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Question 31277
JEE Physics · Hard

Sound Waves: Doppler Application MCQ

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Doppler ApplicationHardQuestion 31277

Question

The whistle of a stationary train is 1000 Hz. An observer approaches at 34 m/s (v_sound = 340 m/s). Observed frequency is:
  1. A
    900 Hz
  2. B
    1100 Hz
    Correct
  3. C
    1050 Hz
  4. D
    1000 Hz

Correct answer

1100 Hz

Explanation

Observer moving toward stationary source: f = f₀(v+v₀)/v = 1000×(340+34)/340 = 1000×1.1 = 1100 Hz.

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