The fundatmental 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

In case of vibration of a string,fundamental frequency is given by
`f = (1)/(2L) sqrt((T)/(mu))`
So if length of given wire is kept constant
`(f'//f) = (T'//T) 1//2`
and as here `f' = f + 6` and `T' = T + 0.44 T = 1.44 T`
`((f + 6))/(f) = sqrt((1.44T)/(T))`
f = 30 Hz
Now if keeping the original tension (7), the length wire is changed, we have
`(f'')/(f) = (l)/(l'') = (1)/(1.20)`
`f'' = (30)/(1.2) = 25 Hz`
`Deltaf = f'' - f`= 25 - 30 = - 5Hz`
Thus,fundamental frequencywilldecreaseby 5 Hz