An observer standing on a railway crossing receives frequencies 2.2 kHz and 1.8 kHz when the tran approaches and recedes from the observer. Find the velocity of the train (speed of sound in air is 300 m/s).

Explanation

Let `v=300(m)/(s)` be the velocity of sound and u the velocity of train. When the train approaches the observer, the frequency heard is
`n_1=(v)/(v-u)n` ..(i)
When the train recedes from the observer, the frequency heard is
`n_2=(v)/(v+u)n` ..(ii)
Dividing Eqs. (i) and (ii) we get
`(n_1)/(n_2)=(v+u)/(v-u)impliesu=(n_1n_2)/(n_1+n_2)v`
Given `n_1=2.2kHz=2.2xx10^3Hz`.
`u=((2.2-1.8)xx10^3)/((2.2+1.8)xx10^3)xx300(m)/(s)=(0.4xx300)/(4)=30(m)/(s)`