A ball is suspended by a thread of length L at the point O on a wall which in indeed to the vertical by `alpha`. The thread with the ball is displaced by a small angle `beta` away from the vertical and also away from the wall. If the ball is released, the period of observation of the pendulum when `beta gt alpha` will be...........

Options

A. `sqrt((L)/(g)) [pi + 2 sin ^(-1) (alpha/beta)]`
B. `sqrt((L)/(g)) [pi - 2 sin ^(-1) (alpha/beta)]`
C. `sqrt((L)/(g)) [2 sin ^(-1) (alpha/beta)-pi]`
D. `sqrt((L)/(g)) [2sin ^(-1) (alpha/beta)+pi]`

Answer

A

Explanation

Angular position `theta= theta_(0) sin omega t" and " omega = sqrt((g)/(L))`
`therefore T= 2pi sqrt((L)/(g))`
When `beta propto alpha`, time taken by pendulum from B to C and C to B
`t_(1)= (T)/(2)= (1)/(2)xx 2pi sqrt((L)/(g))= pi sqrt((L)/(g))" ""........"(1)`
Time taken by pendulum from B to A and A to B,
`t_(2)= 2t = (2)/(omega)sin^(-1) (alpha/beta)" "(" using " theta = theta_(0) sin omega t)`
`therefore alpha= beta sin omega t" or "t= (1)/(omega) sin^(-1) (alpha/beta) = sqrt((L)/(g)) sin^(-1) (alpha)/(beta)`
`t_(2)= 2t = 2sqrt((L)/(g)) sin^(-1) (alpha/beta)" ""........"(2)`
Time period of motion,
`T= t_(1)+t_(2)`
`= sqrt((L)/(g)) [pi + 2 sin^(-1) (alpha)/(beta)]`.