Question

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

• A. $\frac{M g A^{2}}{2 L}$
• B. $ \, \frac{M g A}{2 L}$
• C. $\frac{M g A^{2}}{L}$
• D. $\frac{2 M g A^{2}}{L}$

Answer 1

Answer: (A)

Potential energy $=\frac{1}{2}M\omega ^{2}A^{2}$
$=\frac{1}{2}M.\frac{g}{L}.A^{2} \, \left(\because \, \, \omega = \sqrt{\frac{g}{L}}\right)$