The fundamental frequency of a sonometer wire increases by `6 Hz` if its tension is increased by `44%` keeping the length constant. Find the change in the fundamental frequency of the sonometer when the length of the wire is increased by `20%` keeping the original tension in the wire.

Explanation

Fundamental frequency of vibrations of string
` n = (1)/( 2 l) sqrt ((T)/(m))`
At constant length , `n prop sqrt(T) or (n)/ ( sqrt(T)) = constant`
i. New tension , `T' = T + ( 44)/( 100) T = 1.44 T`
If `n'` is new frequency , then `(n')/(sqrt( T')) = (n) /( sqrt(T))`
` n' = ( sqrt ((T')/(T)))n` (ii)
` :. n' = n + 6 `
From Eq. (iii)
` n + 6 = ( sqrt((1.44 T)/(T)))n`
or ` n + 6 = 1.2 n`
` 0.2 n = 6 or n = (6)/( 0.2) = 30 Hz`
ii. As tension is constant
` n prop (1)/(l) or nl = constant ` (iii)
When length increase by ` 20%`
New length ` l' = l + (20)/( 100) l = 1.2 l`
As ` nl = constant `
Therefore , ` nl = n' l'`
` n' = (l)/( l') n = (l)/( 1.2 l) xx 30 = 25 Hz`
Change in fundamental frequency
` Delta n = n' - n = 25 - 30 = - 5 Hz`
Therefore , ` Deltan = 5 Hz( decrease)`