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

Question

When the observer moves towards a stationary source with velocity V 1 ​ , the apparent frequency of emitted note is F 1 ​ . When 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 ​ is equal to

TestHub · Accepted Answer

Using the Doppler effect formula for a stationary source and moving observer:

When the observer moves towards the source, the apparent frequency is:

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

When the observer moves away from the source, the apparent frequency is:

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

Given that $\frac{F _{1}}{F _{2}} = 2$ :

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

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

$3 V_{1} = V$

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

Answer: 3

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