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 $\frac{F_{1}}{F_{2}} = 2$ then, $\frac{V}{V_{1}}$ ?

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Solution

The correct option is A 3
Let the original frequency of the source be $F_{o}$ .
From Doppler effect, apparent frequency heard by observer when it moves towards stationary source,
$F_{1} = F_{o} [\frac{v_{s o u n d} + V_{o b s e r v e r}}{v_{s o u n d}}]$
OR $F_{1} = F_{o} [\frac{V + V_{1}}{V}]$ .........(1)
From Doppler effect, apparent frequency heard by observer when it moves away from stationary source,
$F_{2} = F_{o} [\frac{v_{s o u n d} - V_{o b s e r v e r}}{v_{s o u n d}}]$
OR $F_{2} = F_{o} [\frac{V - V_{1}}{V}]$ .........(2)
Dividing (1) and (2) we get $\frac{F_{1}}{F_{2}} = \frac{V + V_{1}}{V - V_{1}}$
Or $2 = \frac{V + V_{1}}{V - V_{1}}$
Or $2 V - 2 V_{1} = V + V_{1}$
Or $V = 3 V_{1}$
$⟹$ $\frac{V}{V_{1}} = 3$

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When the observer moves towards the stationary source with velocity, 'V1', the apparent frequency of emitted note is 'F1'. When the observer moves away from the source with velocity 'V1', the apparent frequency is 'F2'. If 'V' is the velocity of sound in air and F1F2=2 then, VV1?