Two source A and B are producing notes of frequency 680 Hz. A listener moves from A to B with a constant velcoity v. if the speed of sound in air is 340 `ms^(-1)` , the value of y so that the hears 10 beats per second is

Options

A. `2.0 ms^(-1)`
B. `2.5 ms^(-1)`
C. `3.0 ms^(-1)`
D. `3.5 ms^(-1)`

Answer

B

Explanation

Given
notes of frequency produced by the sources A and B is 680 Hz.
i.e., `f_(A) ` and `f_(B) = 680 Hz`
Velocity of listener moves from A to B is constant = v speed of sound `v_(s) = 340 m//s`
and betats per second n=10
Now beats per second from point A to B is given as
`n= f_(A) .((v_(s) + u )/( v_(s)) ) - f_(B) ((v_(s) - u)/(v_(s)))`
`10 = 680 [ ((340+ u)/( 340)) - ((340 - u)/( 340))]`
`1= 68 ((2u )/( 340))`
`rArr 2u = (340)/(68) = (340)/( 68 xx 2)`
`u = 2.5 m//s`