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`
`[f_1 ="apparent frequency when velocity" `v_1` "is 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`
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)`
Rightarrow `(V+V_1)/(V-V_1)=2 Rightarrow 2V-2V_1=V+V_1` lt brgt `Rightarrow V= 3V_1 Rightarrow V/V_1=3`