A simple pendulum of bob mass m and length l is displaced from its mean position O to a point A and then released . If v is the velocity of the bob at O, h is the height of string of pendulum when bob passes through point O is /are (neglect friction )

Options

A. `(2mgh)/(l)`
B. `2mg""(1+(h))/(l)`
C. `mg""(1+(2h))/(l)`
D. `mg""(1-(2h))/(l)`

Answer

C

Explanation

P.E. of pendulum at A = mgh
P.E. will be completely converted to K.E. when bob reaches its mean position O.
`therefore (1)/(2) mv^(2) = mgh`
`rArr v^(2) = 2gh`
Also , T -mg = centripetal force `(A + 0) = (mv^(2))/(l)`
`rArr T = mg + (mv^(2))/(l) = mg + (2mgh)/(l)`
`T = mg + ((1 + 2h)/(l))`