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Standard X

Physics

Simple Pendulum

A simple pend...

Question

A simple pendulum bob has a mass $m$ and length $L$ . The bob is drawn aside such that the string is horizontal and then it is released. The velocity of the bob while it crosses the equilibrium position is (Take acceleration due to gravity $= g$ )

\sqrt{g L}

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\sqrt{2 g L}

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\sqrt{5 g L}

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\sqrt{3 g L}

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Solution

The correct option is A $\sqrt{2 g L}$
At the horizontal position, since the bob is at rest it only has potential energy. We take the equilibrium position as the reference point.
So the potential energy attained at horizontal point is $m g l$ .
When, it crosses the equilibrium position, all its potential energy gets converted to kinetic energy.
Thus, applying law of conservation of energy,
$\frac{1}{2} m v^{2} = m g l$
This gives, $v = {(2 g l)}^{0.5}$

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A simple pendulum bob has a mass m and length L. The bob is drawn aside such that the string is horizontal and then it is released. The velocity of the bob while it crosses the equilibrium position is (Take acceleration due to gravity =g)

A simple pendulum bob has a mass $m$ and length $L$ . The bob is drawn aside such that the string is horizontal and then it is released. The velocity of the bob while it crosses the equilibrium position is (Take acceleration due to gravity $= g$ )