Two tuning forks with natural frequencies of `340 Hz` each move relative to a stationary observer. One fork moves away form the observer, while the other moves towards him at the same speed. The observer hears beats of frequency `3 Hz`. Find the speed of the tuning fork.

Answer

1.5

Explanation

The apparem\nt frequency from tuning fork `T_1` as heard by observed will be `f_1=f_0((v)/(v-u))`….(i)
The apparent frequency from tuning fork `T_2` as heard by the observer will be `f_2=f_0((v)/(v+u))`…..(ii)
Given that beat frequency is `3,f_1-f_2=3`
`f_0((v)/(v-u))-f_0((v)/(v+u))=3`
`f_0v[(v+u-v+u)/(v^(2)-u^(2))]=3 implies f_0. ((v2u))/(v^(2)-u^(2))=3`
`Asgtgtu` Hence `f_0.(2uv)/(v^(2))=3implies u=(3v)/(2 f_0)=1.5m//s`