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 (a)the thread is displacement througha small angle away from the vertical and (b) the ball is released. Find the period of obcillation 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

If `alpha gt beta`, the ball does not collide which the wall and it performed full oscillation like simple pendulum. Thus,
`Period = 2 pi sqrt((l)/(g))`

b. If `alpha lt beta`, the ball collide with the wall and rebounds with same speed the motion of ball from `A and Q` is one part of a simple pendulum time Period of ball`= 2 (t_(AQ)`
Consider A as the starting point `(t = 0)`, equition of motion is `x(t) = A cos omega t`.
`x (t) = l beta cos omega t`, because amplitude `= A = l beta`
Time from A to Q is the time t when x becomes - l alpha`
`implies - l alpha - l beta cos omega t implies t = t_(AQ)` = l// omega cos^(-1) ((-alpha)/(beta))`
The return path from `Q and A` will involve the same time interval. Hence time period of ball `= 2 t_(AQ)`
`= (2)/(omega) cos^(-1) (-(alpha)/(beta)) = 2 sqrt((t)/(g)) cos^(-1) ((-alpha)/(beta))`
`= 2 pi sqrt((l)/(g)) - 2 sqrt ((l)/(g)) cos ^(-1) ((alpha)/(beta))`