Question

The bob of a simple pendulum is displaced from its equilibrium position $O$ to a position $Q$ which is at heaght $h$ above $O$ and the bob is than released. Assuming themass of the bob to be $m$ and time period of osccilation to be $2s$,The tension in the strimg when the bob passes through $O$ is

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

Answer 1

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

Tension in the string ob passes through the lowest point
$T=mg+\frac{m{v}^2}{r}=mg+mv\omega$

$\therefore v=r\omega$

Substitute
$v=\sqrt{2gh}$ and $\omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi$

we get
$T=m(g+\pi \sqrt{2gh})$