Time period of a simple pendulum of length L is `T_(1)` and time period of a uniform rod of the same length L pivoted about an end and oscillating in a vertical plane is `T_(2)`. Amplitude of osciallations in both the cases is small. Then `T_(1)/T_(2)` is

Options

A. `1/(sqrt(3))`
B. `1`
C. `sqrt(4/3)`
D. `sqrt(3/2)`

Answer

D

Explanation

Time period of simple pendulum is given by
`T_(1)=2pisqrt(l/g)`
and time period of uniform rod in given position is given by
`T_(2)=2pisqrt(("inertia factor")/("spring factor"))`

Here inertia factor moment of inertia of rod at one end
`=(ml^(2))/12+(ml^(2))/4=(ml^(2))/3`
Spring factor =restojring torque per unit angular displacement
`=mgxxl/2(sin theta)/(theta)`
`=mgxxl/2` (if `theta` is small)
`:.T_(2)=2pisqrt((ml^(2)//3)/(mgl//2))=2pisqrt(2/3" "l/g)`
Hence `(T_(1))/(T_(2))=sqrt(3/2)`