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

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$m (g + π \sqrt{2 g h})$

$m (g + \sqrt{π^{2} g h})$

$m (g + \sqrt{\frac{π^{2}}{2} g h})$

$m (g + \sqrt{\frac{π^{2}}{3} g h})$

answer is A.

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Detailed Solution

Tension in the string when bob passes through lowest point $T = mg + \frac{{mv}^{2}}{r} = mg + mvω$ ( $∵ v = rω$ )
putting $v = \sqrt{2 gh}$ and $ω = = \frac{2 π}{T} = \frac{2 π}{2} = π$
we get $T = m (g + π \sqrt{2 gh})$

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