A stationary source is emitted sound at a fixed frequency $f_0$, which is reflected by two cars approaching the source. The difference between the is frequencies of sound reflected from the car is $1.2\%$ of $f_0$. What is the difference in the speed of the cars (in $\mathrm{m} {\mathrm{s}}^{-1}$)? The cars are moving at constant speeds much smaller than the speed of sound which is $330 \mathrm{m} {\mathrm{s}}^{-1}$

Solution

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,

$f_1=f_o\left(\frac{v+v_1}{v-v_1}\right)$

$f_2=f_o\left(\frac{v+v_2}{v-v_2}\right)$

$\therefore f_1-f_2=\left(\frac{1.2}{100}\right)f_o$

$=f_o\left[\frac{v+v}{v+v_1}-\frac{v+v_2}{v-v_2}\right]$

Or

$\left(\frac{1.2}{100}\right)f_o=\frac{2v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)}{f}_o$

$as v_{1}and v_2$ are very very less than v.

We can write, $\left(v-v_1\right)or \left(v-v_2\right)\approx v.$

$\therefore \left(\frac{1.2}{100}\right)f_o=\frac{2\left(v_1-v_2\right)}{v}{f}_o$

$Or \left(v_1-v_2\right)=\frac{v\times 1.2}{200}$

$=\frac{300\times 1.2}{200}=1.98\mathrm{m} {\mathrm{s}}^{-1}$

Hence, the correct answer is 1.98.