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(D) Given: Velocity of source $V_s = 30 \,m/s$,frequency of source $n_0 = 256 \,Hz$,and speed of sound $V = 330 \,m/s$.When the source is approaching

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$6:5$

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(D) Given: Velocity of source $V_s = 30 \,m/s$,frequency of source $n_0 = 256 \,Hz$,and speed of sound $V = 330 \,m/s$.When the source is approaching the stationary observer,the observed frequency $n_1$ is given by $n_1 = n_0 \left( \frac{V}{V - V_s} \right)$.When the source is moving away after crossing the observer,the observed frequency $n_2$ is given by $n_2 = n_0 \left( \frac{V}{V + V_s} \right)$.The ratio of the frequencies is $\frac{n_1}{n_2} = \frac{n_0 \left( \frac{V}{V - V_s} \right)}{n_0 \left( \frac{V}{V + V_s} \right)} = \frac{V + V_s}{V - V_s}$.Substituting the values: $\frac{n_1}{n_2} = \frac{330 + 30}{330 - 30} = \frac{360}{300} = \frac{6}{5}$.

$A$ source of sound is moving with a constant velocity of $30 \,m/s$ emitting a note of frequency $256 \,Hz$. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him is [speed of sound in air $= 330 \,m/s$].

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