A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed $v~\text{ms}^{-1}$. The velocity of sound in air is $300~\text{ms}^{-1}$. If the person can hear frequencies upto a maximum of $10,000~\text{Hz}$, the maximum value of v upto which he can hear the whistle is

(B) $15\sqrt{2}~\text{ms}^{-1}$

(C) $15/\sqrt{2}~\text{ms}^{-1}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Waves & Sound

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 upto a maximum of

10, 000 Hz

, the maximum value of v upto which he can hear the whistle is

Answer: (D)

With the source approaching a stationary listener, each emitted frequency

f

is heard raised to

f^{'} = f \frac{v _{s}}{v _{s} - v}

, where

v_{s} = 300 m/s

is the speed of sound and

v

is the speed of the whistle. The listener's hearing cuts off at

10000 Hz

, and the lowest frequency the whistle emits is

9500 Hz

; the limiting case is when this

9500 Hz

is shifted exactly up to

10000 Hz

. Setting

10000 = 9500 \frac{300}{300 - v}

gives

300 - v = 285

, so

v = 15 m/s

. Beyond this speed even the lowest emitted tone is pushed above the audible limit.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option