
length by a full wavelength λ = 0.863 m to produce destructive interference again.
518
CHAPTER 14
14.31
At point D, the distance of the ship from point A is
d1 = d22 + ( 800 m ) =
2
( 600 m ) 2 + ( 800 m ) 2
A
800 m
= 1000 m
d1
Since destructive interference occurs for the first time when
the ship reaches D, it is necessary that d1 − d2 = λ 2 , or
B
d2
D
λ = 2 ( d1 − d2 ) = 2 (1000 m − 600 m ) = 800 m
14.32
The wavelength of the sound produced by the speakers is
λ=
v 345 m s
=
= 0.690 m
f
500 Hz
(a) To produce destructive interference, the speaker should be moved back a distance
of d =
λ
2
= 0.345 m .
(b) The speakers will now be separated by a full wavelength and
constructive interference will again occur.
14.33
The wavelength of the sound is
v 345 m s
λ= =
= 0.500 m .
f
690 Hz
Speaker 2
0.700 m
(a) At the first relative maximum
(constructive interference),
Speaker 1
d1 = d2 + λ = d2 + 0.500 m
Using the Pythagorean theorem,
( d2 + 0.500 m) 2 = d22 + ( 0.700 m) 2 , giving d2 =
0.240 m
(b) At the first relative minimum (destructive interference),
d1 = d2 + λ 2 = d2 + 0.250 m
Therefore, the Pythagorean theorem yields
( d2 + 0.250 m) 2 = d22 + ( 0.700 m) 2 , or d2 =
0.855 m
d1
d2
Observer
Sound
14.34
In the fundamental mode of vibration, the wavelength of waves in the wire is
λ = 2 L = 2 ( 0.700 0 m ) = 1.400 m
If the wire is to vibrate at f = 261.6 Hz , the speed of the waves must be
v = λ f = (1.400 m )( 261.6 Hz ) = 366.2 m s
With µ =
m 4.300 × 10 -3 kg
=
= 6.143 × 10 -3 kg m , the required tension is given by
L
0.700 0 m
v = F µ as
(
)
F = v 2 µ = ( 366.2 m s ) 6.143 × 10 -3 kg m = 824.0 N
2
14.35
In the third harmonic, the string forms a standing wave of three loops, each of length
λ 8.00 m
=
= 2.67 m . The wavelength of the wave is then λ = 5.33 m .
2
3
(a) The nodes in this string fixed at each end will occur at distances of
0, 2.67 m, 5.33 m, and 8.00 m from the end. Antinodes occur halfway between
each pair of adjacent nodes, or at 1.33 m, 4.00 m, and 6.67 m from the end.
(b) The linear density is µ =
m 40.0 × 10 -3 kg
=
= 5.00 × 10 -3 kg m
L
8.00 m
F
and the wave speed is v =
Thus, the frequency is f =
µ
v
λ
=
=
49.0 N
= 99.0 m s
5.00 × 10 -3 kg m
99.0 m s
= 18.6 Hz
5.33 m
519
520
CHAPTER 14
14.36
With antinodes at each end and a single node located at the center of the rod, the length of
the rod is one-half wavelength, or
λ = 2L = 2 (1.00 m ) = 2.00 m
The speed of sound in aluminum is v = 5 100 m s (see Table 14.1 in the textbook), so the
frequency of the resonance in the rod is
f=
14.37
v
λ
=
5 100 m s
= 2.55 × 10 3 Hz = 2.55 kHz
2.00 m
The facing speakers produce a standing wave in the space between them, with the
spacing between nodes being
dNN =
λ
2
=
1  v  343 m s
=
= 0.214 m
2  f  2 ( 800 Hz )
If the speakers vibrate in phase, the point halfway between them is an
antinode of pressure, at
1.25 m
= 0.625 m from either speaker.
2
Then there is a node at 0.625 m −
0.214 m
= 0.518 m ,
2
a node at
0.518 m − 0.214 m = 0.303 m ,
a node at
0.303 m − 0.214 m = 0.0891 m ,
a node at
0.518 m + 0.214 m = 0.732 m ,
a node at
0.732 m + 0.214 m = 0.947 m , and
a node at
0.947 m + 0.214 m = 1.16 m from one speaker.
Sound
14.38
521
In a wire of length A is fixed at both ends, the wavelength of the fundamental mode of
vibration is λ1 = 2A . The speed of transverse waves in the wire is v = F µ , where F is
the tension in the wire and µ is the mass per unit length of the wire. The fundamental
frequency for the wire is then
f1 =
v
λ1
=
1
2A
F
µ
If we have two wires with the same mass per unit length, one of length L and under
tension F while the second has length 2L and tension 4F, the ratio of the fundamental
frequencies of the two wires is
f1, long
f1, short
=
(1 2L )
(1 L)
4F µ
Fµ
=
1
4 =1
2
or the two wires have the same fundamental frequency of vibration. If this frequency is
f1 = 60 Hz , then the frequency of the second harmonic for both wires is
f 2 = 2 f1 = 2 ( 60 Hz ) = 120 Hz
14.39
d
(a) From the sketch at the right, notice that when
d = 2.0 m , L =
and
5.0 m − d
= 1.5 m ,
2
ur
F
 d /2 
= 42°
θ = sin 
 L 
−1
Then evaluating the net vertical force on the lowest bit of string,
ΣFy = 2F cos θ − mg = 0 gives the tension in the string as
F=
(
L
)
(12 kg ) 9.80 m s2 = 79 N
mg
=
2cos θ
2cos ( 42°)
q
ur
F
ur
mg
522
CHAPTER 14
(b) The speed of transverse waves in the string is
v=
F
µ
=
79 N
= 2.8 × 10 2 m s
0.001 0 kg m
For the pattern shown, 3 ( λ 2) = d , so λ =
Thus, the frequency is f =
14.40
v
λ
=
2d 4.0 m
=
3
3
(
3 2.8 × 10 2 m s
4.0 m
)=
2.1 × 10 2 Hz
L 2.0 m
 λ
(a) For a standing wave of 6 loops, 6   = L , or λ = =
 2
3
3
The speed of the waves in the string is then
 2.0 m 
v=λf =
150 Hz ) = 1.0 × 10 2 m s
(

 3 
(
)
Since the tension in the string is F = mg = ( 5.0 kg ) 9.80 m s 2 = 49 N
v = F µ gives µ =
F
49 N
=
2
v
1.0 × 10 2 m
(
(
)
2
= 4.9 × 10 −3 kg m
)
(b) If m = 45 kg , then F = ( 45 kg ) 9.80 m s 2 = 4.4 × 10 2 N , and
v=
4.4 × 10 2 N
= 3.0 × 10 2 m s
4.9 × 10 −3 kg m
Thus, λ =
v 3.0 × 10 2 m s
=
= 2.0 m
f
150 Hz
and the number of loops is n =
L
2.0 m
=
= 2
λ 2 1.0 m
Sound
(
523
)
(c) If m = 10 kg , the tension is F = (10 kg ) 9.80 m s 2 = 98 N , and
98 N
= 1.4 × 10 2 m s
4.9 × 10 −3 kg m
v=
v 1.4 × 10 2 m s
=
= 0.94 m
f
150 Hz
Then, λ =
L
2.0 m
=
is not an integer,
λ 2 0.47 m
and n =
so no standing wave will form .
14.41
The speed of transverse waves in the string is
F
v=
=
µ
50.000 N
= 70.711 m s
1.000 0 × 10 −2 kg m
The fundamental wavelength is λ1 = 2 L = 1.200 0 m and its frequency is
f1 =
v
λ1
=
70.711 m s
= 58.926 Hz
1.200 0 m
The harmonic frequencies are then
f n = nf1 = n ( 58.926 Hz ) , with n being an integer
The largest one under 20 000 Hz is f 339 = 19 976 Hz = 19.976 kHz
14.42
The distance between adjacent nodes is one-quarter of the circumference.
dNN = dAA =
λ
2
=
20.0 cm
= 5.00 cm
4
so
λ = 10.0 cm=0.100 m ,
and
f=
v
λ
=
900 m s
= 9.00 × 10 3 Hz = 9.00 kHz
0.100 m
The singer must match this frequency quite precisely for some interval of time to feed
enough energy into the glass to crack it.
524
CHAPTER 14
14.43
Assuming an air temperature of T = 37°C = 310 K , the speed of sound inside the pipe is
v = ( 331 m s )
T
310 K
= ( 331 m s )
= 353 m s
273 K
273 K
In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at
one end is λ = 4 L . Thus, for the whooping crane
λ = 4 ( 5.0 ft ) = 2.0 × 101 ft
f=
and
14.44
v
λ
=
( 353 m s)  3.281 ft  =


2.0 × 101 ft  1 m 
58 Hz
(a) In the fundamental resonant mode of a pipe open at both ends, the distance
between antinodes is dAA = λ 2 = L .
Thus, λ = 2 L = 2 ( 0.320 m ) = 0.640 m
f=
and
(b) dAA =
14.45
λ
=
1 v 
=  =
2 2 f 
λ
340 m s
= 531 Hz
0.640 m
1  340 m s 

 = 0.042 5 m = 4.25 cm
2  4 000 Hz 
Hearing would be best at the fundamental resonance, so λ = 4 L = 4 ( 2.8 cm )
and
14.46
v
f=
v
λ
=
340 m s  100 cm 
3

 = 3.0 × 10 Hz = 3.0 kHz
1m
4 ( 2.8 cm )
For a closed box, the resonant frequencies will have nodes at both sides, so the permitted
nλ nv
=
, ( n = 1, 2, 3, . . .) .
wavelengths will be L =
2
2f
Thus,
fn =
nv
. With L = 0.860 m and L′ = 2.10 m , the resonant frequencies are:
2L
f n = n ( 206 Hz ) for L = 0.860 m for each n from 1 to 9
and
f n = n ( 84.5 Hz ) for L′ = 2 .10 m for each n from 2 to 23
Sound
14.47
(a) The speed of sound is 331 m s at 0 °C, so the fundamental wavelength of the pipe
open at both ends is
λ1 = 2 L =
331 m s
v
v
giving L =
=
= 0.552 m
f1
2 f1 2 ( 300 Hz )
(b) At T = 30 ° C = 303 K , v = ( 331 m s )
f1 =
and
14.48
525
v
λ1
=
T
303 K
= ( 331 m s )
= 349 m s
273 K
273 K
349 m s
v
=
= 316 Hz
2 L 2 ( 0.552 m )
For a pipe open at both ends, the frequency of the nth harmonic is, f n = n ( v 2 L) . Thus,
the difference between two successive resonant frequencies is
 v 
 v 
v
∆f = f n + 1 − f n = ( n + 1)   − n   =
 2 L
 2 L 2 L
In this case, L = 2.00 m and ∆f = 492 Hz − 410 Hz = 82 Hz . Thus, the speed of sound in
the pipe is
v = 2 L ( ∆f ) = 2 ( 2.00 m )( 82 Hz ) = 3.3 × 10 2 m s
14.49
Since the lengths, and hence the wavelengths of the first harmonics, of the strings are
identical, the ratio of their fundamental frequencies is
f1′ v ′ λ1 v ′
 v′ 
=
= , or f1′ = f1  
 v
f1 v λ 1
v
Thus, the beat frequency heard when the two strings are sounded simultaneously is
f beat = f1 − f1′ = f1 (1 − v ′ v )
From v = F µ , the speeds of transverse waves in the two strings are
v=
200 N
µ
and v ′ =
(
196 N
µ
)
, so
v′
=
v
196 N
= 0.980
200 N
Therefore, f beat = ( 523 Hz ) 1 − 0.980 = 5.26 Hz
526
CHAPTER 14
14.50
By shortening her string, the second violinist increases its fundamental frequency. Thus,
f1′ = f1 + fbeat = (196 + 2.00 ) Hz = 198Hz .
Since the tension and the linear density are both identical for the two strings, the speed
of transverse waves, v = F µ , has the same value for both strings.
Thus, λ1′ f1′ = λ1 f1 , or λ1′ = λ1 ( f1 f1′) . Since the fundamental wavelength of a string fixed
at both ends is λ = 2L , this yields
 f 
 196 
L′ = L  1  = ( 30.0 cm ) 
= 29.7 cm
 198 
 f1′
14.51
If the second train is moving toward the stationary observer with speed vS , the Doppler
 v 
effect gives fO = fS 
 > fS
 v − vS 
Therefore, fO = fS + f beat = 180 Hz + 2 Hz = 182 Hz , and the speed of the train is

f 
180 

vS = v  1 − S  = ( 345 m s )  1 −
 = 3.79 m s

fO 
182 

and the velocity of the train is vS = 3.79 m s toward the station
(
)
If the second train is moving away from the stationary observer vS = − vS , then
 v 
fO = f S 
 < fS , giving fO = fS − f beat = 180 Hz − 2 Hz = 178 Hz
 v + vS 
Thus,
 f

 180 
vS = v  S − 1 = ( 345 m s ) 
− 1 = 3.88 m s
 178 
 fO

and the velocity of the train is vS = 3.88 m s away from the station
Sound
14.52
527
The temperatures of the air in the two pipes are T1 = 27°C = 300 K and
T2 = 32°C = 305 K . The speed of sound in the two pipes is
v1 = ( 331 m s )
T1
273 K
and v2 = ( 331 m s )
T2
273 K
Since the pipes have the same length, the fundamental wavelength, λ = 4 L , is the same
for them. Thus, from f = v λ , the ratio of their fundamental frequencies is seen to be
f 2 f1 = v2 v1 , which gives f 2 = f1 ( v2 v1 ) .
 T

v

The beat frequency produced is then f beat = f 2 − f1 = f1  2 − 1 = f1  2 − 1
 v1

 T1

or
14.53
 305 K

f beat = ( 480 Hz ) 
− 1 = 3.98 Hz
 300 K

(a) First consider the wall a stationary observer receiving sound from an approaching
source having velocity va . The frequency received and reflected by the wall is
 v 
.
f reflect = fS 
 v − va 
Now consider the wall as a stationary source emitting sound to an observer
approaching at velocity va . The frequency of the wave heard by the observer is
 v   v + va 
 v + va 
 v + va 
fO = f reflect 
= fS 
=
f



S
 v − v 
 v 
 v − va   v 
a
Thus, the beat frequency between the tuning fork and its echo is
 v + va

 2 va 
 2 (1.33) 
− 1 = fS 
= ( 256 Hz ) 
= 1.98 Hz
f beat = fO − fS = fS 

 345 − 1.33 
v
v
v
v
−
−



a
a
528
CHAPTER 14
(b) When the student moves away from the wall, va changes sign so the beat frequency
heard is
(
)
 2 − va  2 fS va
v f beat
f beat = fS 
, giving va =
=
2 fS − fbeat
 v − − va  v + v a
(
)
The receding speed needed to observe a beat frequency of 5.00 Hz is
va =
14.54
( 345 m s) ( 5.00 Hz)
2 ( 256 Hz ) − 5.00 Hz
= 3.40 m s
The extra sensitivity of the ear at 3000 Hz appears as downward dimples on the curves
in Figure 14.29 of the textbook.
At T = 37°C = 310 K , the speed of sound in air is
v = ( 331 m s )
T
310 K
= ( 331 m s )
= 353 m s
273 K
273 K
Thus, the wavelength of 3 000 Hz sound is λ =
v 353 m s
=
= 0.118 m
f 3 000 Hz
For the fundamental resonant mode in a pipe closed at one end, the length required is
L=
14.55
λ
4
=
0.118 m
= 0.0294 m = 2.94 cm
4
At normal body temperature of T = 37° C = 310 K , the speed of sound in air is
v = ( 331 m s )
T
310 K
= ( 331 m s )
= 353 m s
273 K
273 K
and the wavelength of 20 000 Hz sound is
λ=
353 m s
v
=
= 1.76 × 10 −2 m = 1.76 cm
f 20 000 Hz
Thus, the diameter of the eardrum is 1.76 cm
Sound
14.56
529
(a) If a source emits sound of frequency fS (as detected by an observer stationary
relative to the source), the frequency detected by the observer when the source
and/or the observer is in motion is fO = fS ( v + vO ) ( v − vS ) where v is the velocity of
sound in air, vO is the velocity of the observer, and vS is the velocity of the source.
In the given situation, fS = 320 Hz, vO = 0, and when the train is approaching the
observer, vS = + 40 m s . Thus, the frequency heard by the observer is


345 m s + 0
fO , a = ( 320 Hz ) 
 = 362 Hz
 345 m s − 40.0 m s 
(b) When the train is receding from the stationary observer, vS = − 40.0 m s and the
detected frequency will be

345 m s + 0
fO , b = ( 320 Hz ) 
 345 m s − ( − 40.0 m s )

c)
The wavelengths measured by the observer in each of the 2 cases above are
λa =
14.57
v
fO , a
=
345 m s
345 m s
v
= 0.953 m and λb =
=
= 1.20 m
fO , b
362 Hz
287 Hz
This situation is very similar to the fundamental resonance of an organ pipe that is open
at both ends. The wavelength of the standing waves in the crystal is λ = 2 t , where t is
the thickness of the crystal, and the frequency is
f=
14.58

 345 m s 
 = ( 320 Hz ) 
 = 287 Hz
 385 m s 

v
λ
=
3.70 × 10 3 m s
(
2 7.05 × 10 -3 m
)
= 2.62 × 10 5 Hz = 262 kHz
The distance from the balcony to the man's head is
∆y = 20.0 m − 1.75 m=18.3 m
The time for a warning to travel this distance is t1 =
18.3 m
= 0.0529 s , so the total
345 m s
time needed to receive the warning and react is t1′ = t1 + 0.300 s = 0.353 s .
The time for the pot to fall this distance, starting from rest, is
t2 =
2 ( ∆y )
ay
=
2 ( −18.3 m )
= 1.93 s
−9.80 m s 2
530
CHAPTER 14
Thus, the latest the warning should be sent is at
t = t2 − t1′ = 1.93 s − 0.353 s = 1.58 s
into the fall. In this time interval, the pot has fallen
∆y =
(
)
1 2 1
2
g t = 9.80 m s 2 (1.58 s ) = 12.2 m
2
2
and is h = 20.0 m − 12.2 m= 7.8 m above the sidewalk.
14.59
On the weekend, there are one-fourth as many cars passing per minute as on a week
day. Thus, the intensity, I 2 , of the sound on the weekend is one-fourth that, I1 , on a
week day. The difference in the decibel levels is therefore:
β1 − β 2
so,
14.60
=
I 
I 
I 
10 log  1  − 10 log  2  = 10 log  1  = 10 log(4)=6 dB
 I2 
 Io 
 Io 
β2 = β1 − 6 dB = 70 dB − 6 dB = 64 dB
The length of the air column when the first resonance is heard is L1 = λ a 4 , where λ a is
the wavelength of the sound in air. Thus, λ a = 4L1 = 4 ( 0.340 m ) = 1.36 m
The frequency of the sound wave, and hence the vibrating wire producing the sound, is
f=
vsound
λa
=
340 m s
=250 Hz
1.36 m
When the wire vibrates in its third harmonic, its length is Lw = ( 3 2) λ w where λ w is the
wavelength of the waves traveling in the wire.
2 Lw 2 (1.20 m )
=
= 0.800 m , and the speed of transverse waves in the
3
3
vw = λ w f = ( 0.800 m )( 250 Hz ) = 200 m s
Therefore, λ w =
wire is
Sound
14.61
531
The maximum speed of the oscillating block and speaker is
vmax = Aω = A
20.0 N m
k
= ( 0.500 m )
= 1.00 m s
m
5.00 kg
When the speaker (sound source) moves toward the stationary observer, then vS = + vmax
and the maximum frequency heard is
(f )
O max



345 m s
v 
= fS 
= ( 440 Hz ) 
= 441 Hz

 v − vmax 
 345 m s − 1.00 m s 
When the speaker moves away from the stationary observer, the source velocity is
vS = − vmax and the minimum frequency heard is



345 m s
v
 = 439 Hz
 = ( 440 Hz ) 
 v + vmax 
 345 m s + 1.00 m s 

( fO )min = fS 
14.62
(a) At T = 20° C = 293 K , the speed of sound in air is
v = ( 331 m s )
T
293 K
= ( 331 m s )
= 343 m s
273 K
273 K
The first harmonic or fundamental of the flute (a pipe open at both ends) is given by
343 m s
v
λ1 = 2L = =
= 1.31 m . Therefore, the length of the flute is
f1 261.6 Hz
L=
λ1
2
=
1.31 m
= 0.655 m
2
532
CHAPTER 14
(b) In the colder room, the length of the flute and hence its fundamental wavelength is
essentially unchanged (that is, λ1′ = λ1 = 1.31 m ). However, the speed of sound and
thus the frequency of the fundamental will be lowered. At this lower temperature,
the frequency must be
f1′ = f1 − fbeat = 261.6 Hz − 3.00 Hz=258.6 Hz
The speed of sound in this room is
v ′ = λ1′ f1′ = (1.31 m )( 258.6 Hz ) = 339 m s
From v = ( 331 m s) T 273 K , the temperature in the colder room is given by
2
2


 339 m s 
v
T = ( 273 K ) 
= ( 273 K ) 
= 286 K = 13.0°C

 331 m s 
 331 m s 
14.63
The frequency heard from the first train, moving toward the stationary observer at
( vS )1 = + 30.0 m s , is
(f )
O 1




345 m s
v
= fS 
= 328.6 Hz
 = ( 300 Hz ) 

 345 m s − 30.0 m s 
 v − ( vS )1 
The second train moves toward the observer at ( vS ) 2 = + v2 > 30.0 m s . The frequency
heard from this train must be
(f ) =(f )
O 2
O 1
+ f beat = 328.6 Hz + 3.0 Hz=331.6 Hz


 v 
v
Then, from ( fO ) 2 = fS 
, the speed of the approaching second train
 = fS 
 v − v2 
 v − ( vS ) 2 
must be

f 
300 Hz 

v2 = v  1 − S  = ( 345 m s )  1 −
 = 32.9 m s

331.6 Hz 
( fO ) 2 

Sound
14.64
(a) The wavelength of the original note is
345 m s
v
=
= 0.233 m
f 1.480 × 10 3 Hz
λ=
(b) The wavelength of the transposed note is
λ′ =
345 m s
v
=
= 0.247 m
f ′ 1.397 × 10 3 Hz
and the increment of change is
∆λ = λ ′ − λ = ( 0.247 − 0.233) m=0.014 m= 14 mm
14.65
The frequency heard from the stationary speaker the student is approaching
vO = + vO is
(
)
(f )
O 1
 v + vO 
= fS 

 v 
The frequency heard from the stationary speaker the student moves away from
vO = − vO is
(
)
(f )
O 2
 v − vO 
= fS 

 v 
Thus, the beat frequency heard is
f beat = ( fO )1 − ( fO ) 2 =
2 fS vO
v
With vO = 1.50 m s , this gives
f beat =
2 ( 456 Hz ) ( 1.50 m s )
345 m s
= 3.97 Hz
533
534
CHAPTER 14
14.66
(a) The wavelength of the sound generated by the speakers is
λ=
v 344 m s
=
= 16.0 m
f
21.5 Hz
At point A, the distance from the observer to one speaker exceeds the distance to
the other speaker by
∆d = d1 − d2 = 9.00 m − 1.00 m = 8.00 m
Observe that this path difference is ∆d = λ 2 . Thus, waves arriving at point A
from the two speakers meet out of phase and undergo destructive interference .
(b) The coordinates of points on the path should be so the path difference
∆d = d1 − d2 =
2
2
( x + 5.00 m ) + y 2 − ( x − 5.00 m ) + y 2
will always equal λ 2 .
( x + 5.00 m ) 2 + y 2
Thus, we write
=
( x − 5.00 m ) 2 + y 2
+ 8.00 m
Square both sides of this equation and simplify to find
5x
− ( 4.00 m ) =
4
( x − 5.00 m ) 2 + y 2
Again, square both sides and simplify to obtain 9x 2 − 16 y 2 = 144 m 2 ,
or
14.67
x2 y2
−
= 1.00 m 2
16 9
This is the equation of a hyperbola.
The speeds of the two types of waves in the rod are
vlong =
Y
ρ
=
Y ( A ⋅ L)
Y
=
and vtrans =
mV
m
Thus, if vlong = 8 vtrans , we have
(
F
µ
=
F
=
mL
F⋅L
m
Y ( A ⋅ L)
 F ⋅ L
, or the required tension is
= 64 
 m 
m
) (
)
2
6.80 × 1010 N m 2 π 0.200 × 10 −2 m 
Y⋅A

 = 1.34 × 10 4 N
F=
=
64
64
Sound
14.68
535
The speed of transverse waves in the wire is
F
v=
µ
=
F⋅L
=
m
( 400 N )( 0.750 m )
2.25 × 10 −3 kg
= 365 m s
When the wire vibrates in its third harmonic, λ = 2L 3 = 0.500 m , so the frequency of the
vibrating wire and the sound produced by the wire is
f=
v
λ
=
365 m s
= 730 Hz
0.500 m
Since both the wire and the wall are stationary, the frequency of the wave reflected from
the wall matches that of the waves emitted by the wire. Thus, as the student approaches
the wall at speed vO , he approaches one stationary source and recedes from another
stationary source, both emitting frequency fS = 730 Hz . The two frequencies that will be
observed are
(f )
O 1
 v + vO 
 v − vO 
= fS 
and ( fO ) 2 = fS 


 v 
 v 
 v + vO − ( v − vO
The beat frequency is f beat = ( fO )1 − ( fO )2 = fS 

v

so
14.69
f 
 8.30 Hz 
vO =  beat  v = 
 ( 340 m s ) = 1.93 m s
 2 fS 
 2 ( 730 Hz ) 
The speed of the trailing ship (the source) is
 0.278 m s 
vS = ( 64.0 km h ) 
= 17.8 m s
 1 km h 
and that of the leading ship (observer) is
 0.278 m s 
vO = ( 45.0 km h ) 
= 12.5 m s
 1 km h 
)  = 2 f


S
vO
v
536
CHAPTER 14
The trailing ship (source) moves towtard the leading ship (observer), so vS = + vS . The
observer is moving away from the source, so vO = − vO . The observed frequency is
therefore
 v − vO 
 1 520 m s − 12.5 m s 
 v + vO 
fO = f S 
f
=
= (1 200 Hz ) 
= 1 204 Hz
S



 v − vS 
 1 520 m s − 17.8 m s 
 v − vS 
14.70
(a) If the source and the observer are moving away from each
other along the same line, the relevant angles as defined
in the sketch at the right are θS = 180° and θO = 180° .
Thus,
fS
 v + vO cos θO 
fO = 
 fS
 v − vS cos θS 
vS
qS
vO
qO
fO
 v − vO 
becomes fO = 
 fS
 v + vS 
which is the same as Equation 14.12 in the text when both vO and vS are negative
(that is, when source and observer move away from each other).
(b) When the train is 40.0 m to the left of the crossing and the car is 30.0 m from the
 30.0 m 
tracks, θS = tan −1 
= tan −1 ( 0.750 ) = 36.9° . Then, with vO = 0 , the observed
 40.0 m 
frequency is




343 m s
v
fO = 
f
=
S

 ( 500 Hz ) = 531 Hz
 v − vS cos θS 
 343 m s − ( 25.0 m s) cos 36.9° 
14.71
(a) The distance the sound traveled in 2.00 s is
t=0
d = vsound t = ( 345 m s ) ( 2.00 s ) = 690 m
Observe the sketch at the right and apply the
Pythagorean theorem to obtain
2
 h
h2 +   = d2
 2
or
5 2
h = d2
4
Thus, the altitude of the plane is h =
2d 2 ( 690 m )
=
= 617 m
5
5
h
t = 2.00 s
h¤2
d
Sound
v=
(b) The speed of the plane is
14.72
537
h2
617 m
=
= 154 m s
t
2 ( 2.00 s )
(a) When the observer is stationary with a source emitting sound of frequency
fS = 1.80 kHz moving toward it, the frequency detected by the observer is
 v+0
 v + vO 
fO = f S 
 = fS 
 v − vS 
 v − vS



where vS is the speed of the source, and v = 343 m s is the speed of sound in the
air. Thus, if the detected frequency is fO = 2.15 kHz , the speed of the source is

f 
1.80 kHz 

vS = v  1 − S  = ( 343 m s )  1 −
 = 55.8 m s
f
2.15 kHz 

O 

(b) In this case, the skydiver is a moving observer with velocity vO = + 58.8 m s ,
moving toward a stationary source (the ground) that is emitting (reflecting) sound
of frequency fS = 2.15 kHz . The frequency detected by the skydiver will be
 v + vO
fO = f S 
 v − vS
 343 m s + 55.8 m s 

 = 2.50 kHz
 = ( 2.15 kHz ) 
343 m s − 0


