Question

A simple pendukum with a bob of mass m swings with angular amplitude of ${60}^o$. When its angular displacement ${30}^o$ , find the tension of string.

• A. $\frac{mg}{2}$
• B. $\frac{3\sqrt{3mg}}{2}$
• C. $\frac{mg}{2}(3\sqrt{3}-2)$
• D. $mg(\sqrt{3}-\frac{1}{2})$

Answer 1

The correct option is D $mg(\sqrt{3}-\frac{1}{2})$

Suppose 'l' is the lenght of the string. Then the height of the mass from the ground when the pendulum is at ${60}^0$ is $l\cos 60=\frac{1}{2}$ and when at ${30}^0$ then height is $l\cos 30=\frac{\sqrt{3}l}{2}$.

So change in height is $\frac{l}{2}(\sqrt{3}-1)$.

Thus cahnge in potential energy $\Delta P.E=mg\frac{l}{2}(\sqrt{3}-1)$ is

So the change in potentila energy will be equal to the change in kinetic energy.

Since at the extreme position the kinetic energy is 0. The total kinetic energy at the point will be equal to the loss of potential energy.

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

$v{]}^{[}2=lg(\sqrt{3}-1)$

Now the outward force due to centrifugal force is

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

Now the weight of the mass is mg and is radially outward component at that point is $mg\sin 30=\frac{mg}{2}$

Now the tension in the string will be equal to the sum of the above two forces, hence tension $T=mg(\sqrt{3}-1)+\frac{mg}{2}=mg(\sqrt{3}-\frac{1}{2})$