Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Given,


Frequency of string A, fA = 324 Hz


Let the frequency of string B be fB.


Beat’s frequency, n = 6 Hz


Beat's Frequency is given as:


n=|fAfB|


6 = 324 fB


fB = 318 Hz or 330 Hz


Frequency decreases with a decrease in the tension in a string because frequency is directly proportional to the square root of tension as f T


Hence, the beat frequency cannot be 330 Hz.


So, fB = 318 Hz


NOTE: A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies.


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