1) 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) $\frac{MgA^2}{2L}$ B) $\frac{MgA}{2L}$ C) $\frac{MgA^2}{L}$ D) $\frac{2MgA^2}{L}$ Answer: Option AExplanation:Potential energy of a simple pendulum is given as, $=\frac{1}{2}M\omega^{2}A^{2}=\frac{1}{2}M.\frac{g}{L}.A^{2}$ $\left(\because \omega = \sqrt{\frac{g}{L}}\right)$
