STANDARD Physics Beat Frequency MCQ Question
Two sitar strings A and B are slightly out of tune and produce beats of frequency 5 Hz. When the tension in the string B is slightly increased, the beat frequency is found to reduce to 3 Hz. If the frequency of string A is 427 Hz, the original frequency of string B is
422 Hz
424 Hz
430 Hz
432 Hz
Correct Answer
Detailed Explanation
The frequency of string A, vₐ = 427 Hz. Let original frequency of string B be vᵦ. vᵦ = (vₐ ± 5) Hz = 432 Hz or 422 Hz. Increase in the tension of a string B, increases its frequency. If vᵦ = 432 Hz, a further increase in vᵦ would increase the beat frequency, which is not given. If vᵦ = 422 Hz, a further increase in vᵦ decreases the beat frequency, which matches the question. Thus, the original frequency of string B is 422 Hz.
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