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JEE Main 2024 (1 Feb, Shift 1) · Beats

Beats question from JEE Main 2024 (1 Feb, Shift 1)

Q36JEE Main 2024 · 1 FEB · SHIFT 1Medium

Beats

A tuning fork resonates with a sonometer wire of length 1 m1\text{ m} stretched with a tension of 6 N6\text{ N}. When the tension in the wire is changed to 54 N54\text{ N}, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is ________________ Hz\text{Hz}.

Numerical answer type. Work it out, then open the answer.

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Answer: 6

  1. f∝Tf \propto \sqrt{T}; at 6 N the wire is in unison with the fork, so f=nf = n.
  2. At 54 N: f′=n54/6=3nf' = n\sqrt{54/6} = 3n.
  3. 3n−n=12⇒n=6 Hz3n - n = 12 \Rightarrow n = 6\text{ Hz}
  4. So the frequency of the fork is 6 Hz6\text{ Hz}.
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