Question

The potential energy of a particle varies with position $x$ according to the relation $U\left(x\right)=2x^4-27x$ . The point $x=\frac{3}{2}$ is point of

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

Answer 1

Answer: (B)

$U=2x^4-27x$
$\frac{dU}{dx}=8x^3-27$
At $x=\frac{3}{2},\frac{dU}{dx}=0$ , therefore it is an equilibrium position.
$\frac{d^{2}U}{d{x}^2}=24x^2$
At $x=\frac{3}{2},\frac{d^{2}U}{d{x}^2}>0$ , therefore it is a stable equilibrium position.