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 then, the bob is then released. Calculate the tension in the string if the bob passes through $O$ if mass of the bob is $m$ and time period of oscillations is $2s$ .

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

Answer 1

Answer: (A)

Tension in the string when bob passes through lowest point
$T=mg+\frac{m{v}^2}{r}=mg+mv\omega$ $\left[\because v=r\omega \right]$
Putting $v=\sqrt{2gh}$ and $\omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi$
We get $T=m\left(g+\pi \sqrt{2gh}\right)$