An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

Options

A. Zero
B. `0.5 %`
C. `5%`
D. `20%`

Answer

D

Explanation

Given: `v_(0) = (v)/(5)`
`rArr v_(0) = (320)/(5) = 64 m//s`
When observer moves towards the stationary source, then
`n' = ((v + v_(0))/(v)) n = ((320 + 64)/(320)) n = ((384)/(320))n`
`(n')/(n) = (384)/(320)`
Hence, percentage increase
`((n' - n)/n)) = ((384 - 320)/(320) xx 100)%`
`= ((64)/(320) xx 100)% = 20%`