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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
$m\left( {g + \pi \sqrt {2gh} } \right)$
$m\left( {g + \sqrt {{\pi ^2}gh} } \right)$
$m\left( {g + \sqrt {\frac{{{\pi ^2}}}{2}gh} } \right)$
$m\left( {g + \sqrt {\frac{{{\pi ^2}}}{3}gh} } \right)$
Solution
Tension in the string when bob passes through lowest poin
$\mathrm{T}=\mathrm{mg}+\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{mg}+\mathrm{mv} \omega(\because \mathrm{v}=\mathrm{r} \omega)$
putting $\mathrm{v}=\sqrt{2 \mathrm{gh}}$ and $\omega=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{2}=\pi$
we get $\mathrm{T}=\mathrm{m}(\mathrm{g}+\pi \sqrt{2 \mathrm{gh}})$
