63. The speed of a wave is determined by the properties of the medium and does not change with wavelength or frequency.
However, the wave speed is equal to the product of the frequency and the wavelength. Therefore, if the wavelength of a sound
wave is doubled, the frequency of the waves will decrease (in fact, they will be cut in half) while the wave speed will stay the
same.
64.
Picture the Problem: A 245-Hz tone and a 240-Hz tone are played together and produce a beat frequency. A 140-Hz tone and a
145-Hz tone produce beats in a like manner.
Strategy: The beat frequency is the difference between the tone frequencies.
Solution: 1. (a) The beat frequency for the first two tones is 245 – 240 Hz = 5 Hz. The beat frequency for the second two tones
is 145 – 140 Hz = 5 Hz. We conclude that the beat frequency for the first two tones is equal to the beat frequency produced by
the second two tones.
2. (b) The best explanation is A. The beat frequency is determined by the difference between the frequencies and is independent
of their actual values. Statements B and C are both false.
Insight: For instance, the beat frequency produced by tones of 123 Hz and 128 Hz is larger than the beat frequency produced by
two tones of frequency 12,600 Hz and 12,602 Hz.
68.
Picture the Problem: The wavelength of sound in air can be calculated from its frequency.
Strategy: Solve the equation v   f for the wavelength, using 343 m/s for the speed of sound in air.
Solution: 1. (a) Calculate the wavelength:

v 343 m/s

 0.807 m
f
425 Hz
2. (b) Wavelength is inversely related to frequency, so if the frequency is increased the wavelength will decrease.
3. (c) Calculate the wavelength at 475 Hz:

343 m/s
 0.722 m
475 Hz
Insight: As predicted, an increase in frequency corresponds to a decrease in wavelength.
71. The sliding part of a trombone varies the length of the vibrating air column that produces the trombone’s sound. By adjusting this
length, the player controls the resonant frequencies of the instrument. This, in turn, varies the frequency
of sound produced by the trombone.
73.
Picture the Problem: Blowing across the opening of a partially filled two-liter plastic bottle produces a tone.
Strategy: A standing wave is formed inside the plastic bottle, the frequency of which depends upon the length of the column of
air trapped in the bottle.
Solution: 1. (a) Pouring out some of the water lengthens the vibrating column of air, which means that the wavelength of the
sound is longer as well. Because the speed of sound is the same, it follows that the new tone will have a lower frequency than
the first tone.
2. (b) The best explanation is C. A lower level of water results in a longer column of air and hence a lower frequency.
Statements A and B are both false.
Insight: You can also produce a slightly lower tone by cooling the air in the bottle, reducing the speed of sound.
75.
Picture the Problem: An organ pipe that is L = 3.5 meters long and closed at one end resonates at its
fundamental frequency.
Strategy: Because the pipe is closed at one end, use f n  nv 4L with n = 1 to calculate the fundamental
frequency:
Solution: Calculate the fundamental frequency:
f1 
nv 1343 m/s 

 25 Hz
4L
4 3.5 m
Insight: When a pipe is closed at one end the fundamental wavelength is equal to 4 times the length of
the pipe, or 14 meters in this problem.
84. When a person travels toward a stationary source of sound, she is a moving observer that intercepts more wavefronts per second
than she would if she stood still. Therefore, because of the Doppler effect the frequency of the sound detected by the person is
greater than when the person is at rest. Likewise, the wavelength of the sound detected by the person is less than the wavelength
the person would measure if she were at rest.
87.
Picture the Problem: The train, a moving source, sounds its horn, and a person
standing near the tracks observes the frequency of the sound.
Strategy: Use the Doppler effect for a moving source to calculate the frequency
observed by the person. Choose the minus sign because the source is moving toward
the observer.
Solution: Calculate f observer with the Doppler effect equation
for a moving source, choosing the minus sign for a source
that moves toward the observer:
f observer 
fsource
 vsource 
1  v

sound

136 Hz
 150 Hz
 31.8 m/s 
1



343 m/s 
Insight: If the train were moving away from the observer, he would hear a frequency of 124 Hz.
88.
Picture the Problem: A stationary person sounds a horn as a train approaches
him, and a person on the train observes the frequency of the sound.
Strategy: Use the Doppler effect for a moving observer to calculate the
frequency observed by the person. Choose the positive sign because the observer
is moving toward the source.
Solution: Calculate f observer with the Doppler effect
equation for a moving observer, choosing the plus sign for
an observer that moves toward the source:
 v

f observer  fsource 1  observer 
vsound 

 31.8 m/s 
 136 Hz  1 
  149 Hz

343 m/s 
Insight: The frequency is slightly lower than the frequency found in the previous problem, where the source was moving toward
an observer who was at rest.
94. Sound spreads out over a larger area as it moves away from its source. The energy from a point source of sound is spread out
over the surface of the expanding sphere of the sound wave, and the surface area of a sphere is 4  r 2 . Therefore, the intensity of
a point source of sound decreases with the square of the distance r from the source.