Question

A ball is suspended by a thread of length l at the point O on the wall, forming a small angle $\beta (\beta >\alpha )$ with the vertical (see the figure). Then the thread with the ball was deviated through a small angle $\beta$ and set free. Assuming the collision of the ball against the wall to be perfectly elastic, the period of oscillation of this pendulum is:

• A. $T=2\sqrt{\frac{l}{g}}[\frac{\pi }{2}+si{n}^{-1}\left(\frac{\alpha }{\beta }\right)]$
• B. $T=2\pi \sqrt{\frac{l}{g}}[\frac{\pi }{2}+si{n}^{-1}\left(\frac{\alpha }{\beta }\right)]$
• C. $T=2\sqrt{\frac{l}{g}}[\frac{\pi }{2}+co{s}^{-1}\left(\frac{\alpha }{\beta }\right)]$
• D. $T=2\pi \sqrt{\frac{l}{g}}[\frac{\pi }{2}+co{s}^{-1}\left(\frac{\alpha }{\beta }\right)]$

Answer 1

The correct option is C $T=2\sqrt{\frac{l}{g}}[\frac{\pi }{2}+co{s}^{-1}\left(\frac{\alpha }{\beta }\right)]$

We have,
$\theta =\beta cos\left(\sqrt{\frac{g}{l}}t\right)$ [since at t = 0, $\theta =\beta$ ]
$\Rightarrow$ when $\theta =\alpha$
$t=\sqrt{\frac{l}{g}}co{s}^{-1}\left(\frac{\alpha }{\beta }\right)$
The time period is,
$T=2t+\frac{1}{2}{T}_{SHM}$
$=2\sqrt{\frac{l}{g}}o{s}^{-1}\left(\frac{\alpha }{\beta }\right)+\pi \sqrt{\frac{l}{g}}$