Two tuning forks with natural frequencies 340 Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards the observer at the same speed. The observer hears beats of frequency 3 Hz. Find the speed of the tuning forks (speed of sound is `340(m)/(s)`).

Explanation

Let v be the speed of sound and `v_S` the speed of forks. The apparent frequency of fork which moves towards the observer is
`n_1=(v)/(v-v_S)n`
The apparent frequency of fork which moves away from observer is
`n_2=(v)/(v+v_S)n`
If x is the number of beats heard per second the
`x=n_1-n_2impliesx(v)/(v-v_S)n-(v)/(v+v_S)n`
or `x=(v+v_S-(v-v_S))/(v^2-v_S^2)(vn)`
or `(2vv_Sn)/(v^2-v_S^2)=x`
`2((v_S)/(v))n=x`
or `v_S=(vx)/(2n)`
Given `v=340(m)/(s)`,`x=3`,`n=340Hz`
`v_S=(340xx3)/(2xx340)=1.5(m)/(s)`