A simple pendulum of length `1m` is allowed to oscillate with amplitude `2^(@)`. it collides at `T` to the vertical its time period will be (use `g = pi^(2)`)

Options

A. `2//3 sec`
B. `4//3 sec`
C. `2 sec`
D. none of these

Answer

B

Explanation

Time period for half part
`T = 2pi sqrt((1)/(g)) = 2pi sqrt((1)/(g)) = 2pi sqrt((1)/(g)) = (2pi)/(g) = 2 sec`
So `2^(@)` part will be covered in a time `t = (T)/(2) = 1 sec`
For the half `1^(@)` part
`theta = theta_(2) sin (omega t)`
`1^(@) = 2^(@) sin ((2pi)/(T) xx t) rArr (t)/(2) = sin ((2pi)/(T) xx t)`
`rArr (pi)/(6) = pi xx 1 rArr t = 1//6sec`.
Total time `rArr (T)/(2) + 2t`
`rArr 1 + 2 xx (1)/(6) = 1 + (1)/(3) = (4)/(3) sec`