Two sitar stings A and B playing the note 'ga' are slightly out of tune and produce beats of frequency 6Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to by 3Hz. If the original frequency of a is 324 Hz, then what is the frequency of B ?

Explanation

Given, `f_(1) ~ f_(2) =6Hz`
Since `f prop sqrt( T)`, and upon reducing the tension, beat frequency reduces,
`f_(1) gt f_(2)`
i.e., ` f_(1)-f_(2)=6`
and `f_(2) = f_(1) - 6 = 324 - 6 = 318 Hz`
`f _(2) = 318 Hz`
Natural frequency of B is 318 Hz
Frequency of A under lower tension will be
`f'_(A) = f_(B) + f'_(b) = 318 + 3 = 321 Hz`.