Question

A ball is suspended by a thread of length l at the point O on the wall PQ which is inclined 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, find the period of oscillation of the pendulum when $\beta >\alpha$. Assume the collision to be perfectly elastic:

• A. $\sqrt{\frac{L}{g}}[\pi +8{\sin }^{-1}\frac{\alpha }{\beta }]$
• B. $\sqrt{\frac{L}{g}}[\pi +2{\sin }^{-1}\frac{\alpha }{\beta }]$
• C. $\sqrt{\frac{L}{g}}[\pi +{\sin }^{-1}\frac{\alpha }{\beta }]$
• D. $\sqrt{\frac{L}{g}}[\pi +4{\sin }^{-1}\frac{\alpha }{\beta }]$

Answer 1

Answer: (B)

$\sqrt{\frac{L}{g}}[\pi +2{\sin }^{-1}\frac{\alpha }{\beta }]$

When $\beta >\alpha$, there are two sections of the motion.

1) The motion of ball with displacement of $\beta$ from the the vertical position.

2) The motion of ball with displacement of $\alpha$ from the vertical position.

In the motion of part (1), the time taken to go from vertical position upto $\beta$ and back to vertical=$\frac{T_{\beta }}{2}=\pi \sqrt{\frac{L}{g}}$

where $T_{\beta }$ is the time period of ball with maximum displacement $\beta$.

In the motion of part (2), let the equation of motion be $\theta =\beta sin\left(\omega t\right)$

where $\omega =\sqrt{\frac{g}{L}}$

Thus time taken to reach upto $\alpha =\frac{1}{\omega }si{n}^{-1}\frac{\alpha }{\beta }=\sqrt{\frac{L}{g}}si{n}^{-1}\frac{\alpha }{\beta }$

Thus, time period of the motion (2)$=2\sqrt{\frac{L}{g}}si{n}^{-1}\frac{\alpha }{\beta }$

Time period of motion=$\sqrt{\frac{L}{g}}[\pi +2si{n}^{-1}\frac{\alpha }{\beta }]$