NEET 2020 Physics Beats MCQ Question
In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be:
536 Hz
537 Hz
523 Hz
524 Hz
Correct Answer
Detailed Explanation
Beat frequency = |fA − fB| = 6 Hz. So fB = 530 + 6 = 536 Hz or fB = 530 − 6 = 524 Hz. When tension in B is decreased, frequency of B decreases. If fB were 536 Hz, decreasing it would bring it closer to 530, reducing the beat frequency. But the beat frequency increases to 7 Hz, so fB must be below fA. If fB = 524 Hz and it decreases further, |530 − fB| increases, giving 7 Hz. Therefore, the original frequency of B is 524 Hz (option D).
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