Question

A simple pendulum of length $1m$ has mass $10g$ and oscillates freely with amplitude of $5cm$. Calculate its potential energy at extreme position.

Answer 1

Given $L=1m$
$m=10g=10\times {10}^{-3}g$
$a=5cm=5\times {10}^{-2}m$
We know that, $t=2\pi \sqrt{\frac{l}{g}}$ & $w=\frac{2\pi }{t}$

$\Rightarrow w=\frac{2\pi }{2\pi \sqrt{\frac{l}{g}}}=\sqrt{\frac{g}{l}}$

At extreme position
$PE=\frac{1}{2}m{w}^2{a}^2$

$=\frac{1}{2}m\frac{g}{l}{a}^2$

$=\frac{1}{2}\times \frac{10\times {10}^{-3}\times 9.8\times (5\times {10}^{-2}{)}^2}{1}$

$=0.0001225$
$=1.225\times {10}^{-4}J$

$\therefore$ Potential energy at extreme position
$=1.225\times {10}^{-4}J$