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. $15\sqrt{2}\,ms^{-1}$
• B. $15/\sqrt{2}\,ms^{-1}$
• C. $15\, ms^{-1}$
• D. $30\, ms^{-1}$

Answer 1

Answer: (C)

Velocity of sound in air $= 300\, m/s$
If a source of sound is moving towards a stationaiy listener, the frequency heard by the listener would be different from the actual frequency of the source, this apparent frequency is given by
$f_{app.}=\left(\frac{v_{sound\,in\,air}}{v_{_{sound\,in\,air}}\pm\,v_{source}}\right),$ where symbols have their usual meanings.
In the denominator +ve sign would be taken _ when source is receding away from the listener, while -ve sign would be taken when source is approaching the listener.
Let v be the maximum value of source velocity for which the person is able to hear the sound, then
$10000=f_{app.}=\left(\frac{300}{300-v}\right)\times9500$
$\Rightarrow v=15\,m/s$