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?

#### Solution 1

Frequency of string A, *f*_{A} = 324 Hz

Frequency of string B = *f*_{B}

Beat’s frequency, *n* = 6 Hz

Beat's Frequency is given as:

`n = |f_A +- f_B|`

`6 = 324 +- f_B`

`F_B = 330 "Hz" or 318 "Hz"`

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:

`v prop sqrtT`

Hence the beat frequency cannot be 330 Hz

`:.f_n = 318Hz`

#### Solution 2

Let υ_{1} and υ_{2} be the frequencies of strings A and B respectively.

Then, υ_{1} = 324 Hz, υ_{2} = ?

Number of beats, b = 6

υ_{2} = υ_{1} ± b = 324 ± 6 !.e., υ_{2} = 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 υ_{1} will be reduced i.e., number of beats will increase if υ_{2} = 330 Hz. This is not so because number of beats become 3.

Therefore, it is concluded that the frequency υ_{2} = 318 Hz. because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with υ_{2} = 318 Hz.