Q. The bob of a simple pendulum is displaced from its equilibrium position $O$ to a position $Q$ which is at height $h$ above $O$ and the bob is then released. Assuming the mass of the bob to be $m$ and time period of oscillations to be $2.0\ s,$ the tension in the string when the bob passes through $O$ is
A) $m(g + \pi\sqrt{2gh})$
B) $m(g + \sqrt{\pi^{2}gh})$
C) $m\left( g + \sqrt{\frac{\pi^{2}}{2}gh} \right)$
D) $m\left( g + \sqrt{\frac{\pi^{2}}{3}gh} \right)$
Explanation
(a) Tension in the spring when bob passes through lowest point
$T = mg + \frac{mv^{2}}{r} = mg + mv\omega\ \ \ \ \lbrack\because v = r\omega\rbrack$
putting $v = \sqrt{2gh}\ $and $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi$
we get$\ T = m(g + \pi\sqrt{2gh})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.