Question

The angular amplitude of a simple pendulum is ${\theta }_0$. The maximum tension in its string will be

• A. $mg(1-{\theta }_0)$
• B. $mg(1+{\theta }_0)$
• C. $mg(1-{\theta }_0^2)$
• D. $mg(1+{\theta }_0^2)$

Answer 1

Answer: (D)

$mg(1+{\theta }_0^2)$

The loss in gravitational energy of bob when it reaches the bottom most point $=mgl(1-cos{\theta }_0)$

$=mgl\left(2si{n}^2\right(\frac{{\theta }_0}{2}\left)\right)$

$=mgl\left(2\right(\frac{{\theta }_0}{2}{)}^2)$

$=mgl\frac{{\theta }_0^2}{2}$

This energy converts to kinetic energy $=\frac{1}{2}m{v}^2$

$⟹v=\sqrt{gl}{\theta }_0$

The centripetal force that acts on the bob downwards at this point $=\frac{m{v}^2}{l}$ $=mg{\theta }_0^2$

Hence, tension in the string at the point $=mg+mg{\theta }_0^2$

$=mg(1+{\theta }_0^2)$