When the observer moves towards the stationary source with velocity, `v_1`, the apparent frequency of emitted note is `f_1`. When the observer moves away from the source with velocity `v_1`, the apparent frequency is `f_2`. If v is the velocity of sound in air and `f_1/f_2` = 2,then `v/v_1` = ?

Options

A. 2
B. 3
C. 4
D. 5

Answer

B

Explanation

According to question,
`f_(1)/f_(2)=2` (given)
[`f_(1)=` apparent frequency when velocity is `v_(1)` towards the observer. `f_(2)=` apparent frequency when velocity `v_(1)` is away from the observer].
Now, the apparent frequency of sound when observer moves towards the source is given by `f_(1)=(V/(V-V_(1)))f_(0)` ...(i)
[Symbols have their usual meanings]
Similarly when observer moves away from the source, apparent frequency is given by
`f_(2) =(V/(V+V_(1)))f_(0)` ...(ii)
From Eqs. (i) and (ii), we get
`f_(1)/f_(2)=((V/(V-V_(1)))f_(0))/((V/(V+V_(1)))f_(0))=(V+V_(1))/(V-V_(1))`
`implies (V+V_(1))/(V-V_(1))=2 implies 2V-2V_(1)=V+V_(1)`
`implies V=3V_(1) implies V/V_(1)=3`