Question

A system has potential energy

$U\left(x\right)=x+\sin \left(\left(2rad/m\right)x\right)$

as a particle moves over the range$0\mathrm{m}\le x\le \pi \mathrm{m}$.

a. Where are the equilibrium positions in this range?

b. For each, is it a point of stable or unstable equilibrium?

Short answer

(a) The equilibrium position in the range $0\mathrm{m}\le x\le \pi \mathrm{m}x=\frac{\pi }{3},\frac{2\pi }{3}$.

(b) The equilibrium position $x=\frac{\mathrm{\pi }}{3}$is unstable and $x=\frac{2\mathrm{\pi }}{3}$is a stable equilibrium.

Step-by-step solution

Given information (part a)

A system has potential energy $U\left(x\right)=x+\sin \left(2x\right(\mathrm{rad}\left)\right)$, where x is in m,as the particle moves over the range $0\mathrm{m}\le x\le \pi \mathrm{m}$.

Explanation (part a)

To find the equilibrium positions over the range $0\mathrm{m}\le x\le \pi \mathrm{m}$,

$$\begin{gathered} U\left(x\right)=x+\sin \left(2x\right) \\ -\frac{dU\left(x\right)}{dx}=0 \\ -\frac{d[x+\sin (2x\left)\right]}{dx}=0 \\ 1+2\cos \left(2x\right)=0 \\ \cos \left(2x\right)=-\frac{1}{2} \\ 2x=\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3} \\ x=\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3} \end{gathered}$$

Given information (part b)

A system has potential energy $U\left(x\right)=x+\sin \left(2x\right(\mathrm{rad}\left)\right)$, wherexis inm, as the particle moves over the range$0\mathrm{m}\le x\le \pi \mathrm{m}$.

Explanation (part b)

To determine the stability,

$$\begin{gathered} -\frac{d^{2}U}{d{x}^2}=-\frac{d^2[x+\sin (2x\left)\right]}{d{x}^2} \\ -\frac{d^{2}U}{d{x}^2}=-\frac{d[1+2\cos (2x\left)\right]}{dx} \\ -\frac{d^{2}U}{d{x}^2}=-[-4\sin (2x\left)\right] \\ -\frac{d^{2}U}{d{x}^2}=4\sin \left(2x\right) \\ -{\left.\frac{d^{2}U}{d{x}^2}\right|}_{x=\frac{\pi }{3}}=4\sin \left(\frac{2\pi }{3}\right) \\ -{\left.\frac{d^{2}U}{d{x}^2}\right|}_{x=\frac{\pi }{3}}=2\sqrt{3}>0 \end{gathered}$$

And,

$$\begin{gathered} -\frac{d^{2}U}{d{x}^2}=4\sin \left(2x\right) \\ -{\left.\frac{d^{2}U}{d{x}^2}\right|}_{x=\frac{\pi }{3}}=4\sin \left(\frac{4\pi }{3}\right) \\ -{\left.\frac{d^{2}U}{d{x}^2}\right|}_{x=\frac{\pi }{2}}=-2\sqrt{3}<0 \end{gathered}$$