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