MHT CET · Physics · Waves and Sound
A source of sound is moving with constant velocity of \(30 \mathrm{~m} / \mathrm{s}\) emitting a note of
frequency \(256 \mathrm{~Hz}\). The ratio of frequencies observed by a stationary observer
while the source is approaching him and after it crosses him is
[speed of sound in air \(=330 \mathrm{~m} / \mathrm{s}]\)
- A \(8: 9\)
- B \(9: 8\)
- C \(5: 6\)
- D \(6: 5\)
Answer & Solution
Correct Answer
(D) \(6: 5\)
Step-by-step Solution
Detailed explanation
A source of sound is moving with constant velocity of \(30\) mis emitting a note of frequency \(256 \mathrm{~Hz}\). The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him is \(6: 5\).
\(V_{S}=30 \mathrm{~m} / \mathrm{s} \quad\mathrm{n}_{0}=256 \mathrm{~Hz}\)
\(n_{1}=n_{0} \frac{V}{V-V_{s}}\)
\(n_{2}=n_{0} \frac{V}{V+V_{s}}\)
\(\therefore \frac{n_{1}}{n_{2}}=\frac{V+V_{s}}{V-V_{s}}=\frac{360}{300}=\frac{6}{5}\)
\(V_{S}=30 \mathrm{~m} / \mathrm{s} \quad\mathrm{n}_{0}=256 \mathrm{~Hz}\)
\(n_{1}=n_{0} \frac{V}{V-V_{s}}\)
\(n_{2}=n_{0} \frac{V}{V+V_{s}}\)
\(\therefore \frac{n_{1}}{n_{2}}=\frac{V+V_{s}}{V-V_{s}}=\frac{360}{300}=\frac{6}{5}\)
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