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MHT CET · Physics · Waves and Sound

When an observer moves towards a stationary source with velocity ' \(\mathrm{V}_1\) ', the appearent frequency of emitted note is ' \(\mathrm{F}_1\) '. When observer moves away from stationary source with velocity ' \(\mathrm{V}_1\) ' the appearent frequency is ' \(F_2\) '. If ' \(v\) ' is velocity of sound in air and \(\frac{F_1}{F_2}=2\), then \(\frac{V}{V_1}\) is equal to

  1. A \(6\)
  2. B \(5\)
  3. C \(3\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

\(F_1 = F_0 \left( \frac{v + V_1}{v} \right)\) \(F_2 = F_0 \left( \frac{v - V_1}{v} \right)\)
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