Question

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

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

Answer 1

The correct option is A $\sqrt{\frac{L}{g}}[\pi +2si{n}^{-1}\frac{\alpha }{\beta }]$

The motion of the string can be represented as an angular displacement from the mean position,given as: $\theta =\beta sin\left(\omega t\right)$

Time taken from displacement from mean position to $\beta$ and back to mean position=$t_1=\frac{T}{2}$

where $T$ is the time period of the oscillation without the wall barrier.

Thus $t_1=\frac{1}{2}\times 2\pi \sqrt{\frac{L}{g}}=\pi \sqrt{\frac{L}{g}}$

Time taken to displace from mean position to $\alpha$ can be found by:
$\alpha =\beta sin\left(\omega t\right)$

$⟹t=\frac{1}{\omega }si{n}^{-1}\frac{\alpha }{\beta }$

$=\sqrt{\frac{L}{g}}si{n}^{-1}\frac{\alpha }{\beta }$

Thus time taken to displace from mean position to $\alpha$ and back to mean position = $t_2=2\sqrt{\frac{L}{g}}si{n}^{-1}\frac{\alpha }{\beta }$

Thus the time period of the oscillation = $t_1+t_2=\sqrt{\frac{L}{g}}[\pi +2si{n}^{-1}\frac{\alpha }{\beta }]$