When an observer moves towards a stationary source with velocity ' V 1 ', the appearent frequency of emitted note is ' F 1 '. When observer moves away from stationary source wi

Question

When an observer moves towards a stationary source with velocity ' V 1 ​ ', the appearent frequency of emitted note is ' F 1 ​ '. When observer moves away from stationary source with velocity ' V 1 ​ ' the appearent frequency is ' F 2 ​ '. If ' v ' is velocity of sound in air and F 2 ​ F 1 ​ ​ = 2 , then V 1 ​ V ​ is equal to

TestHub · Accepted Answer

Let the actual frequency of the source be $F$ .

When the observer moves towards the stationary source with velocity $V_{1}$ , the apparent frequency is given by Doppler's effect:

$F_{1} = F (\frac{v + V _{1}}{v})$

When the observer moves away from the stationary source with velocity $V_{1}$ , the apparent frequency is:

$F_{2} = F (\frac{v - V _{1}}{v})$

Given that $\frac{F _{1}}{F _{2}} = 2$ , we can write:

$\frac{v + V _{1}}{v - V _{1}} = 2$

$v + V_{1} = 2 (v - V_{1})$

$v + V_{1} = 2 v - 2 V_{1}$

$3 V_{1} = v$

$\frac{v}{V _{1}} = 3$

Answer: $3$

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