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 :-

Options

A. `360`
B. `330`
C. `320`
D. `340`

Answer

A

Explanation

When a listener moves towards a stationary source apparent frequency
`n'=((v+v_0)/(v))n=200`
When listener moves away from the same source
`n' =((v-v_O))/(v)n=160`
From (i) and (ii)
`(v+v_O)/(v-v_O)=(200)/(160)implies(v+v_O)/(v-v_O)=(5)/(4)implies360 m//sec`