The apparent frequency of a note, when a listener moves towards a stationary source, with velocity of $40\ m/s$ is $200\ Hz.$ When the 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 is (in $m/s$)

Question

Dr. Abubacker · Accepted Answer

(a) When a listener moves towards a stationary source apparent frequency
$n^{'} = \left( \frac{v + v_{O}}{v} \right)n = 200$ …(i)
When listener moves away from the same source
$n" = \ \frac{(v - v_{O})}{v}n = 160$ …(ii)
From (i) and (ii)
$\frac{v + v_{O}}{v - v_{O}} = \frac{200}{160} \Rightarrow \frac{v + v_{O}}{v - v_{O}} = \frac{5}{4} \Rightarrow v = 360m/sec$

Q. The apparent frequency of a note, when a listener moves towards a stationary source, with velocity of $40\ m/s$ is $200\ Hz.$ When the 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 is (in $m/s$)

Explanation

Reviewed by Dr. Abubacker

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