the fundamental frequency of a sonometer wire of length is `f_(0)`.A bridge is now introduced at a distance of `Deltal`from the centre of the wire `(Deltal lt lt l )`. The number of beats heard if their fundamental mode are

Options

A. `(8f_(0)Deltal)/(l)`
B. `(f_(0)Deltal)/(l)`
C. `(2f_(0)Deltal)/(l)`
D. `(4f_(0)Deltal)/(l)`

Answer

C

Explanation

`f_(0)=(V)/(2l)`, Now beats/s `=f_(1)-f_(2)`
`=(V)/(2)[(1)/(((l)/(2)-Deltal))-(1)/(((l)/(2)+Deltal))]`
`2f_(0)l[(l+2Deltal-l+2Deltal)/(l^(2)-4Deltal^(2))]~-(8f_(0)Deltal)/(l)`