Question

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 $2\text{s}$, 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. $m(g+\sqrt{\frac{\pi }{2}gh})$
• D. $m(g+\sqrt{\frac{{\pi }^2}{3}}gh)$

Answer 1

The correct option is A $m(g+\pi \sqrt{2gh})$

Using principle of conservation of energy,
$\Delta KE+\Delta PE=0$
$\frac{1}{2}m{v}^2=mgh$
$v=\sqrt{2gh}$
Angular velocity of motion is $\omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi$

Tension in the string when bob passes through the lowest point is
$T=mg+\frac{m{v}^2}{r}=mg+mv\omega (\because v=r\omega )$
Substituting,
$v=\sqrt{2gh},\omega =\pi$
we get,
$T=m(g+\pi \sqrt{2gh})$