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 , `upsilon = (upsilon)/(5) rArr upsilon_(0) rArr upsilon_(0) d = (320)/(5) = 64 m//s`
When observer moves towards the stationary source, then
`n' = ((upsilon + upsilon_(o))/(upsilon))n`
`n' = ((320 + 64)/(320))n`
`n' = ((384)/(320))n`
`(n')/(n) = (384)/(320)`
Hence, percentage increase.
`((n' - n)/(n)) = ((384 - 320)/(320) xx 100)% = ((64)/(320) xx 100)% = 20%`