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Time period of a simple pendulum of leng...

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

A

`(1)/(sqrt3)`

B

1

C

`sqrt((4)/(3))`

D

`sqrt((3)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
D

Time period of a simple pendulum of length l
`T_(1)=2pisqrt((l)/(g))`
Time period of a uniform rod
`T_(2)=2pisqrt(("Inertia factor")/("Spring factor"))`
Moment of inertia `=(ml^(2))/(12)+(ml^(2))/(4)=(ml^(2))/(3)`
and Spring factor=Restoring torque per unit angular displacement.
`T_(2)=2pisqrt(((ml^(2))/(3))/((mgl)/(2)))=2pisqrt((2l)/(3g)),(T_(1))/(T_(2))=sqrt(3/2)`.
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