JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2019
- Easy
- Numerical
A source of sound \(S\) is moving with the velocity of \(50\,m/s\) towards a stationary observer. The observer measures the frequency of the sound as \(1000\,Hz.\) What will be the apparent frequency of the source when it is moving away from the observer after crossing him ... \(Hz\) ? (Take velocity of sound in air is \(350\,m/s\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(750\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}_{\mathrm{a}}=\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}} \mathrm{f}_{\mathrm{o}}=1000 \mathrm{Hz}\)
\(\mathrm{f}_{\mathrm{a}}^{\prime}=\frac{\mathrm{V}}{\mathrm{V}+\mathrm{V}_{\mathrm{s}}} \mathrm{f}_{\mathrm{o}}\)
\(\frac{f_{a}^{\prime}}{f_{a}}=\frac{V-V_{s}}{V+V_{s}}=\frac{350-50}{350+50}=\frac{300}{400}=\frac{3}{4}\)
\(\mathrm{f}_{\mathrm{a}}^{\prime}=\frac{3}{4} \times 1000=750 \mathrm{Hz}\)
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