A simple pendulum of length ' $L$ ' has mass ' $M$ ' and it oscillates freely with amplitude 'A'. At extreme…
A simple pendulum of length ' $L$ ' has mass ' $M$ ' and it oscillates freely with amplitude 'A'. At extreme position, its potential energy is
$\frac{\mathrm{MgA}^2}{\mathrm{~L}}$
$\frac{2 \mathrm{MgA}^2}{\mathrm{~L}}$
$\frac{\mathrm{MgA}}{2 \mathrm{~L}}$
$\frac{\mathrm{MgA}^2}{2 \mathrm{~L}}$
Solution
At extreme position, the potential energy of simple pendulum,
$\mathrm{PE}=\frac{1}{2} \mathrm{~mA}^2 \omega$
where, $\omega=$ angular frequency $=\frac{2 \pi}{\mathrm{~T}}$
For simple pendulum, $T=2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}$
$\therefore \omega=\frac{2 \pi}{2 \pi} \frac{\sqrt{\mathrm{~g}}}{\mathrm{~L}}=\frac{\sqrt{\mathrm{g}}}{\mathrm{~L}} \quad \ldots\left[\because \mathrm{~T}=\frac{2 \pi}{\omega}\right]$
On putting value of $\omega$ in Eq. (i), we get
$\begin{aligned}
& \mathrm{PE}=\frac{1}{2} \mathrm{mM}^2\left(\sqrt{\frac{\mathrm{~g}}{\mathrm{~L}}}\right)^2 \\
& =\frac{\mathrm{mgA}}{}{ }^2 \\
& 2 \mathrm{~L}
\end{aligned}$