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Question 31284
JEE Physics · Hard

Sound Waves: Beats Advanced MCQ

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Beats AdvancedHardQuestion 31284

Question

A tuning fork of frequency f is held near an open pipe of length L. As pipe length slowly increases from L₀, beats are first heard then become zero at length L₁ = v/(2f). This means:
  1. A
    Pipe has gone out of resonance
  2. B
    Pipe resonates at f — beats disappear at resonance
    Correct
  3. C
    Fork frequency changed
  4. D
    Temperature changed

Correct answer

Pipe resonates at f — beats disappear at resonance

Explanation

When L₁ = v/(2f), the pipe's fundamental matches the fork. Pipe vibrates in resonance with fork — single frequency, no beats. Beats disappear at resonance.

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