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 be at frequency increases to $7\,Hz$. If the frequency of A is $530\,Hz$, the original frequency of B will be :

• A. 523 Hz
• B. 524 Hz
• C. 536 H z
• D. 537 H z

Answer 1

The Correct Option is B

To determine the original frequency of string B, we must first understand the concept of beats, which occur when two sound waves of slightly different frequencies interfere with each other. The beat frequency is given by the absolute difference between these two frequencies.

Given that the frequency of string A is \(530\,Hz\), and the beat frequency with string B is initially \(6\,Hz\), this means that the frequency of string B could be either:

• \(530\,Hz + 6\,Hz = 536\,Hz\)
• \(530\,Hz - 6\,Hz = 524\,Hz\)

When the tension in string B is slightly decreased, its frequency decreases as well. The beat frequency with string A increases to \(7\,Hz\). This indicates that the frequencies are now even closer together, suggesting that the original frequency of B must have been \(524\,Hz\).

Explanation: Since decreasing tension results in a reduced frequency, if the original frequency had been \(536\,Hz\), decreasing the tension would result in B's frequency being closer to that of A's frequency of \(530\,Hz\), reducing the beat frequency. However, the beat frequency increased, confirming that the original frequency was the lower value, \(524\,Hz\).

Thus, the original frequency of string B was 524 Hz.