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 2.0 sec, the tension in the string when the bob passes through O is :

• A. $m(g+2{\pi }^{2}h)$
• B. $m(g+{\pi }^{2}h)$
• C. $m(g+\frac{{\pi }^2}{2}h)$
• D. $m(g+\frac{{\pi }^2}{3}h)$

Answer 1

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

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