Physics - SHM Question with Solution | TestHub

The restoring force for a simple pendulum is given by $F=-Mg\sin \theta$.

For small angular displacements, $\sin \theta \approx \theta =\frac{x}{L}$, where $x$ is the linear displacement.

Thus, $F=-\frac{Mg}{L}x$.

This represents simple harmonic motion with an effective force constant $k=\frac{Mg}{L}$.

The potential energy of the pendulum at displacement $x$ is $U=\frac{1}{2}k{x}^2$.

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

Therefore, the potential energy at the extreme position is $U=\frac{1}{2}\left(\frac{Mg}{L}\right)A^2=\frac{Mg{A}^2}{2L}$.

Answer: $\frac{{\mathrm{MgA}}^2}{2\mathrm{L}}$