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JEE Main 2026 (28 Jan, Shift 2) · Beats

Beats question from JEE Main 2026 (28 Jan, Shift 2)

Q7JEE Main 2026 · 28 JAN · SHIFT 2Easy

Beats

Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________ Hz.

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

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Answer: 384

Beat frequency: $8/2 = 4\text{ Hz}$ , then $4/2 = 2\text{ Hz}$ .
$n_A = 380 \pm 4 = 384$ or $376\text{ Hz}$ .
Wax lowers $n_A$ and the beats fall, so $n_A > 380$ : $384 \to 382\text{ Hz}$ .
So $n_A = 384\text{ Hz}$ .

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