24 April 2011

Problems in Sound

1. An automatic focus camera is able to focus on objects by use of an ultrasonic sound wave. The camera sends out sound waves that reflect off distant objects and return to the camera. A sensor detects the time it takes for the waves to return and then determines the distance an object is from the camera. If a sound wave (speed = 340 m/s) returns to the camera 0.150 seconds after leaving the camera, how far away is the object?

Answer = 25.5 m

The speed of the sound wave is 340 m/s. The distance can be found using d = v • t resulting in an answer of 25.5 m. Use 0.075 seconds for the time since 0.150 seconds refers to the round-trip distance.

2. On a hot summer day, a pesky little mosquito produced its warning sound near your ear. The sound is produced by the beating of its wings at a rate of about 600 wing beats per second.
a. What is the frequency in Hertz of the sound wave?
b. Assuming the sound wave moves with a velocity of 350 m/s, what is the wavelength of the wave?

Part a Answer: 600 Hz (given)
Part b Answer: 0.583 meters 

Let = wavelength. Use v = f • where v = 350 m/s and f = 600 Hz. Rearrange the equation to the form of
= v / f. Substitute and solve.

3. Doubling the frequency of a wave source doubles the speed of the waves.
a. True b. False

Answer: B 

Doubling the frequency will halve the wavelength; speed is unaffected by the alteration in the frequency. The speed of a wave depends upon the properties of the medium.
4. Playing middle C on the piano keyboard produces a sound with a frequency of 256 Hz. Assuming the speed of sound in air is 345 m/s, determine the wavelength of the sound corresponding to the note of middle C.

Answer: 1.35 meters (rounded)

Let = wavelength. Use v = f • where v = 345 m/s and f = 256 Hz. Rearrange the equation to the form of = v / f. Substitute and solve.

5. Most people can detect frequencies as high as 20 000 Hz. Assuming the speed of sound in air is 345 m/s, determine the wavelength of the sound corresponding to this upper range of audible hearing.

Answer: 0.0173 meters (rounded)

Let = wavelength. Use v = f • where v = 345 m/s and f = 20 000 Hz. Rearrange the equation to the form of = v / f. Substitute and solve.
6. An elephant produces a 10 Hz sound wave. Assuming the speed of sound in air is 345 m/s, determine the wavelength of this infrasonic sound wave.

Answer: 34.5 meters 

Let = wavelength. Use v = f • where v = 345 m/s and f = 10 Hz. Rearrange the equation to the form of = v / f. Substitute and solve.

7. Determine the speed of sound on a cold winter day (T=3 degrees C).

Answer: 332.8 m/s 

The speed of sound in air is dependent upon the temperature of air. The dependence is expressed by the equation:
v = 331 m/s + (0.6 m/s/C) • T
where T is the temperature in Celsius. Substitute and solve.
v = 331 m/s + (0.6 m/s/C) • 3 C v = 331 m/s + 1.8 m/s
v = 332.8 m/s
8. Miles Tugo is camping in Glacier National Park. In the midst of a glacier canyon, he makes a loud holler. He hears an echo 1.22 seconds later. The air temperature is 20 degrees C. How far away are the canyon walls?

Answer = 209 m 

The speed of the sound wave at this temperature is 343 m/s (using the equation described in the Tutorial). The distance can be found using d = v • t resulting in an answer of 343 m. Use 0.61 second for the time since 1.22 seconds refers to the round-trip distance.

9. Two sound waves are traveling through a container of unknown gas. Wave A has a wavelength of 1.2 m. Wave B has a wavelength of 3.6 m. The velocity of wave B must be __________ the velocity of wave A.
a. one-ninth b. one-third
c. the same as d. three times larger than

Answer: C 

The speed of a wave does not depend upon its wavelength, but rather upon the properties of the medium. The medium has not changed, so neither has the speed.
10. Two sound waves are traveling through a container of unknown gas. Wave A has a wavelength of 1.2 m. Wave B has a wavelength of 3.6 m. The frequency of wave B must be __________ the frequency of wave A.
a. one-ninth b. one-third
c. the same as d. three times larger than

Answer: B 

Since Wave B has three times the wavelength of Wave A, it must have one-third the frequency. Frequency and wavelength are inversely related.