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))xxn=((340)/(340-20))xx1000=1063Hz`
When source is receding the frequency heard
`n_r=((v)/(v+v_S))xxn=(340)/(340-20)xx1000=944`
`impliesn_a:n_r=9:8`
Short tricks : `(n_a)/(n_r)=(v+v_S)/(v-v_S)=(340+20)/(340-20)=(9)/(8)`
(speed of sound `v=340m//s`).