A whistle is whirled around in a horizontal circle 120 times a minute at the end of a piece of string 1.5 m in length. If the standard frequency of the whistle is 550 Hz, calculate the difference between the highest and lowest frequencies heard by a stationary observer assume the speed of sound in air is 343 m/s).
A) 58 Hz
B) 64 Hz
C) 61 Hz
D) 32 Hz
E) 29 Hz

Question

Quizplus · Accepted Answer

The difference between the highest and lowest frequencies heard by a stationary observer is given by the Doppler equation:

Δf = (f0 v/c) * (1 ± v/u)

where f0 is the standard frequency of the whistle, v is the speed of the whistle in a circular path, c is the speed of sound in air, and u is the speed of the observer (assumed to be stationary). The sign in the second term depends on whether the observer is moving towards or away from the source.

In this case, the whistle is whirled around in a circle, so its speed is given by:

v = 2πr f = 2π(1.5 m)(120 rev/min)/60 = 18π m/s

Plugging in the numbers, we get:

Δf = (550 Hz * 18π m/s)/(343 m/s) * (1 ± 0)

The second term is zero since the observer is assumed to be stationary. Evaluating the expression, we get:

Δf = 61 Hz

So the answer is C).

A Whistle Is Whirled Around in a Horizontal Circle 120