The potential energy of a particle of mass 0.1kg.  Moving along the x-axis, is given by   $U=5x\left(x-4\right)J$ where x is in meter. It can be concluded that

$m=0.1kg$  Moving $x-axis$

Potential energy  $\text{U}=5x\left(x-4\right)J$

$F=\frac{-dU}{dx}=-\frac{d}{dx}\left[5x\left(x-4\right)\right]$

$F=-\frac{d}{dx}\left(5x^2-20x\right)$

$F=-10x+20$

$\therefore$ Particle execute SHM about 20m,

At $x=2$ then $F=0$

It is mean position velocity is maximum $\omega =\sqrt{\frac{k}{m}=}\sqrt{\frac{10}{0.1}}=10$

$T=\frac{2\pi }{\omega }=\frac{\pi }{5}\sec$