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The bob of simple pendulum having length...

The bob of simple pendulum having length 1, is displaced from mean position to an angular position o with respect to vertical. If it is released, then velocity of bob at equilibrium position is

A

`sqrt(2g(1- cos theta))`

B

`sqrt(2g(1+cos theta))`

C

`sqrt(2gl cos theta)`

D

`sqrt(2gl)`

Text Solution

Verified by Experts

The correct Answer is:
A
In `triangleOAC, cos theta = OA//l` or, `OA = l cos theta`
`therefore AB = 1(1- cos theta) =h`
At point C, the velocity of bob =0.
The vertical acceleration =g
`therefore v^(2) = 2gh` or, `v=sqrt(2gl(1- cos theta))`
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