The distance of point of a compound pendulum form its centre of gravity is `l`, the time period of oscillation relative to this point `T`. If `g = pi^(2)`, the relation between `l` and `T` will be :-

Options

A. `l^(2)-[T^(2)/4]l+k^(2)=0`
B. `l^(2)+[T^(2)/4]l+k^(2)=0`
C. `l^(2)-[T^(2)/4]l-k^(2)=0`
D. `l^(2)+[T^(2)/4]l-k^(2)=0`

Answer

A

Explanation

`tau = -mgsin theta l rArr lalpha = -mglsin theta`
`rArr m = - (gl theta)/((l^(2) + k^(2))) = -omega^(2)theta (sin~ theta)`
`rArr alpha = - (g l theta)/((l^(2) + k^(2))) = -omega^(2) theta rArr omega = sqrt((gl)/(l + k^(2)))`
`rArr T = (2pi)/(omega) = 2pi sqrt((l^(2) + k^(2))/(gl)) = 2 sqrt((l^(2) + k^(2))/(l)) [g = pi^(2)]`
`rArr l^(2) - ((T^(2))/(4)) l + k^(2) = 0`