Answer
(a) $x = \frac{\pi}{3}, \frac{2\pi}{3}$
(b) $x = \frac{\pi}{3}$ is an unstable equilibrium.
$x = \frac{2\pi}{3}$ is a stable equilibrium.
Work Step by Step
(a) We can find the positions where $F = 0$:
$F = -\frac{dU}{dx} = 0$
$-1-2~cos~2x = 0$
$cos~2x = -0.5$
$2x = cos^{-1}~(-0.5)$
$2x = \frac{2\pi}{3}, \frac{4\pi}{3}$
$x = \frac{\pi}{3}, \frac{2\pi}{3}$
(b) We can find an expression for $\frac{d^2U}{dx^2}$:
$U(x) = x+sin~2x$
$\frac{dU}{dx} = 1+2~cos~2x$
$\frac{d^2U}{dx^2} = -4~sin~2x$
When $x = \frac{\pi}{3}$:
$\frac{d^2U}{dx^2} = -4~sin~2x$
$\frac{d^2U}{dx^2} = -4~sin~[(2)(\frac{\pi}{3})]$
$\frac{d^2U}{dx^2} = -3.46$
When $x = \frac{\pi}{3},$ then $\frac{d^2U}{dx^2} \lt 0$, so this is a local maximum.
$x = \frac{\pi}{3}$ is an unstable equilibrium.
When $x = \frac{2\pi}{3}$:
$\frac{d^2U}{dx^2} = -4~sin~2x$
$\frac{d^2U}{dx^2} = -4~sin~[(2)(\frac{2\pi}{3})]$
$\frac{d^2U}{dx^2} = 3.46$
When $x = \frac{2\pi}{3},$ then $\frac{d^2U}{dx^2} \gt 0$, so this is a local minimum.
$x = \frac{2\pi}{3}$ is a stable equilibrium.
