Question

The potential energy of a particle is given by, $U=2x^3-3x^2$. (in SI unit) The nature of equilibrium at $x=0$ & $x=1$ are

• A. Stable, Unstable
• B. Unstable, Stable
• C. Unstable, Stable
• D. Stable, Unstable

Answer 1

The correct option is B Unstable, Stable

It is given that, potential energy
$U=2x^3-3x^2$
$\frac{dU}{dx}=(6x^2-6x)$

Under equilibrium,
$\Rightarrow 6x(1-x)=0$
$\Rightarrow (x=0)\&(x=1)$

$\frac{d^{2}U}{d{x}^2}=\frac{d}{dx}(6x^2-6x)=12x-6$

$\frac{d^{2}U}{d{x}^2}{|}_{\text{at x = 0}}=-6<0\to$ (Unstable equilibrium)
$\frac{d^{2}U}{d{x}^2}{|}_{x=1}=6>0\to$(Stable equilibrium)

Hence, option B is the correct answer.