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(A) The Doppler effect formula for a moving source and stationary observer is given by $n' = n \left( \frac{v}{v \mp v_s} \right)$,where $v$ is the sp

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$2 : 1$

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(A) The Doppler effect formula for a moving source and stationary observer is given by $n' = n \left( \frac{v}{v \mp v_s} ight)$,where $v$ is the speed of sound and $v_s$ is the speed of the source.When the source is approaching the observer,the observed frequency $n_1$ is:$n_1 = n \left( \frac{330}{330 - 110} ight) = n \left( \frac{330}{220} ight) = \frac{3}{2} n$When the source is receding from the observer,the observed frequency $n_2$ is:$n_2 = n \left( \frac{330}{330 + 110} ight) = n \left( \frac{330}{440} ight) = \frac{3}{4} n$The ratio of the frequencies is:$\frac{n_1}{n_2} = \frac{(3/2)n}{(3/4)n} = \frac{3}{2} 	imes \frac{4}{3} = \frac{2}{1}$Thus,the ratio is $2 : 1$.

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? (Speed of sound $= 330\, m/s$)

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