Question

The potential energy of a particle varies with position $x\text{(in m)}$ according to the relation $U\left(x\right)=(x^2-4x)\text{J}$. The point $x=2\text{m}$ is a point of

• A. stable equilibrium
• B. unstable equilibrium
• C. neutral equilibrium
• D. none of the above

Answer 1

The correct option is A stable equilibrium

Given:
$U\left(x\right)=(x^2-4x)\text{J}$
Conservative force is given by,
$F=-\frac{dU}{dx}$
At equilibrium position, $F=0$
$\Rightarrow \frac{dU}{dx}=0$
$\Rightarrow 2x-4=0$
$\Rightarrow x=2\text{m}$
Thus, $x=2\text{m}$ is a point of equilibrium.
Again, $\frac{d^{2}U}{d{x}^2}=\frac{d}{dx}(2x-4)=2$
As $\frac{d^{2}U}{d{x}^2}>0$, hence, potential energy is minimum at $x=2\text{m}$.
$\therefore x=2\text{m}$ is a point of stable equilibrium.