Two tuning forks with natural frequencies of 340 Hz move relative to a stationary oberserver. One fork moves away from the obsorver, while the other mover towards at the same speed. The observer hears beats of freqeuency 3 Hz. Find the speed of the tuning forks assuming the velocity of sound in air to be 340 m/s.

Answer

1.5 m/s.

Explanation

To solve the problem we should makes use of the expression for the frequency of the signal from a moving source as measured by a stationary observer :
`v_(1)=(v_(0))/(1+x)andv_(2)=(v_(0))/(1-x)`
where x = v/u is the ratio of the velocity of the source to the wave velocity. The beat frequency is equal to the difference in frequencies :
`v-v_(1)-v_(2)=(2xv_(0))/(1-x^(2))=(2v v_(0))/(u(1-v^(2)//u^(2)))`
We obtain a quadratic equation `vx^(2) + 2v_(0)x - x` 0, which gives
`x=(-v_(0)+sqrt(v_(0)^(2)+v^(2)))/(v)=(v_(0)^(2)+v^(2)-v_(0)^(2))/(v(v_(0)+sqrt(v_(0)^(2)+v^(2))))=(v)/(v_(0)+sqrt(v_(0)^(2)+v^(2)))approx (v)/(2v_(0))`
since the beat frequency v is much lower than the natural frequency of the tuning fork, `v_(0)`. Hence
`(v)/(u)approx(v)/(2v_(0)),` from which `v approx (u v)/(2v_(0))`