Question

The potential energy of a conservative system is given by $U=a{x}^2-bx$
where a and b are positive constants. Find the equilibrium position and discuss whether the equilibrium is stable, unstable or neutral.

• A. $x=-\frac{b}{2a}$ is the stable equilibrium position.
• B. $x=\frac{b}{2a}$ is the stable equilibrium position
• C. $x=0$ is the stable equilibrium position
• D. $x=\frac{b}{2a}$ is a unstable equilibrium position

Answer 1

Answer: (B)

$x=\frac{b}{2a}$ is the stable equilibrium position

In a conservative field
$F=-\frac{dU}{dx}$
$\therefore F=-\frac{d}{dx}(a{x}^2-bx)=b-2ax$
For equilibrium F=0
or $b-2ax=0\therefore x=\frac{b}{2a}$
From the given equation we can see that $\frac{d^{2}U}{d{x}^2}=2a$ (positive), i.e., U is minimum.
Therefore, $x=\frac{b}{2a}$ is the stable equilibrium position.