Two masses m and `(m)/(2)` are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is `tau=ktheta` for angular displacement `theta`. If the rod is rotated by `theta_(0)` and released, the tension in it when it passes through its mean position will be :

Options

A. `(3 k theta_(0)^(2))/l`
B. `(ktheta_(0)^(2))/(2l)`
C. `(2k theta_(0)^(2))/l`
D. `(ktheta_(0)^(2))/l`

Answer

D

Explanation

`I alpha =-k theta`
`[m/2((2l)/3)^(2)+m(l/3)^(2)] alpha=-k theta`
`(ml^(2))/3 alpha=-k theta :. omega^(2)=(3k)/(ml^(2))`
Angular velocity at man `omega_(0)=theta_(0) omega =theta_(0) sqrt((3k)/(ml^(2)))`
`T=m omega_(0)^(2)l/3=m theta_(0)^(2) (3k)/(ml^(2)) l/3=(k theta_(0)^(2))/l`