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

here 200 = `(v_(s) + 40)/(v_(s)) xx n`
160 ` = (v_(s) - 40)/(v_(s)) xx n`
Dividing ` (200)/(160) = (v_(s) + 40)/(v_(s)- 40) `
this gives `v_(s) = 360 ms^(-1)`
correct choice is (a) .