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Q. The frequency of whistle of an engine appears to be $(4 / 5)^{\text {th }}$ of initial frequency when it crosses a stationary observer. If the velocity of sound is $330 \,m / s$, then the speed of engine will be

BITSATBITSAT 2021

Solution:

$n'=\frac{n v}{v-v_{s}} \ldots \ldots \ldots(1)$
$n''=\frac{n v}{v+v_{s}} \ldots \ldots . .(2)$
From (1) and (2),
$\frac{n'}{n''}=\frac{v+v_{s}}{v-v_{s}}$ .... (3)
According to question, $\frac{n'}{n'}=\frac{5}{4}$
$v_{s}=? v=330 \,m / s$ ... (4)
From eq. (3) and (4)
$\frac{5}{4}=\left[\frac{330+v_{s}}{330-v_{s}}\right]$
$ \Rightarrow 9 v_{s}=330 $
$\therefore v_{s}=36.6\, m / s$