Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 `ms^(-1)` and the man behind walks at a speed `2.0 ms^(-1)`. A third mad is standing at a height 12m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330 `ms^(-1)`. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is

Answer

`5.00`

Explanation

Let source `S_(1)` be moving with velocity 2 m/s and `S_(2)` be moving with velocity 1 m/s.
`f_(1) = f((upsilon)/(upsilon - 2 cos theta)) , f_(2) = f((upsilon)/(upsilon+cos theta))`
Beat frequency
`Delta f = f_(1) - f_(2)`
`rArr " "Delta f = (f upsilon xx 3 cos theta)/((upsilon - 2 cos theta) (upsilon + cos theta))`
In the denominator, `cos theta and 2 cos theta` can be neglected in comparison to `upsilon`, hence we can write as follows :
`rArr" "Delta f ~~ (3 f upsilon cos theta)/(upsilon^(2)) = (3 f cos theta)/(upsilon)`
`rArr" "Delta f = (3 xx 1430 xx (5)/(13))/(330)`
`rArr" "Delta f = 5`