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Physics-

General

Easy

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 $2.0 blank s comma$ the tension in the string when the bob passes through $O$ is

$m (g + π \sqrt{2 g h})$
$m (g + \sqrt{π^{2} g h})$
$m (g + \sqrt{\frac{π^{2}}{2} g h})$
$m (g + \sqrt{\frac{π^{2}}{3} g h})$

The correct answer is: $m left parenthesis g plus pi square root of 2 g h end root right parenthesis$

Tension in the spring when bob passes through lowest point
$T equals m g plus fraction numerator m v to the power of 2 end exponent over denominator r end fraction equals m g plus m v omega blank left square bracket because v equals r omega right square bracket$
putting $v equals square root of 2 g h end root$ and $omega equals fraction numerator 2 pi over denominator T end fraction equals fraction numerator 2 pi over denominator 2 end fraction equals pi$
we get $T equals m left parenthesis g plus pi square root of 2 g h end root right parenthesis$

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