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

Question

When the observer moves towards a stationary source with velocity ' V 1 ​ ', the apparent frequency of the 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 speed of sound in air and F 1 ​ / F 2 ​ = 2 , then V / V 1 ​ =

TestHub · Accepted Answer

Using the Doppler effect formula, the apparent frequency when the observer moves towards the stationary source is given by

$F_{1} = F_{0} (\frac{V + V _{1}}{V})$

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

$F_{2} = F_{0} (\frac{V - V _{1}}{V})$

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

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

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

$3 V_{1} = V$

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

Answer: $3$

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