Question

A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position $(seefig.)$ and released.The ball hits the wall, the coefficient of restitution being $\frac{2}{\sqrt{5}}$. What is the minimum number of collision after which the amplitude of oscillations becomes less than 60 degrees?

Answer 1

Given: A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position (see fig.) and released.The ball hits the wall, the coefficient of restitution being $\frac{2}{\sqrt{5}}$

To find the minimum number of collision after which the amplitude of oscillations becomes less than 60 degrees

Solution:

The pendulum bob is left from the position A. When it is at position C, the angular position amplitude is ${60}^{\circ }$

In $\Delta OCM$ as shown in above figure,

$\cos {60}^{\circ }=\frac{OM}{l}⟹OM=\frac{l}{2}$

The velocity of the bob at B, $V_B$ before first collision is

$mgl=\frac{1}{2}m{V}_B^2⟹V_B=\sqrt{2gl}$

Let after n collisions, the angular amplitude is ${60}^{\circ }$ when the bob again moves towards the wall from C, the velocity $V_B^{\prime }$ before collision is

$mg\frac{l}{2}=\frac{1}{2}m{V}_B^{\prime 2}⟹V_B^{\prime }=\sqrt{gl}$

This means that the velocity of the bob should reduce from $\sqrt{2gl}$ to $\sqrt{gl}$ due to collisions with walls.

The final velocity after n collisions is $\sqrt{gl}$

Therefore, $e^n\left(\sqrt{2ggl}\right)=\sqrt{gl}$

where $e$ is te coefficient of restitution.

${\left(\frac{2}{\sqrt{5}}\right)}^n\times \sqrt{2gl}=\sqrt{gl}⟹{\left(\frac{2}{\sqrt{5}}\right)}^n=\frac{1}{\sqrt{2}}$

Taking log on both sides we get

$n\log \left(\frac{2}{\sqrt{5}}\right)=\log \frac{1}{\sqrt{2}}⟹n=4$

Therefore, number of collisions will be 4