Two sources P and Q produce notes of frequency 660Hz each. A listener moves from P to Q with a speed of $1 m{s}^{-1}$. If the speed of sound is 330 $m{s}^{-1}$, then the number of beats heard by the listener per second will be

The situation as shown in the figure.

$$\begin{gathered} Here, \\ \text{ Speed of listener, }{v}_L=1{ \mathrm{ms}}^{-1} \\ \text{ Speed of sound, }v=330{ \mathrm{ms}}^{-1} \\ \text{ Frequency of each source, }\nu =660 \mathrm{Hz} \\ \text{ Apparent frequency due to }\mathrm{P}\text{ , } \\ {\mathrm{v}}^{\text{'}}=\frac{\mathrm{v}\left(\mathrm{v}-{\mathrm{v}}_{\mathrm{L}}\right)}{\mathrm{v}} \\ \text{ Apparent frequency due to }\mathrm{Q}\text{ , } \\ {\mathrm{v}}^{\text{'}\text{'}}=\frac{\mathrm{v}\left(\mathrm{v}+{\mathrm{v}}_{\mathrm{L}}\right)}{\mathrm{v}} \\ \text{ Number of beats heard by the listener per second is } \\ {\mathrm{v}}^{\text{'}\text{'}}-{\mathrm{v}}^{\text{'}}=\frac{\mathrm{v}\left(\mathrm{v}+{\mathrm{v}}_{\mathrm{L}}\right)}{\mathrm{v}}-\mathrm{v}\frac{\left(\mathrm{v}-{\mathrm{v}}_{\mathrm{L}}\right)}{\mathrm{v}} \\ =\frac{2{\mathrm{vv}}_{\mathrm{L}}}{\mathrm{v}}=\frac{2\times 660\times 1}{330}=4 \end{gathered}$$