A stationary source is emitting sound at a fixed frequency \(f_0\), which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is \(1.2 \%\) of \(f_0\). What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is \(330 \mathrm{~ms}^{-1}\).

Firstly, car will be treated as an observer which is approaching the source. Then, it will be treated as a source, which is moving in the direction of sound.

Hence,
\(f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)\)
\(f_2 =f_0\left(\frac{v+v_2}{v-v_2}\right) \)
\( \therefore f_1-f_2 =\left(\frac{1.2}{100}\right) f_0 \)
\( =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right] \)
\( \text { or }\left(\frac{1.2}{100}\right) f_0 =\frac{2 v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)} f_0\)
As \(v_1\) and \(v_2\) are very very less than \(v\). We can write, \(\left(v-v_1\right)\) or \(\left(v-v_2\right) \approx v\)
\(
\therefore \left(\frac{1.2}{100}\right) f_0=\frac{2\left(v_1-v_2\right)}{v} f_0
\)
or \(\left(v_1-v_2\right)=\frac{v \times 1.2}{200}\)
\(
\begin{aligned}
& =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\
& =7.128 \mathrm{kmh}^{-1}
\end{aligned}
\)
\(\therefore\) The nearest integer is 7 .
Hence, \(\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)\)
\(f_2 =f_0\left(\frac{v+v_2}{v-v_2}\right) \)
\( \therefore f_1-f_2 =\left(\frac{1.2}{100}\right) f_0 \)
\( =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right] \)
\( \text { or }\left(\frac{1.2}{100}\right) f_0 =\frac{2 v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)} f_0\)
As \(v_1\) and \(v_2\) are very very less than \(v\). We can write, \(\left(v-v_1\right)\) or \(\left(v-v_2\right) \approx v\)
\(
\therefore \left(\frac{1.2}{100}\right) f_0=\frac{2\left(v_1-v_2\right)}{v} f_0
\)
or \(\left(v_1-v_2\right)=\frac{v \times 1.2}{200}\)
\(
\begin{aligned}
& =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\
& =7.128 \mathrm{kmh}^{-1}
\end{aligned}
\)
\(\therefore\) The nearest integer is 7 .