JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2020
- Hard
- MCQ
A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears \(2\; beats/sec\). The oscillation frequency of each tuning fork is \(v_{0}=1400 \;\mathrm{Hz}\) and the velocity of sound in air is \(350\; \mathrm{m} / \mathrm{s}\). The speed of each tuning fork is close to
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Answer & Solution
Correct Answer
(D) \(\frac{1}{4}\; \mathrm{m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(v_{1}=\left(\frac{c}{c-v}\right) v_{0}\)
\(v_{2}=\left(\frac{c}{c+v}\right) v_{0}\)
beat frequency \(=v_{1}-v_{2}\)
\(=\mathrm{cv}_{0} \quad\left(\frac{1}{\mathrm{c}-\mathrm{v}}-\frac{1}{\mathrm{c}+\mathrm{v}}\right)\)
\(=c v_{0}\left(\frac{c+v-c+v}{c^{2}-v^{2}}\right)=\frac{2 c v_{0}^{2} v}{c^{2}-v^{2}}\)
\(\approx \frac{2 \mathrm{cv}_{0} \mathrm{v}}{\mathrm{c}^{2}}=\frac{2 \mathrm{v}_{0} \mathrm{v}}{\mathrm{c}}=2\)
\(\Rightarrow \frac{2 \times 1400 \times v}{350}=2\)
\(\Rightarrow \mathrm{v}=\frac{1}{4} \mathrm{m} / \mathrm{s}\)
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