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A ball is suspended by a thread of length $L$ at the point $O$ on a wall 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, the period of oscillation of the pendulum when $\beta>\alpha$ will be
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The correct answer is:
$\sqrt{\frac{L}{g}}\left[\pi+2 \sin ^{-1} \frac{\alpha}{\beta}\right]$
$\theta=\theta_0 \sin \omega t, \quad \omega=\sqrt{\frac{g}{L}}$
$\because T=2 \pi \sqrt{\frac{L}{g}}$
When $\beta>\alpha$, time taken by pendulum from $B$ to $C$ and $C$ to $B$
$t_1=\frac{T}{2}=\frac{1}{2} \times 2 \pi \sqrt{\frac{L}{g}}=\pi \sqrt{\frac{L}{g}}$
Time taken by pendulum from $B$ to $A$ and $A$ to $B$ $t_2=2 t=\frac{2}{\omega} \sin ^{-1}\left(\frac{\alpha}{\beta}\right)$ using $\theta=\theta_0 \sin \omega \mathrm{t}$ $\alpha=\beta \sin \omega t$ or $t=\frac{1}{\omega} \sin ^{-1}\left(\frac{\alpha}{\beta}\right)$
$\therefore$ Time period of motion
$T=t_1+t_2=\sqrt{\frac{L}{g}}\left[\pi+2 \sin ^{-1} \frac{\alpha}{\beta}\right]$
$\because T=2 \pi \sqrt{\frac{L}{g}}$
When $\beta>\alpha$, time taken by pendulum from $B$ to $C$ and $C$ to $B$
$t_1=\frac{T}{2}=\frac{1}{2} \times 2 \pi \sqrt{\frac{L}{g}}=\pi \sqrt{\frac{L}{g}}$
Time taken by pendulum from $B$ to $A$ and $A$ to $B$ $t_2=2 t=\frac{2}{\omega} \sin ^{-1}\left(\frac{\alpha}{\beta}\right)$ using $\theta=\theta_0 \sin \omega \mathrm{t}$ $\alpha=\beta \sin \omega t$ or $t=\frac{1}{\omega} \sin ^{-1}\left(\frac{\alpha}{\beta}\right)$
$\therefore$ Time period of motion
$T=t_1+t_2=\sqrt{\frac{L}{g}}\left[\pi+2 \sin ^{-1} \frac{\alpha}{\beta}\right]$
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