Question

A simple pendulum with a bob of mass m swings with an angular amplitude of ${60}^{\circ }$, when its angular displacement is ${30}^{\circ }$, what is the tension of string?

Answer 1

Suppose $l$ is the length of string. Then the height of the mass from the ground when the pendulum is at ${60}^o$ is $l\cos {60}^o=\frac{l}{2}$ and when at ${30}^o$ then height is $l\cos {30}^o=\frac{\sqrt{3}l}{2}$

Change in height $=\frac{\sqrt{3}l}{2}-\frac{l}{2}=\frac{l}{2}(\sqrt{3}-1)$

Change in potential energy $mg\frac{1}{2}(\sqrt{3}-1)$

So change in potential energy will be equal to change in K.E

since at the extreme position KI is zero

$\therefore$ total KE at point will be due to loss of PE only

$\frac{1}{2}m{v}^2=mg\frac{1}{2}(\sqrt{3}-1)$

$v^2=lg\sqrt{3}-1$

Now actual force is centrifugal force

$\frac{m{v}^2}{l}=\frac{mdg(\sqrt{3}-1)}{l}=mg(\sqrt{3}-1)$

Weight of mass in $mg$ and its radially outward component

$mg\sin {30}^o=\frac{mg}{2}$

$\therefore$ Tension in string will be equal to sum two force

$T=mg(\sqrt{3}-1)+\frac{mg}{2}=mg(\sqrt{3}-\frac{1}{2})$

$\therefore$ $T=mg(\sqrt{3}-\frac{1}{2})$