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A car is moving with a speed of $72 \mathrm{km}\mathrm{h}^{-1}$ towards a roadside source that emits sound at a frequency of $850 \mathrm{Hz}$. The car driver listens to the sound while approaching the source and again while moving away from the source after crossing it. If the velocity of sound is $340 \mathrm{ms}^{-1}$, the difference of the two frequencies, the driver hears is
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Verified Answer
The correct answer is:
$100 \mathrm{Hz}$
By Doppler's Effect When observer is moving with velocity $v_{0}$ towards a source at rest then approach frequency
$$
N_{\text {Approach }}=N\left(\frac{v+v_{0}}{v}\right)
$$
where $N=850 \mathrm{Hz}, v=340 \mathrm{ms}^{-1}, v_{0}=72 \mathrm{kmh}^{-1}$
$$
=20 \mathrm{ms}^{-1}=850\left(\frac{340+20}{340}\right)
$$
Similarly. when observer is moving away from source, then
$$
N_{\text {Soparation }}=N\left(\frac{v-v_{0}}{v}\right)=850\left(\frac{340-20}{340}\right)
$$
The different of the two frequency.
$$
\begin{array}{l}
N_{\text {Approuch }}-N_{\text {Soparation }} \\
\qquad \begin{aligned}
&=850\left(\frac{360}{340}\right)-850\left(\frac{320}{340}\right) \\
&=\frac{850}{340} \times 40=\frac{850}{8.5}=100 \mathrm{Hz}
\end{aligned}
\end{array}
$$
$$
N_{\text {Approach }}=N\left(\frac{v+v_{0}}{v}\right)
$$
where $N=850 \mathrm{Hz}, v=340 \mathrm{ms}^{-1}, v_{0}=72 \mathrm{kmh}^{-1}$
$$
=20 \mathrm{ms}^{-1}=850\left(\frac{340+20}{340}\right)
$$
Similarly. when observer is moving away from source, then
$$
N_{\text {Soparation }}=N\left(\frac{v-v_{0}}{v}\right)=850\left(\frac{340-20}{340}\right)
$$
The different of the two frequency.
$$
\begin{array}{l}
N_{\text {Approuch }}-N_{\text {Soparation }} \\
\qquad \begin{aligned}
&=850\left(\frac{360}{340}\right)-850\left(\frac{320}{340}\right) \\
&=\frac{850}{340} \times 40=\frac{850}{8.5}=100 \mathrm{Hz}
\end{aligned}
\end{array}
$$
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