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

• Correct answer
  A

  4
• B

  8
• C

  2
• D

  zero

Solution

$$ \text { According to the question } $$
Speed of listener $V_L=1 \mathrm{~ms}^{-1}$ Speed of sound $V=330 \mathrm{~ms}^{-1}$ Frequency of each source $v=600 \mathrm{~Hz}$ $\therefore$ Apparent frequency due to $\mathrm{P}=\mathrm{V}^{\prime}$ $=\frac{v\left(\mathrm{~V}-\mathrm{V}_{\mathrm{L}}\right)}{\mathrm{V}}$ Apparent frequency due to $\mathrm{Q}=\mathrm{V}^{\prime \prime}=$ $\frac{v\left(\mathrm{~V}-\mathrm{V}_{\mathrm{L}}\right)}{\mathrm{V}}$ $\therefore$ No. of heat heard by the listener per second. $$ \begin{aligned} & \mathrm{V}^{\prime \prime}-\mathrm{V}^{\prime}=\frac{v\left(\mathrm{~V}-\mathrm{V}_{\mathrm{L}}\right)}{\mathrm{V}}-\frac{v\left(\mathrm{~V}_0-\mathrm{V}_{\mathrm{L}}\right)}{\mathrm{V}} \\ & =\frac{2 v_{\mathrm{L}}}{v}=\frac{2 \times 660 \times 1}{330}=4 \end{aligned} $$

Hence, the correct answer is option: 'A'.