An observer standing at station observes frequency `219 Hz` when a train approaches and `184 Hz` when train goes away from him. If velocity of sound in air is `340m//s`, then velocity of train and actual frequency of whistle will be

Options

A. `15.5ms_(1), 200Hz`
B. `19.5ms^(-1), 205Hz`
C. `29.5ms^(-1), 200Hz`
D. `32.5ms^(-1), 200Hz`

Answer

C

Explanation

When train is approaching frequency heard by the observer is
`n_a=n((v)/(v-v_S))implies219=n((340)/(340-v_S))` .(i)
When train is receding (goes away), frequency heard by the observer is
`n_r=n((v)/(v+v_S))implies184=n((340)/(340-v_S))` .(ii)
On solving equation (i) and (ii) we get `n=200Hz` and `v_S=29.5m//s`.