A band playing music at a frequency `f_(0)` is moving towareds a wall at a speed `v_(0)`. A motorist is following the band with a speed `v_(m)`. If v be the speed of the sound the expression for beat frequency heard by motorist is

Options

A. `(v+v_(m))/(v+v_(b))f`
B. `(v+v_(m))/(v-v_(b))f`
C. `(2v_(b)(v+v_(m)))/(v^(2)-v_(b)^(2))f`
D. `(2v_(m)(v+v_(b)))/(v^(2)-v_(m)^(2))f`

Answer

C

Explanation

Motorist hears two frequencies, one directly from the band and second reflected from the wall.
`rarr` from band
`f_(1)=f[(v+v_(m))/(v+v_(b))]`
Frequency of sound reaching the wall
`f..=f((v)/(v-v_(b)))`
The frequency of reflected sound from the wall, heard by motorist
`f_(2)=f..((v+v_(m))/(v)),f_(2)=f((v+v_(m))/(v-v_(b)))`
Now, `f_("beat")=f_(2)-f_(1)`
`=f[(v+v_(m))/(v-v_(b))-((v+v_(m))/(v+v_(b)))]`
`implies f_("beat")=f((v+v_(m))/(v^(2)-v_(b)^(2)))xx2v_(b)`.