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…

Question

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 $330 \mathrm{~ms}^{-1}$.

Aicharya · Accepted Answer

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, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$ $ \begin{aligned} f_2 & =f_0\left(\frac{v+v_2}{v-v_2}\right) \\ \therefore \quad 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 \end{aligned} $ 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 \quad\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, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$ $ \begin{aligned} f_2 & =f_0\left(\frac{v+v_2}{v-v_2}\right) \\ \therefore \quad 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 \end{aligned} $ 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 \quad\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 . :

A stationary source is emitting sound at a fixed frequency $f_0$, which is reflected by two cars approaching…

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