Question:
A source of sound S emitting waves of frequency 100Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4m s⁻1 at an angle of \(60^\circ\) with the source–observer line as shown. The observer is at rest. The apparent frequency observed is (velocity of sound in air =330m s⁻1):
A source of sound S emitting waves of frequency 100Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4m s⁻1 at an angle of \(60^\circ\) with the source–observer line as shown. The observer is at rest. The apparent frequency observed is (velocity of sound in air =330m s⁻1):
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Only the component of velocity along the line of sight affects Doppler shift.
Updated On: Apr 2, 2026
- \(103\,\text{Hz}\)
- \(106\,\text{Hz}\)
- \(97\,\text{Hz}\)
- 100Hz
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The Correct Option is A
Solution and Explanation
Step 1: Component of source velocity towards observer:
\( v_s = 19.4 \cos 60^\circ = 9.7 \,\text{m s}^{-1} \)
Step 2: Doppler formula:
\( f' = \frac{f v}{v - v_s} \)
\( = \frac{100 \times 330}{330 - 9.7} \approx 103 \,\text{Hz} \)
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