# SHM with elastic collision.

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A ball suspended by a thread of length $l$ at the point $O$ on the wall, forming a small angle $\alpha$ with the vertical (Fig.). Then the thread with the ball deviated through a small angle $\beta$ $(\beta>\alpha)$ and set free. Assuming the collision of the ball with the wall to be perfectly elastic, find the oscillation period of such a pendulum. From fig $2\theta$ part of SHM will be skipped over, $\cos\theta=\dfrac{\alpha}{\beta}\implies\cos2\theta=\dfrac{2\alpha^2-\beta^2}{\beta^2}$. So time it would be skpping over will be $t=\dfrac{1}{\omega}\cos^{-1}\bigg(\dfrac{2\alpha^2-\beta^2}{\beta^2}\bigg)$. On substrating from total time $T=2\pi\sqrt{\dfrac{l}{g}}$ gives me $T'=2\sqrt{\dfrac{l}{g}}\bigg[\dfrac{\pi}{2}+\sin^{-1}\bigg(\dfrac{2\alpha^2-\beta^2}{\beta^2}\bigg)\bigg]$. But in official answer argument of sin inverse is $\dfrac{\alpha}{\beta}$.

Please help.

asked Nov 16, 2018
edited Nov 16, 2018

## 1 Answer

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The free oscillation (without the wall) can be described by $\phi=\beta\cos\omega t$ where angle $\phi$ made with the vertical is +ve to the right, and $\omega=\sqrt{\frac{g}{\ell}}$.

The time $t_1$ taken from $\phi=+\beta$ to $\phi=-\alpha$ is given by $-\alpha=\beta\cos\omega t_1$. So $$\frac{\alpha}{\beta}=-\cos\omega t_1=\sin(\omega t_1-\frac{\pi}{2})$$ $$\omega t_1=\frac{\pi}{2}+\sin^{-1}\frac{\alpha}{\beta}$$

The period of oscillation is $$T=2t_1=\frac{2}{\omega}(\frac{\pi}{2}+\sin^{-1}\frac{\alpha}{\beta})=2\sqrt{\frac{\ell}{g}}(\frac{\pi}{2}+\sin^{-1}\frac{\alpha}{\beta})$$

answered Nov 16, 2018 by (28,746 points)
selected Nov 19, 2018 by n3