Question

A whistle producing sound waves of frequency $9500Hz$ and above is approaching a stationary person with speed v m/s. The velocity of sound in air is $300m/s$ . If the person can hear frequencies up to maximum of $10000Hz$ , the maximum value of $v$ up to which he can hear the whistle is

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

Answer 1

Answer: (D)

Given that the velocity of sound in air $=300m/s$. If a source of sound is moving toward a stationary listener, the frequency heard by listener would be different from the actual frequency of the source, this apparent frequency is given by
$f_{app}=\left(\frac{V_{\text{sound in air}}}{V_{\text{sound in air}}\pm V_{\text{source}}}\right)$
where symbols have their usual meanings. In the denominator positive sign would be taken when source is recending away from the listener, while negative 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
$1000=\left(\frac{300}{300-v}\right)\times 9500$
$\Rightarrow v=15m/s$