Question

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

• A. $\frac{Mg{A}^2}{2L}$
• B. $\frac{MgA}{2L}$
• C. $\frac{Mg{A}^2}{L}$
• D. $\frac{2Mg{A}^2}{L}$

Answer 1

The correct option is A $\frac{Mg{A}^2}{2L}$

$PE=\frac{m{\omega }^2{x}^2}{2}$

$\omega =\sqrt{\frac{g}{L}}$ , at extreame point $x=A$

So $PE=\frac{Mg{A}^2}{2L}$