Question

A ball suspended by a thread of length $l$ at the point $O$ on an inclined was as shown. The inclination of the wall with the vertical is $\alpha$. The thread is displaced through a small angle $\beta (>\alpha )$ away from the vertical and the ball is released. If the collision between the wall and the ball is elastic, find the period of oscillation of pendulum.

Answer 1

Obviously for small $\beta$ the ball execute part of S.H.M. Due to the perfectly elastic collision the velocity of ball simply reversed. As the ball in S.H.M. $(|\theta |<\alpha$ on the left) its motion law is differential from can be written as
$\ddot{\theta }=-\frac{g}{l}\theta =-{\omega }_0^2\theta \left(1\right)$
If we assume that ball is released from the extreme position, $\theta =\beta$ at $t=0$, the solution of differential equation would be taken in the form$\theta =\beta {\omega }_{0}t=\beta \cos \sqrt{\frac{g}{l}}t\left(2\right)$
If $t^{{}^{\prime }}$ be the time taken by the ball to go from the extreme position $\theta =\beta$ to the wall .i.e.
$\theta =-\alpha$, then Eqn $\left(2\right)$ can be rewritten as
$-\alpha =\beta \cos \sqrt{\frac{g}{l}}{t}^{{}^{\prime }}$
or $t^{{}^{\prime }}=\sqrt{\frac{l}{g}}{\cos }^{-1}(-\frac{\alpha }{\beta })=\sqrt{\frac{l}{g}}(\pi -{\cos }^{-1}\frac{\alpha }{\beta })$
Thus the sought time $T=2t^{{}^{\prime }}=2\sqrt{\frac{l}{g}}(\pi -{\cos }^{-1}\frac{\alpha }{\beta })$
$=2\sqrt{\frac{l}{g}}(\frac{\pi }{2}+{\sin }^{-1}\frac{\alpha }{\beta })$, [because ${\sin }^{-1}+{\cos }^{-1}=\pi /2$]