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 ) rArr v _ o = ( 320 ) /( 5 ) `
` = 64 m//s `
when observer moves towards the stationary source, then
` n' = (( v + v _ o ) / ( v ) ) n or n' = ((320 + 64 )/(320)) n `
or `n' = (( 384 ) /( 320)) n or ( n' ) /( n ) = (384 ) /( 320) `
Hence, percentage increase
` (( n' - n ) /( n ) ) = (( 384 - 320)/(320 ) xx 100 ) % `
` = ((64 )/( 320 ) xx 100) % = 20 % `