$y=a \sin (\omega t+\phi)$

velocity $=\frac{\mathrm{dy}}{\mathrm{dt}}=\omega \mathrm{a} \cos (\omega \mathrm{t}+\phi)$

The velocity is maximum when the particle passes through the mean position i.e...

$\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)_{\max }=\omega \mathrm{a}$

The kinetic energy at this instant is given by

$\frac{1}{2} \mathrm{m}\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)_{\max }^{2}=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{a}^{2}=8 \times 10^{-3} \mathrm{joule}$

$\text { or } \quad \frac{1}{2} \times(0.1) \omega^{2} \times(0.1)^{2}=$

$8 \times 10^{-3}$

Solving we get $\omega=\pm 4$

Substituting the values of a, $\omega$ and $\phi$ in the equation of $S.H.M$., we get

$y=0.1 \sin (\pm 4 t+\pi / 4)$ metre