A simple pendulum of length $L$ has mass $m$ and it oscillates freely with amplitude A. At extreme position,…
A simple pendulum of length $L$ has mass $m$ and it oscillates freely with amplitude A. At extreme position, its potential energy is ( $\mathrm{g}=$ acceleration due to gravity)
$\frac{m g A}{2 L}$
$\frac{\mathrm{mgA}^2}{\mathrm{~L}}$
$\frac{\mathrm{mgA}}{\mathrm{L}}$
$\frac{\mathrm{mgA}^2}{2 \mathrm{~L}}$
Solution
Potential energy of particle at extreme position is,
$\begin{aligned}
\text {,P.E. } & =\frac{1}{2} m \omega^2 A^2 \\
& =\frac{1}{2} m \times \frac{g}{L} \times A^2 \quad \ldots\left(\because \omega=\sqrt{\frac{g}{l}}\right)
\end{aligned}$