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NEET PhysicsHard

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:

1

523 Hz

2

524 Hz

3

536 Hz

4

537 Hz

Step-by-Step Solution

Difference of fAf_A and fBf_B is 6 Hz. If tension decreases, fBf_B decreases to fB′f_B'. New difference ∣fA−fB′∣=7 Hz|f_A - f_B'| = 7 \text{ Hz}. Since the beat frequency increased, fA>fBf_A > f_B. Thus fA−fB=6 Hz  ⟹  530−fB=6  ⟹  fB=524 Hzf_A - f_B = 6 \text{ Hz} \implies 530 - f_B = 6 \implies f_B = 524 \text{ Hz}.

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