A source sound is moving with constant velocity of `20m//s` emitting a note of frequency `1000 Hz`. The ratio of frequencies observed by a stationary observer while the source approaching him and after it crosses him will be source is approaching him and after it crosses him will be

Options

A. `9 : 8 `
B. `8 : 9 `
C. `1 : 1 `
D. `9 : 10 `

Answer

A

Explanation

When source is approaching the observer, the frequency heard
`n_a=-(v/(v-v_S)) xx n =(340/(340-20)) xx 1000` = 1063 Hz
When source is receding, the frequency heard
`n_r=(v/(v-v_S))xx n = 340/(340-20)xx 1000` = 944
`rArr n_a : n_r = 9:8`
Short tricks : `n_a/n_r=(v+v_S)/(v-v_S)=(340+20)/(340-20)=9/8`