Question

A ball is suspended by a thread of length Lat the point O on the wall PQ which is inclined to the vertical through an angle $\alpha$ . The thread with the ball is now displaced through a small angle $\beta$ away from the vertical and the wall. If $\beta <\alpha ,$ then the time period of oscillation of the pendulum will be-

• A. $2\pi \sqrt{\frac{L}{g}}$
• B. $2\pi \sqrt{\frac{L}{g}}[\pi +2{\sin }^{-1}\left(\frac{\alpha }{\beta }\right)]$
• C. $2\sqrt{\frac{L}{g}}[\frac{\pi }{2}+{\sin }^{-1}\left(\frac{\alpha }{\beta }\right)]$
• D. None of the above

Answer 1

The correct option is C $2\sqrt{\frac{L}{g}}[\frac{\pi }{2}+{\sin }^{-1}\left(\frac{\alpha }{\beta }\right)]$

Obviously for small $\beta$ the ball execute part of SHM. Due to the perfectly elastic collision the velocity of the ball simply reversed. As the ball is in SHM ($\theta <\alpha onitsleft$)of its motion law in differential from can be given as:

$\theta =-\frac{g}{l}\theta =-{\omega }_0^2\theta$

If we assumethat the ball is relased from the extreme position,

$\theta =\beta cos{\omega }_{0}t=Bcos\sqrt{\frac{g}{l}}t$

If ${}^{\prime }{t}^{\prime }$ be the time taken by the ball to go from the extreme position$\theta =\beta$ to the wall.

i.e. $\theta =-\alpha ,$ then equation can be given as

$-\alpha =\beta cos\sqrt{\frac{g}{l}}{t}^{\prime }$

or

$t^{\prime }=\sqrt{\frac{l}{g}}co{s}^{-1}(-\frac{\alpha }{\beta })=\sqrt{\frac{l}{g}}(\pi -co{s}^{-1}\frac{\alpha }{\beta })$

thus the sought line $T=2t^{\prime }=2\sqrt{\frac{l}{g}}(\pi -co{s}^{-1}\frac{\alpha }{\beta })$

$=2\sqrt{\frac{l}{g}}(\frac{\pi }{2}+si{n}^{-1}\frac{\alpha }{\beta }),because[si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}]$