A stationary observer receives sonic oscillations from two tuning forks one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears the beats of frequency `f = 2.0 H_(Z)`. Find the velocity of each tuning fork if their oscillation frequency is `f_(o) = 680 H_(Z)` and the velocity of sound in air is `upsilon = 340 m//s`.

Explanation

Let the velocity of tuning fork is `u`. Thus frequency reaching to the observer due to the tuning fork that approaches the observer
`v^(')=v_(0)(v)/(v-u) [v=` velocity of sound `]`
Frequency reaching the observer due to the tunning fork that receds from the observer
`v^('') =v_(0)(v)/(v+u)`
So, Beat frequency `v=v^('') =v=v_(0)v((1)/(v-u)-(1)/(v+u))`
or, `v=( 2 v_(0)vu)/(v^(2)-u^(2))`
So, `vu^(2)+(2 v v_(0))u-v^(2)v=0`
Hence `u=(-2 v v_(0)+- sqrt(4 v _(0)^(2)v^(2)+4 v^(2) v^(2)))/(2v)`
Hence the sought value of `u`, on simplifying and noting that `u gt 0`
`u-(v v _(0))/( v)(sqrt(1+((v)/( v_(0)))^(2)-1))`