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.

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) • Twhere 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.