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.

Explanation

Let the velocity of each tuning fork with respect to stationary observer be `v_(s)` Apparent frequency ofthetuning fork coming towards the stationary observer is given by
` n_(1) = (nv)/(v -v_(s))`
Apparent frequency of the tuning fork moving away from the stationary observer is given by
`n_(2) = (nv)/(v+v_(s))`
From equations-(6.270) and (6.272)
`n_(1) - n_(2) = nv [(1)/(v-v_(s))-(1)/(v + v_(s))]`
`n_(1) - n_(2) = nv [(2v_(s))/(v^(2) - v_(s)^(2))]`
Substituting the given values, we have
`3 = 340 xx 340 [(2v_(s))/((340)^(2) -v_(s)^(2))]`
`3 [(340)^(2) - v_(s)^(2)] = 2 xx(340)^(2)v_(s)]`
`3 [(340)^(2) - 3v_(s)^(2)] = 2 xx(340)^(2)v_(s)]`
`3v_(s)^(2) + 2 xx (340)^(2)v_(s) - 3 xx (340)^(2) = 0`
`v_(s) = (-2 xx (340)^(2)pm sqrt(4 xx (340)^(4) + 4 xx 3 xx 3 (340)^(2)))/(2xx3)`
Solving we get` v_(s) = 1.5 m/s`