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

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