A simple pendulum of length L has mass m and it oscillates freely with amplitude A. At extreme position, its potential energy is ( g = acceleration due to gravity)

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For a simple pendulum, the restoring force for a small displacement $x$ is given by $F = - m g sin θ \approx - m g θ$ .

Since $θ = \frac{x}{L}$ , the restoring force is $F = - (\frac{m g}{L}) x$ .

This represents simple harmonic motion with force constant $k = \frac{m g}{L}$ .

The potential energy of a simple harmonic oscillator at displacement $x$ is $U = \frac{1}{2} k x^{2}$ .

At the extreme position, the displacement is equal to the amplitude, so $x = A$ .

Substituting the values of $k$ and $x$ , we get $U = \frac{1}{2} (\frac{m g}{L}) A^{2} = \frac{m g A ^{2}}{2 L}$ .

Answer: $\frac{mgA ^{2}}{2 L}$

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