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?

Answer 1

Let v₁ and v₂ be the frequencies of the strings A and B, respectively.

Then, v₁ = 324 Hz, v₂ = ?

Number of beats,

b = 6

v₁ = v₁ ± b = 324 ± 6

i.e., v₂ = 330 Hz or 318 Hz

Since, the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency v₁ will be reduced, i.e., number of beats will increase, if v₂ = 330 Hz. This is not so because number of beats becomes 3.

Therefore, it is concluded that the frequency = v₂ = 318 Hz, because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with v₂ = 318 Hz