Physics - SHM Question with Solution | TestHub

$V=Ax(x-4)Jm=0.02=A{x}^2-4Ax$ $$\begin{array}{l}\mathrm{F}=-\frac{\mathrm{dV}}{\mathrm{dx}}=-2\mathrm{Ax}+4\mathrm{A}\text{ (not constant) }\\ \mathrm{F}=-2\mathrm{A}(\mathrm{x}-2)\end{array}$$ Hence position execute S.H.M mean position is when force $=0$ i.e. $x=2$ and speed is maximum at mean position $${\omega }^2=\frac{2\mathrm{A}}{\mathrm{m}}\mathrm{T}=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{\mathrm{m}}{2\mathrm{A}}}=2\pi \sqrt{\frac{1}{100\mathrm{A}}}=\frac{\pi }{5}\sqrt{\frac{1}{\mathrm{A}}}$$