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. 0.005
C. 0.05
D. 0.2

Answer

D

Explanation

Given that `v_("observer")=(v_("sound"))/(5)`
Applying Doppler's effect, we get
`f'=f[(v+v_(0))/(v)],f'=f[(6v//5)/(v)]=(6)/(5),(f')/(f)=(6)/(5)`
`(f'-f)/(f)=(6)/(5)-1=(1)/(5)=20%`.