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?

Question

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?

Philoid · Accepted Answer

Given,

Frequency of string A, f_A = 324 Hz

Let the frequency of string B be f_B.

Beat’s frequency, n = 6 Hz

Beat's Frequency is given as:

n=|f_Af_B|

⇒ 6 = 324 f_B

⇒ f_B = 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, f_B = 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.