Question:

An observer moves towards a stationary source of sound with a velocity equal to one-fourth of the velocity of sound. Then the percentage increase in the apparent frequency observed by the observer is

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For a stationary source and moving observer, \[ f'=f\left(1+\frac{v_o}{v}\right). \] Hence the percentage increase in frequency is simply \[ \frac{v_o}{v}\times100. \] If \(v_o=\frac{v}{4}\), the increase is \(25\%\).
Updated On: Jul 29, 2026
  • \(25\%\)
  • \(20\%\)
  • \(30\%\)
  • \(50\%\)
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The Correct Option is A

Solution and Explanation

Concept: For a stationary source and an observer moving towards the source, the apparent frequency is given by \[ f' = f\left(\frac{v+v_o}{v}\right), \] where \[ v=\text{velocity of sound}, \qquad v_o=\text{velocity of observer}. \]

Step 1: Write the given information. The observer moves with speed \[ v_o=\frac{v}{4}. \]

Step 2: Calculate the apparent frequency. Using Doppler's formula, \[ f' = f\left(\frac{v+\frac{v}{4}}{v}\right). \] \[ = f\left(\frac{5v}{4v}\right). \] \[ = \frac54 f. \]

Step 3: Find the percentage increase in frequency. Increase in frequency: \[ \Delta f = f'-f = \frac54f-f. \] \[ = \frac14f. \] Therefore, \[ \text{Percentage Increase} = \frac{\Delta f}{f}\times100. \] \[ = \frac14\times100. \] \[ = 25\%. \] Therefore, \[ \boxed{\text{Percentage increase}=25\%} \] \[ \boxed{\text{Answer = (A)}} \]
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