There is an insulator rod of length L and of negligible mass with two small balls of mass m and electric charge Q attached to its ends. The rod can rotate in the horizontal plane around a vertical axis crossing it ata distance `L//4` from one of its ends.
What is the time period of the SHM as mentioned in the above question?

Options

A. `2 pi sqrt((mL)/(QE))`
B. `2 pi sqrt((2 mL)/(3 QE))`
C. `2 pi sqrt((5 mL)/(8QE))`
D. `2 pi sqrt((5mL)/(4QE))`

Answer

D

Explanation

Position (A) is of unstable equilibrium, because if displaced from this postion, resultanat torque will be in the direction of displacement and finally it will attain equilibrium position(B). At positions (B), velocity will be maximum.

`W=DeltaKE`
or, `QE (3L)/(2)-QE(L)/(2)=1/2m[v^(2)+(3v)^(2)]`
or, `v=sqrt((QEL)/(5m))`
Now we will find time period.

`tau=QE(3L)/(4) sin theta-QEL/4 sin theta=-I alpha`
or `QEL/2 sin theta= -m[((L)/(4))^(2)+((3L)/(4))^(2)]alpha`
or `QE(L)/(2)theta=(-10mL^(2))/(16)alpha` or `alpha= -((4)/(5) (QE)/(mL))theta`
`omega^(2)=(4QE)/(5mL) or omega sqrt((4QE)/(5mL))`
`T=(2 pi)/(omega)=2 pi sqrt((5mL)/(4QE))`.