A particle is moving on x- axis has potential energy U $=2-20\mathrm{x}+5{\mathrm{x}}^2$ Joules along x- axis. The particle is released at x = -3. Maximum value of ‘x’ will be _______ m. [x is in meters and U is in joules]

$\begin{array}{l}\mathrm{U}=2-20\mathrm{x}+5{\mathrm{x}}^2\\ \mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}=20-10\mathrm{x}\end{array}$
$\text{ At equilibrium position; }\mathrm{F}=0$
$\begin{array}{l}20-10\mathrm{x}=0\\ \Rightarrow \mathrm{x}=2\end{array}$
$\text{ Since particle is released at }x=-3\text{, therfore amplitude of particle is }5.$

It will oscillate about x = 2
With an amplitude of 5.
$\therefore \text{ maximum value of }\mathrm{x}\text{ will be }7$