Chapter 14 Solutions
1.
Picture the Problem: The image shows a wave with the given wave
dimensions.
Strategy: Set the wavelength equal to the horizontal crest-to-crest
distance, or double the horizontal crest-to-trough distance. Set the
amplitude equal to the vertical crest-to-midline distance, or half the
vertical crest-to-trough distance.
13.
Solution: 1. (a) Double the horizontal crest-to-trough distance:
  2  28 cm   56 cm
2. (b) Halve the vertical crest-to-trough distance:
A  12 13 cm   6.5 cm
Picture the Problem: The string tension is changed until the wave speed doubles.
Strategy: The speed of a wave on a string is given by equation 14-2. Solve the equation for the tension in the string. Then use
a ratio to find the factor by which the tension increases.
Solution: 1. Solve equation 14-2 for the tension:
v  F   F  v2 
2. Divide the tension at higher velocity by the initial tension:
F2 v22   v2   32 m/s 



 4
F v 2   v   16 m/s 
2
2
The tension increases by a factor of 4.
23.
Picture the Problem: We are given the equation describing a wave and wish to determine the amplitude, wavelength, period,
speed, and direction of travel.
2 
 2
Strategy: The general form of a wave is given by y  A cos 
x
t  . Compare this equation to
T 
 
 
 
y  15 cm  cos 
x
t  , the equation given in the problem, to identify the wave parameters. Use equation 14-1 and
12 s 
 5.0 cm
the definition of frequency to calculate the wave speed.
Solution: 1. (a) Identify the amplitude as A:
2. (b) Identify the wavelength as :
A = 15 cm
2



5.0 cm
, so   10 cm  0.10 m
3. (c) Identify the period as T:
2


, so T = 24 s
T
12 s
4. (d) Use equation14-1 to calculate the speed:
vf 

T

10 cm
 0.42 cm/s
24 s
5. (e) The wave travels to the right, because the t-term and x-term have opposite signs.
31.
Picture the Problem: The figure shows a person throwing a rock down an 8.85-m deep
well. The sound of the splash reaches the person’s ear 1.20 seconds after the rock is
thrown.
Strategy: Solve equation 2-11 for the initial velocity of the rock, where the fall time is
equal to the total time minus the sound travel time. The sound travel time ts is the depth
of the well divided by the speed of sound.
d 8.85 m

 0.0258 s
v 343m/s
Solution: 1. Calculate ts :
ts 
2. Subtract ts from the total time:
tf  1.20 s  0.0258 s  1.1742 s
3. Solve equation 2-11 for v0 :
1
y  y0  v0 tf  atf 2
2
y  y0 1
v0 
 atf
tf
2

 8.85 m 1
   9.81 m/s 2  1.1742 s   1.78 m s
1.1742 s 2
The initial velocity of the rock is 1.78 m/s downward.
33.
Picture the Problem: We are given the intensity level in a truck and need to calculate the sound intensity.
Strategy: Use equation 14-8 to convert the sound intensity level to sound intensity.
Solution: Solve equation 14-8 for the sound intensity:
 I 

 I0 
  10 log 
I  I 0 10 /10  1012 W/m 2 1092/10   1.6 mW m 2
44.
Picture the Problem: The image shows a car approaching a stationary
observer. The horn has a frequency of 450 Hz, but the observer hears a
frequency of 470 Hz. We want to calculate the speed of the car.
Strategy: Because the source is moving and the observer is stationary, use
equation 14-10 to calculate the speed of the car. Since the observed
frequency is greater than the initial frequency, the car must be moving
toward the observer, so we use the minus sign.
Solution: 1. Solve equation 14-10 for the
bus speed:
2. Insert the given frequencies:
 1 
f
f
 1 u v 
 f 

f 
 450 Hz 
1  u v     u  1   v  1 
  343 m/s   15 m s


f
f
 470 Hz 
 


 450 Hz 
u  1 
  343 m/s   15 m s
 470 Hz 
46.
Picture the Problem: A stationary person sounds a 136-Hz horn as a
train approaches him at 31.8 m/s. We want to know at what frequency
a passenger on the train hears the horn.
Strategy: This problem has a stationary source and an approaching
observer, the passenger. Use equation 14-9 (with a plus sign) for the
observed frequency.
Solution: Insert the given data into equation 14-9:
f   1  u v  f
 1   31.8 m/s 343 m/s   136 Hz   149 Hz
48.
Picture the Problem: The image shows a motorcyclist and police car
approaching each other.
Strategy: We want to calculate the frequency at which the motorcyclist
hears the police car’s siren. In this problem both the source (police car)
and the observer (motorcycle) are moving toward each other. Use
equation 14-11, with the plus sign in the numerator and the minus sign
in the denominator, to calculate the observed frequency.
 1  uo v 
f
f
 1  us v 
 1  13.0 m/s 343 m/s  

  502 Hz   566 Hz
1   27.0 m/s 343 m/s  
Solution: Insert the speeds and emitted
frequency into equation 14-11:
65.
Picture the Problem: The image shows two out-of-phase speakers separated by 3.5
meters and an observer standing by a wall 5.0 meters away. When the observer
moves 0.84 meters along the wall he goes from the central destructive interference
to the first constructive interference. We want to calculate the frequency of sound
emitted by the speaker.
Strategy: Use the Pythagorean Theorem to calculate the distance of the observer
from each speaker. Calculate the difference in distances to each speaker. Because
the speakers are out of phase, constructive interference occurs when the difference
in distances is equal to a half wavelength. Set the difference in distances equal to a
half wavelength and use equation 14-1 to calculate the resulting frequency.
Solution: 1. Calculate d1 :
 5.0 m 
d1 
2
  12  3.5 m   0.84 m 
2
 5.631 m
5.0 m
  12  3.5 m   0.84 m  5.082 m
2
2. Calculate d 2 :
d2 
3. Set d1  d2  12  :
  2  d1  d2   2 5.631 m  5.082 m =1.098 m
4. Use equation 14-1 to
calculate the frequency:
f 
v


2
343 m/s
 0.31 kHz
1.098 m)
68.
Picture the Problem: The image shows an organ pipe 3.5 meters long and
open at both ends.
Strategy: Because the pipe is open at both ends, use equation 14-15, with
n = 1 to calculate the fundamental frequency:
Solution: Set n = 1 in equation 14-15
and solve for the frequency:
69.
f1 
nv 1 343 m/s 

 49 Hz
2L
2  3.5 m 
Picture the Problem: The image shows a wire of length 1.50 m oscillating at its
fundamental frequency. We are given the mass and tension in the wire and need to
calculate the fundamental frequency.
Strategy: Calculate the speed of the wave on the wire using equation 14-2. Then
use equation 14-13, with n = 1, to calculate the fundamental frequency.
Solution: 1. Solve equation 14-2 for the wave speed:
v
2. Solve equation. 14-13 for f1 :
f1 
F


93 N
 232 m/s
 0.0026 kg  1.5 m 
v
232 m/s

 77 Hz
2 L 2 1.5 m 