There is a rod of length `l` and mass `m`. It is hinged at one end to the ceiling. The period of small oscillation is

Options

A. `T=2pisqrt((2l)/(3g))`
B. `T=pisqrt((l)/(3g))`
C. `T=2pisqrt((l)/(3g))`
D. `T=2pisqrt((l)/(g))`

Answer

A

Explanation

When the rod is at position shown in figure , the restoring torque ono the rod is
`tau=-mgxx(GG^('))= (-mg) (l)/(2) theta` …(i)
`tau prop theta`
` ( :' (mgl)/(2)=` a constant for a given position )
Therefore, if the rod is left free, if will execute angular SHM.
Comparing (i) with `tau=-ktheta`
We have, spring factor, `k=(mgl)/(2)`
Inertia factor `=` moment of indertia of rod about point of suspension as axis
`=(ml^(2))/(12)+m((l)/(2))^(2)=(ml^(2))/(3)`
As, time period `T=2pisqrt((I n ertia fact o r)/(spri ng fact o r ))`
`=2pisqrt((ml^(2)//3)/(mgl//2))=2pisqrt((2l)/(3g))`