A particle executing SHM having amplitude 5cm, mass 0.5 kg and angular frequency 5 rad/s is at 1cm from mean position. Find potential energy and kinetic energy.

Given, $\mathrm{m}=0.5 \mathrm{kg}, \mathrm{\omega }=5 \mathrm{rad}/\mathrm{s},$

               $\mathrm{x}=1 \mathrm{cm} = {10}^{-2}\mathrm{m},$

               $\mathrm{A}=5 \mathrm{cm}=5\times {10}^{-2}\mathrm{m}$

$\therefore$ Potential energy $\mathrm{PE}=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2$

or        $\mathrm{PE}=\frac{1}{2}\times 0.5\times (5{)}^2\times {\left({10}^{-2}\right)}^2$

                 $=\frac{25}{4}\times {10}^{-4} \mathrm{J}=6.25\times {10}^{-4} \mathrm{J}$

$\therefore$ Kinetic energy, $\mathrm{KE}=\frac{1}{2}{\mathrm{m\omega }}^2({\mathrm{A}}^2-{\mathrm{x}}^2)$

or        $\mathrm{KE}=\frac{1}{2}\times 0.5\times \left(5{)}^2\right(25-1)\times {10}^{-4}$

                $=150\times {10}^{-4} \mathrm{J}$