Physics - Sound Wave Question with Solution | TestHub

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, $f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$ $$\begin{array}{rl}{f}_2& =f_0\left(\frac{v+v_2}{v-v_2}\right)\\ \therefore f_1-f_2& =\left(\frac{1.2}{100}\right)f_0\\ & =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right]\\ \text{ or }\left(\frac{1.2}{100}\right)f_0& =\frac{2v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)}{f}_0\end{array}$$ 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 $\left(v_1-v_2\right)=\frac{v\times 1.2}{200}$ $$\begin{array}{rl}& =\frac{330\times 1.2}{200}=1.98{\mathrm{ms}}^{-1}\\ & =7.128{\mathrm{kmh}}^{-1}\end{array}$$ $\therefore$ The nearest integer is 7 . Hence, $f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$ $$\begin{array}{rl}{f}_2& =f_0\left(\frac{v+v_2}{v-v_2}\right)\\ \therefore f_1-f_2& =\left(\frac{1.2}{100}\right)f_0\\ & =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right]\\ \text{ or }\left(\frac{1.2}{100}\right)f_0& =\frac{2v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)}{f}_0\end{array}$$ 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 $\left(v_1-v_2\right)=\frac{v\times 1.2}{200}$ $$\begin{array}{rl}& =\frac{330\times 1.2}{200}=1.98{\mathrm{ms}}^{-1}\\ & =7.128{\mathrm{kmh}}^{-1}\end{array}$$ $\therefore$ The nearest integer is 7 .