The potential energy U of a body of unit mass moving in one dimensional conservative force field is given by $\mathrm{U}={\mathrm{x}}^2-4\mathrm{x}+3$. All units are in SI. For this situation mark out the correct statement(s).

$\mathrm{U}={\mathrm{x}}^2-4\mathrm{x}+3\text{ and }\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}=-(2\mathrm{x}-4)$

At equilibrium position F= 0, so x = 2 m.
Let the particle is displaced by Ax from equilibrium position, i.e., from x= 2, then restoring force on body is, $\mathrm{F}=-2(2+\mathrm{\Delta x})+4=-2\mathrm{\Delta x}$

i.e., $\mathrm{F\alpha }-\mathrm{\Delta x}$, so performs simple harmonic motion about x = 2 m. 

Time period, $\mathrm{T}=2\mathrm{\pi }/\mathrm{\omega }=2\mathrm{\pi }/\sqrt{2}=\sqrt{2}\pi s$

From energy conservation, $\frac{{\mathrm{mv}}_{max}^2}{2}+{\mathrm{U}}_{min}={\mathrm{U}}_{max}$

$\frac{1\times 4^2}{2}+\left(2^2-4\times 2+3\right)=(\mathrm{A}+2{)}^2-4(\mathrm{A}+2)+3$

where A is amplitude.

Solving the above equation, we get $\mathrm{A}=2\sqrt{2}\mathrm{m}$