Physics - Sound Wave Question with Solution | TestHub

The apparent frequency heard by the man is given by

$n_{\mathrm{a}\mathrm{p}\mathrm{p}}=n\left(\frac{v\pm v_{\mathrm{o}}}{v\pm v_{\mathrm{s}}}\right)$

On our case,$v_{\mathrm{o}}=0$

$\therefore n_{\mathrm{a}\mathrm{p}\mathrm{p}}=n\left(\frac{v}{v\pm v_{\mathrm{s}}}\right)$

Given,$v=300 \mathrm{m} {\mathrm{s}}^{-1},\mathrm{ }{v}_{\mathrm{s}}=4 \mathrm{m} {\mathrm{s}}^{-1},\mathrm{ }n=240\mathrm{ }\mathrm{H}\mathrm{z}$

In the first case, the train is approaching the man, so frequency heard

$n_1=n\left(\frac{v}{v\pm v_{\mathrm{s}}}\right)=240\left(\frac{300}{300-4}\right)$

$=\frac{240\times 300}{296}=243.24\mathrm{ }\mathrm{H}\mathrm{z}$

In the second case, the train is going away from the man, so frequency heard

$n_2=n\left(\frac{v}{v+v_{\mathrm{s}}}\right)=240\mathrm{ }\left(\frac{300}{300+4}\right)$

$=\frac{240\times 300}{304}=236.84 \mathrm{H}\mathrm{z}$

Hence, the number of beats heard by the man per second

$=n_1-n_2$

$=243.24-236.84=6.4\approx 6$