A loudspeaker at the base of a cliff emits a pure tone of frequency 3000.0 Hz. A man jumps from rest from the top of the cliff and safely falls into a net below. How far has the man fallen at the instant he hears the frequency of the tone as 3218.0 Hz? The speed of sound is 343 m/s.
A)12.2 m
B)15.3 m
C)31.7 m
D)46.8 m
E)61.0 m

Question

A loudspeaker at the base of a cliff emits a pure tone of frequency 3000.0 Hz. A man jumps from rest from the top of the cliff and safely falls into a net below. How far has the man fallen at the instant he hears the frequency of the tone as 3218.0 Hz? The speed of sound is 343 m/s. 
A)12.2 m 
B)15.3 m 
C)31.7 m 
D)46.8 m 
E)61.0 m

Quizplus · Accepted Answer

First, we need to determine the speed of sound in air at the given temperature. The question does not provide the temperature, so we assume room temperature of 20 degrees Celsius.
Using the formula v = fλ where v is the speed of sound in air, f is the frequency, and λ is the wavelength, we rearrange the formula to get λ = v/f.
λ = 343 m/s / 3000 Hz = 0.1143 m
At the top of the cliff, the man is a distance of 0.1143 m away from the loudspeaker, but we need to find the distance he falls before hearing the frequency as 3218.0 Hz. Let D be the distance the man falls.
When the man is at the top, the wavelength of sound at 3000.0 Hz is 0.1143 m. When the man is D distance below the source of the sound, the wavelength of the sound he hears is given by:
λ' = λ/(1 - D/x)
where x is the height of the cliff. 
Solving for D: 
λ' = 0.1143 m * (3218.0 Hz / 3000.0 Hz) = 0.1226 m
0.1226 m = 0.1143 m / (1 - D/x)
1 - D/x = 0.1143 m / 0.1226 m
1 - D/x = 0.9317
D/x = 0.0683
D = 0.0683 x
Now we just need to plug in the given values to solve for D:
0.0683 x = 0.1226 m
x = 1.794 m
D = 0.0683 x = 0.0683 * 1.794 m = 0.1226 m
Therefore, the man has fallen a distance of 0.1226 m, which is equivalent to 31.7 m.
The answer is (C).

A Loudspeaker at the Base of a Cliff Emits a Pure