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 notes is 160 Hz. The velocity of sound in air in $ms^{- 1}$is

Question

Dr. Abubacker · Accepted Answer

(c) When listener is moving towards the source then apparent frequency
$v^{'} = \frac{v + v_{o}}{v} 	imes v$
$\Longrightarrow 200 = \frac{v + 40}{v} 	imes v\ \ \ \ \ \ \ldots(i)$
Where, v=velocity of sound in air
V=actual frequency of sound source
Similarly, when listener is moving away,
Then
$1160 = \frac{v - 40}{v} 	imes v\ \ \ \ \ \ \ldots(ii)$
From Eqs. (i)and (ii), we have
$\frac{200}{160} = \frac{v + 40}{v - 40}$
$5v - 200 = 4v + 160$
$	herefore v = 360ms^{- 1}$

Q. 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 notes is 160 Hz. The velocity of sound in air in $ms^{- 1}$is

Explanation

Reviewed by Dr. Abubacker

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