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 upto a maximum of 10,000 Hz, the maximum value of V upto which he can hear the
whistle is

• A. $15\sqrt{2}m{s}^{-1}$
• B. $\frac{15}{\sqrt{2}}m{s}^{-1}$
• C. $15m{s}^{-1}$
• D. $30m{s}^{-1}$

Answer 1

Answer: (C)

$15m{s}^{-1}$

Apparent frequency, $f^{\prime }=\left(\frac{v}{v-v_s}\right)f$

$\Rightarrow 10000=\left(\frac{300}{300-v_s}\right)9500$

$\Rightarrow 300-v_s=3\times 95$

$\Rightarrow v_s=15m{s}^{-1}$