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Two tuning forks, A and B, give 4 beats per second when sounded together. The frequency of A is 320 Hz. When some wax is added to B and it is sounded with A, 4 beats per second are again heard. The frequency of B is

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312 Hz

316 Hz

324 Hz

328 Hz

answer is C.

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Detailed Solution

n_A – n_B $↓$ = x (same) ... (i) $\to$ Wrong
n_B $↓$ – n_A = x (same) ... (ii) $\to$ Correct
$\Rightarrow$ n_B = n_A + x = 320 + 4 = 324 Hz.

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