Question

The apparent frequency of a note is $200\, Hz$. When a listener is moving with a velocity of $40\, ms ^{-1}$ towards a stationary source. When he moves away from the same source with the same speed, the apparent frequency of the same note is $160\, Hz$. The velocity of sound in air in $m / s$ is

• A. 340
• B. 330
• C. 360
• D. 320

Answer 1

Answer: (C)

When listener is moving towards the source then apparent frequency
$n'=\frac{v+v_{o}}{v} \times n$
$\Rightarrow \, 200 =\frac{v+40}{v} \times n\,\,\,\,\dots(i)$
where $v=$ velocity of sound in air
$n=$ actual frequency of sound source
Similarly, when listener is moving away,
then $ 160=\frac{v-40}{v} \times n\,\,\,\,\,\dots(ii)$
From Eqs. (i) and (ii), we have
$\frac{200}{160} =\frac{v+40}{v-40} $
$5 v-200 =4 v+160 $
$\therefore V=360 m / s$