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MHT CET 2026 (13 May, Shift 2) · Doppler Effect

Doppler Effect question from MHT CET 2026 (13 May, Shift 2)

Q8MHT CET 2026 · 13 MAY · SHIFT 2Medium

Doppler Effect

Two sirens situated one km apart are producing sound of frequency 330330 Hz. The observer starts moving from one siren to the other with a speed of 22 m/s. If the speed of sound is 330330 m/s, What will be the beat frequency heard by the observer?

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Answer: (b)

  1. Observer moves away from one siren and towards the other with u=2u = 2 m/s.
  2. Towards: f1=fv+uv=330×332330=332f_1 = f\frac{v+u}{v} = 330 \times \frac{332}{330} = 332 Hz
  3. Away: f2=fv−uv=330×328330=328f_2 = f\frac{v-u}{v} = 330 \times \frac{328}{330} = 328 Hz
  4. Beat frequency =f1−f2=4= f_1 - f_2 = 4 Hz
  5. So the answer is (b) 44 Hz.
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