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:
A
524 Hz
B
536 Hz
C
537 Hz
D
523 Hz
Step-by-Step Solution
Given, the frequency of string A is fA=530 Hz.
The beat frequency produced by strings A and B is 6 Hz.
Therefore, the original frequency of string B can be either:
fB=fA+6=536 Hz
or
fB=fA−6=524 Hz
The frequency of a stretched string is directly proportional to the square root of its tension (f∝T). When the tension in string B is decreased, its frequency fB decreases.
Now, let's analyze both cases with the new decreased frequency fB′:
Case 1: If fB was 536 Hz, a decrease in tension would lower the frequency (e.g., to 535 Hz). The new beat frequency would be ∣530−535∣=5 Hz. Here, the beat frequency decreases, which contradicts the given information.
Case 2: If fB was 524 Hz, a decrease in tension would lower the frequency (e.g., to 523 Hz). The new beat frequency would be ∣530−523∣=7 Hz. Here, the beat frequency increases to 7 Hz, which exactly matches the given condition.
Therefore, the original frequency of string B must be 524 Hz.
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NEET PHYSICS: "In a guitar, two strings $A$ and $B$ made of same material are slightly out of tune and produce beats of frequency $6 \t..." — Solved MCQ | TopperSquare