Question

A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms^-1. The velocity of sound in air is 300 ms^-1. If the person can hear frequencies up to 10000 Hz, the maximum value of v upto which he can the whistle is

• A.
• B. 15 ms^-1
• C. 30 ms^-1
• D.

Answer 1

The correct option is B

15 ms^-1

The apparent frequency (n^') in terms of all possible variables is given by

n^' =n(v+w+v_L)/(v+w-v_S) where 'n' is the real frequency of sound, 'v' is the velocity of sound, 'w' is the velocity of wind, 'v_L' is the velocity of listener and 'v_S' is the velocity of the source of sound. Note that in the above expression all velocities are in the same direction and the source is behind the listener and is therefore approaching the listener.

Since the wind velocity is zero and the listener is stationary in the first case, the above expression reduces to:

n^' = nv/(v-v_S) (The apparent frequency is therefore greater than the real frequency).

As the person can hear frequencies upto 10000 Hz only, the maximum value of v_S upto which he can hear the whistle is given by

10000 = 9500x300/(300-v_S)

Therefore, 300 - v_S= 285 so that v_S = 15 ms^-1.