A ball is suspended by a thread of length l at the point O on an incline wall as shown. The inclination of the wall with the vertical is α.The thread is displaced through a small angle β away from the vertical and the ball is released. Find the period of oscillation of pendulum.
Consider both cases
a. `alpha gt beta`
b. `alpha lt beta`
Assuming that any impact between the wall and the ball is elastic.

Explanation

Obviously for small `beta` the ball execute parto for S.H.M. Due to the perfectly elastic collision the velocity of ball simply reversed. As the ball is in S.H.M. `(|theta|lt alpha` n the left) its motion law in differential from can be written as
`ddot (theta)=-(g)/(l)theta=-omega_(0)^(2)theta .....(1)`
If we assume that the ball is released from the extreme position, `theta= beta` at `t=0` , the solout of differential equation would be taken in the form
`theta = beta cos omega_(0)t= beta cos sqrt((g)/(l))t`
If `t^(')` be the time taken by the ball to go from the extreme position `theta= beta` to the wall i.e., ` theta = - alpha`, then Eqn, (2) can be rewritten as
`- alpha=beta cos sqrt((g)/(l))t^(')` or `t^(')=sqrt((l)/(g)) cos^(-1)(-(alpha)/(beta))=sqrt((l)/(g))(pi-cos^(-1)((alpha)/(beta)))`
Thus the sought time `T=2t^(')=2 sqrt((l)/(g))(pi-cos^(-1)((alpha)/(beta)))`
`=2 sqrt((l)/(g))((pi)/(2)+sin ^(-1)((alpha)/(beta)))`, [because`sin ^(-1) x+ cos ^(-1)=pi//2]`