Question

A ball is hung vertically by a thread of length 'l' from a point of an inclined wall that makes an angle with the vertical. The thread with the ball is then deviated through a small angle $\alpha (\alpha >\beta )$ and set free. Assuming the wall to be perfectly elastic, the period of such pendulum is

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

Answer 1

The correct option is A

$2\sqrt{\frac{l}{g}}[\frac{\pi }{2}+si{n}^{-1}\left(\frac{\beta }{\alpha }\right)]$

$T=\frac{T_0}{2}+2t=\pi \sqrt{\frac{l}{g}}+2t\dots \left(i\right)$

$t\to$ time to travel from O to $\beta$ angle ; $\beta =\alpha sin\omega t$

$t=\frac{1}{\omega }si{n}^{-1}\left(\frac{\beta }{\alpha }\right)\Rightarrow t=\frac{T_0}{2\pi }si{n}^{-1}\left(\frac{\beta }{\alpha }\right)$

Here $T_0=2\pi \sqrt{\frac{l}{g}}$

Putting value of t in eqn (1)

$T=2\sqrt{\frac{l}{g}}[\frac{\pi }{2}+si{n}^{-1}\left(\frac{\beta }{\alpha }\right)]$