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

$\text{F}=-\frac{\text{dU}}{\text{d}x}\frac{-\text{d}}{\text{d}x}\left(x^2-4x+3\right)=-\left[2x-4\right]$

At equilibrium $\text{F}=0\text{at}x=2$

$\therefore \text{F}=-2x+4=-2\left[x+2\right]=-2\left[x\right]$

SHM about x = 2. On comparison k = 2

$\omega =\sqrt{\frac{\text{k}}{\text{m}}}$ , m = unit mass

$\omega =\sqrt{2}\Rightarrow \frac{2\pi }{\text{T}}=\sqrt{2}$

$\text{T}=\sqrt{2}\pi$

at mean position velocity is max. i.e.,

${\text{V}}_{\text{max}}=\text{A}\omega =\text{A}\sqrt{2}$

${\text{V}}_{\text{max}}=4\text{then}4=\text{A}\sqrt{2}$

$\text{A}=2\sqrt{2}$