Q121 / 180 · AIIMS 2019 (25 May) · Physics
Two sources of sound $S_1$ and $S_2$ are moving towards and away from a stationary observer with same speed respectively. Observer detects 3 beats per second. Find the speed of sources (approximately).
Given, $v_1=v_2=500 \mathrm{~Hz}$, speed of air $=330 \mathrm{~m} \mathrm{~s}^{-1}$
- Correct answerA$1 \mathrm{~m} \mathrm{~s}^{-1}$
- B$2 \mathrm{~m} \mathrm{~s}^{-1}$
- C$3 \mathrm{~m} \mathrm{~s}^{-1}$
- D$4 \mathrm{~m} \mathrm{~s}^{-1}$
Solution
$v_1=500 \mathrm{~Hz}, v_2=500 \mathrm{~Hz}$
$S_1 \longrightarrow \mathrm{O}$ bserver $S_2 \longrightarrow$
Beats frequency $=3$ beats $\mathrm{sec}^{-1}$
$v_1=\left(\frac{330}{330-v}\right) \times 500, v_2=\left(\frac{330}{330+v}\right) \times 500$
Beats $=\left(v_1-v_2\right)$
$\begin{aligned} & 3=\left(\frac{330}{330-v}\right) \times 500-\left(\frac{330}{330+v}\right) \times 500 \\ & 3=\frac{500}{330^2-v^2} \times 2 v \times 330 \quad\left(\because v^2< < 330^2\right) \\ & 3=\frac{500 \times 330 \times 2 v}{330^2} \\ & v=\frac{3 \times 330}{2 \times 500} \Rightarrow v=0.99 \mathrm{~m} \mathrm{~s}^{-1}=1 \mathrm{~m} \mathrm{~s}^{-1} \text { approx }\end{aligned}$
Hence, the correct answer is option: 'A'.