When the length of the vibrating segment of a sonometer wire is increased by 1%, the percentage change in its frequency is

Options

A. `(100)/(101)`
B. `(99)/(100)`
C. 1
D. 2

Answer

C

Explanation

`n=(1)/(2L)sqrt((T)/(m))`
Take log and differentiate, we get
`(Delta n)/(n)=(Delta L)/(L)+(1)/(2)(Delta T)/(T)+(1)/(2)(Delta m)/(m)` (`because` one takes only absolute values)
`therefore (Delta n)/(n)=(Delta L)/(L)+0+0`
% change in frequency = 1 % increase in length = 1 %.