Physics - SHM Question with Solution | TestHub

The total energy$\mathrm{E}$of a particle vibrating in SHM is given by

$\mathrm{E}=\frac{1}{2}\mathrm{m}{\mathrm{\omega }}^2{\mathrm{a}}^2$.....(1)

The kinetic energy$\mathrm{K}$is given by$\mathrm{K}=\frac{1}{2}\mathrm{m}{\mathrm{\omega }}^2\left({\mathrm{a}}^2-{\mathrm{y}}^2\right)$where$\mathrm{y}=$displacement of the particle but

$\mathrm{K}=\frac{\mathrm{E}}{2}=\frac{1}{2}\left[\frac{1}{2}\mathrm{m}{\mathrm{\omega }}^2{\mathrm{a}}^2\right]$

$\therefore \mathrm{ }\mathrm{ }\frac{1}{2}\left[\frac{1}{2}\mathrm{m}{\mathrm{\omega }}^2{\mathrm{a}}^2\right]=\frac{1}{2}\mathrm{m}{\mathrm{\omega }}^2\left({\mathrm{a}}^2-{\mathrm{y}}^2\right)$

or$\frac{{\mathrm{a}}^2}{2}={\mathrm{a}}^2-{\mathrm{y}}^2$or${\mathrm{y}}^2=\frac{{\mathrm{a}}^2}{2}$

$\therefore \mathrm{ }\mathrm{ }\mathrm{y}=\frac{\mathrm{a}}{\sqrt{2}}$

Hence the kinetic energy is half of the total energy when displacement of the particle is$\frac{\mathrm{a}}{\sqrt{2}}$.

Given that$\mathrm{a}=4 \mathrm{cm}$

$\therefore \mathrm{y}=\frac{4}{\sqrt{2}}=2\sqrt{2} \mathrm{cm}$