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.

Options

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

Answer

D

Explanation

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