Question

A simple pendulum of length l is suspended from a hook mounted on a slanted wall. The
wall makes a small angle theta with the vertical. The pendulum is displaced from the vertical
by a small angle phi(phi>theta)  and released. Assuming that the collision of the bob is elastic,
the time period of oscillation is

Answer 1

Answer on Question#55068 - Physics - Mechanics | Kinematics | Dynamics

A simple pendulum of length $l$ is suspended from a hook mounted on a slanted wall. The wall makes a small angle theta with the vertical. The pendulum is displaced from the vertical by a small angle phi(phi>theta) and released. Assuming that the collision of the bob is elastic, the time period of oscillation is

Solution.

Without wall period of oscillation would be $T_0 = 2\pi \sqrt{\frac{l}{g}}$. Period of oscillation with wall: $T = T_0 - \Delta t$, where $\Delta t$ is time in which pendulum without wall would move from angle $\theta$ to $\varphi$ and return to $\theta$. To find $\Delta t$ we have to solve the equation: $\varphi \sin \sqrt{\frac{g}{l}} = \theta$. We have $t_1 = \sqrt{\frac{l}{g}} \sin^{-1}\left(\frac{\theta}{\varphi}\right)$ (first time pass $\theta$) and $t_2 = \sqrt{\frac{l}{g}} (\pi - \sin^{-1}\left(\frac{\theta}{\varphi}\right))$ (return to $\theta$). Now we find $\Delta t = t_{2-} t_1 = \sqrt{\frac{l}{g}} \left( \pi - 2 \sin^{-1}\left(\frac{\theta}{\varphi}\right) \right)$ and, finally, $T = \sqrt{\frac{l}{g}} \left( \pi + 2 \sin^{-1}\left(\frac{\theta}{\varphi}\right) \right)$.

Answer: $T = \sqrt{\frac{l}{g}} \left( \pi + 2 \sin^{-1} \left( \frac{\theta}{\varphi} \right) \right)$.

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