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:

Explanation:

Given:

String A and B are of the same material and produce a frequency which is 6 Hz.

Therefore, The difference between f_A and f_B is 6 Hz

i.e, fA – fB = 6 Hz ---(1)

Where f_A is the frequency of A and f_B is the frequency of B respectively.

If tension decreases, f_B decreases and becomes f_B.

Now, the difference between f_A and f_B = 7 Hz (increases), which means that the frequency of string A is greater than the frequency of string B.

So, f_A > f_B

As we know, f_A = 530 Hz

Putting the value of f_A in equation (1) we get;

530 – f_B = 6 

⇒ -f_B = 6-530

⇒ -f_B = -524

⇒ f_B = 524 Hz.

Hence, option 4) is the correct answer.