An engine giving a whistle is moving towards | vedclass

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(A) The apparent frequency heard by a stationary observer when the source is moving is given by $n' = n \left( \frac{V}{V \mp V_s} \right)$,where $V$

Vedclass · Accepted Answer

$2 : 1$

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(A) The apparent frequency heard by a stationary observer when the source is moving is given by $n' = n \left( \frac{V}{V \mp V_s} \right)$,where $V$ is the speed of sound and $V_s$ is the speed of the source.When the engine is approaching,the apparent frequency is $n_{app} = n \left( \frac{V}{V - V_s} \right)$.When the engine is receding,the apparent frequency is $n_{rec} = n \left( \frac{V}{V + V_s} \right)$.The ratio of the frequencies is $\frac{n_{app}}{n_{rec}} = \frac{n \left( \frac{V}{V - V_s} \right)}{n \left( \frac{V}{V + V_s} \right)} = \frac{V + V_s}{V - V_s}$.Substituting the given values $V = 330\,m/s$ and $V_s = 110\,m/s$:Ratio $= \frac{330 + 110}{330 - 110} = \frac{440}{220} = \frac{2}{1}$.

An engine giving a whistle is moving towards a stationary observer with a speed of $110\,m/s$. What will be the ratio of the frequency of the whistle heard when the engine is approaching and receding from the observer? (The speed of sound is $330\,m/s$.)

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