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_(o)=(v)/(5)impliesv_(0)=(320)/(5)=64m//s`
When observer moves towards the stationary source, then
`n'((v+v_(0))/(v))n`
`n'=((320+64)/(320))n`
`n'=((384)/(320))n`
`(n')/(n)=(384)/(320)`
Hence, percentage increase
`((n'-n)/(n))=((384-320)/(320)xx100)%`
`=((64)/(320)xx100)%=20%`