Question:

A source of sound of frequency \(500\ \text{Hz}\) is moving towards an observer with velocity \(30\ \text{m/s}\). The speed of sound is \(330\ \text{m/s}\). The frequency heard by the observer will be

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When source moves towards observer, apparent frequency increases; use denominator \(v-v_s\).
  • \(450\ \text{Hz}\)
  • \(550\ \text{Hz}\)
  • \(600\ \text{Hz}\)
  • \(500\ \text{Hz}\)
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The Correct Option is B

Solution and Explanation

Concept: When a source moves towards a stationary observer, apparent frequency increases and is given by: \[ f' = f\left(\frac{v}{v-v_s}\right) \] where \(v\) is speed of sound and \(v_s\) is speed of source.

Step 1:
Given: \[ f=500\ \text{Hz} \] \[ v=330\ \text{m/s} \] \[ v_s=30\ \text{m/s} \]

Step 2:
Apply Doppler effect formula. \[ f'=500\left(\frac{330}{330-30}\right) \] \[ f'=500\left(\frac{330}{300}\right) \] \[ f'=500\left(\frac{11}{10}\right) \] \[ f'=550\ \text{Hz} \] Therefore, \[ \boxed{550\ \text{Hz}} \]
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