Motion in a Two-Dimensional Harmonic Potential

Consider a particle of mass $m$ moving in the two-dimensional harmonic potential

$$U(x,y) = \frac{1}{2}\,k\,r^2,$$ (229)

where $r=\sqrt{x^2+y^2}$, and $k>0$. It follows that the particle is subject to a force,

$${\bf f} = - \nabla U = -k\,(x,\, y) = - k\,{\bf r},$$ (230)

which always points toward the origin, and whose magnitude increases linearly with increasing distance from the origin. According to Newton's second law, the equation of motion of the particle is

$$m\,\frac{d^2{\bf r}}{dt^2} = {\bf f} = - k\,{\bf r}.$$ (231)

When written in component form, the above equation reduces to

$$\frac{d^2 x}{d t^2} = - \omega_0^{\,2}\,x,$$ (232)

$$\frac{d^2 y}{d t^2} = - \omega_0^{\,2}\,y,$$ (233)

where $\omega_0 = \sqrt{k/m}$.

Since Eqs. (232) and (233) are both simple harmonic equations, we can immediately write their general solutions:

$$x = A\,\cos(\omega_0\,t-\phi_1),$$ (234)

$$y = B\,\cos(\omega_0\,t-\phi_2).$$ (235)

Here, $A$, $B$, $\phi_1$, and $\phi_2$ are arbitrary constants of integration. We can simplify the above equations slightly by shifting the origin of time (which is, after all, arbitrary): i.e.,

$$t\rightarrow t' + \phi_1/\omega_0.$$ (236)

Hence, we obtain

$$x = A\,\cos(\omega_0\,t'),$$ (237)

$$y = B\,\cos(\omega_0\,t'-{\mit\Delta}),$$ (238)

where ${\mit\Delta}=\phi_2-\phi_1$. Note that the motion is clearly periodic in time, with period $T=2\,\pi/\omega_0$. Thus, the particle must trace out some closed trajectory in the $x$-$y$ plane. The question, now, is what does this trajectory look like as a function of the relative phase-shift, ${\mit\Delta}$, between the oscillations in the $x$- and $y$-directions?

Using standard trigonometry, we can write Eq. (238) in the form

$$y = B\,\left[\cos(\omega_0\,t')\,\cos{\mit\Delta} + \sin(\omega_0\,t')\,\sin{\mit\Delta}\right].$$ (239)

Hence, using Eq. (237), we obtain

$$\left(\frac{y}{B} - \frac{x}{A}\,\cos{\mit\Delta}\right)^2=\... ...mit\Delta} = \left(1-\frac{x^2}{A^2}\right)\sin^2{\mit\Delta},$$ (240)

which simplifies to give

$$\frac{x^2}{A^2} - 2\,\frac{x\,y}{A\,B} \,\cos{\mit\Delta} + \frac{y^2}{B^2} = \sin^2{\mit\Delta}.$$ (241)

Unfortunately, the above equation is not immediately recognizable as being the equation of any particular geometric curve: e.g., a circle, or an ellipse, or a parabola, etc.

Perhaps our problem is that we are using the wrong coordinates? Suppose that the rotate our coordinate axes about the $z$-axis by an angle $\theta$, as illustrated in Fig. 3. According to Eqs. (93) and (94), our old coordinates ($x$, $y$) are related to our new coordinates ($x'$, $y'$) via

$$x = x'\, \cos\theta - y'\,\sin\theta,$$ (242)

$$y = x'\,\sin\theta + y'\,\cos\theta.$$ (243)

Let us see whether Eq. (241) takes a simpler form when expressed in terms of our new coordinates. Equations (241)-(243) yield

$$x'^{\,2}\left[\frac{\cos^2\theta}{A^2} - \frac{2\,\cos\theta\,\sin\theta\,\cos{\mit\Delta}}{A\,B} + \frac{\sin^2\theta}{B^2}\right]$$

$$+ y'^{\,2}\left[\frac{\sin^2\theta}{A^2} +\frac{2\,\cos\theta\,\sin\theta\,\cos{\mit\Delta}}{A\,B} + \frac{\cos^2\theta}{B^2}\right]$$ (244)

$$+x' y'\left[-\frac{2\,\sin\theta\,\cos\theta}{A^2} + \frac{2\,(\s... ...\theta)\,\cos{\mit\Delta}}{A\,B} + \frac{2\,\cos\theta\,\sin\theta}{B^2}\right] = \sin^2{\mit\Delta}.$$

We can simplify the above equation by setting the term involving $x' y'$ to zero. Hence,

$$-\frac{\sin 2\,\theta}{A^2} - \frac{2\,\cos 2\,\theta\,\cos{\mit\Delta}}{A\,B} + \frac{\sin 2\,\theta}{B^2}= 0,$$ (245)

where we have made use of some simple trigonometric identities. Thus, the $x' y'$ term disappears when $\theta$ takes the special value

$$\theta = \frac{1}{2}\,\tan^{-1}\left(\frac{2\,A\,B\,\cos{\mit\Delta}}{A^2-B^2}\right).$$ (246)

In this case, Eq. (244) reduces to

$$\frac{x'^{\,2}}{a^2} + \frac{y'^{\,2}}{b^2} = 1,$$ (247)

where

$$\frac{1}{a^2} = \frac{1}{\sin^2{\mit\Delta}}\left[\frac{\cos^2\theta}{A^2} - \fra... ...s\theta\,\sin\theta\,\cos{\mit\Delta}}{A\,B} + \frac{\sin^2\theta}{B^2}\right],$$ (248)

$$\frac{1}{b^2} = \frac{1}{\sin^2{\mit\Delta}}\left[\frac{\sin^2\theta}{A^2} +\frac... ...s\theta\,\sin\theta\,\cos{\mit\Delta}}{A\,B} + \frac{\cos^2\theta}{B^2}\right].$$ (249)

Of course, we immediately recognize Eq. (247) as the equation of an ellipse, centered on the origin, whose major and minor axes are aligned along the $x'$- and $y'$-axes, and whose major and minor radii are $a$ and $b$, respectively (assuming that $a>b$).

We conclude that, in general, a particle of mass $m$ moving in the two-dimensional harmonic potential (229) executes a closed elliptical orbit (which is not necessarily aligned along the $x$- and $y$-axes), centered on the origin, with period $T=2\,\pi/\omega_0$, where $\omega_0 = \sqrt{k/m}$.

Figure 25: Trajectories in a two-dimensional harmonic oscillator potential.

Figure 25 shows some example trajectories calculated for $A=2$, $B=1$, and the following values of the phase difference, ${\mit\Delta}$: (a) ${\mit\Delta}=0$; (b) ${\mit\Delta}=\pi/4$; (c) ${\mit\Delta}=\pi/2$; (d) ${\mit\Delta}= 3\pi/4$. Note that when ${\mit\Delta}=0$ the trajectory degenerates into a straight-line (which can be thought of as an ellipse whose minor radius is zero).

Perhaps, the main lesson to be learned from the above study of two-dimensional motion in a harmonic potential is that comparatively simple patterns of motion can be made to look complicated when written in terms of ill-chosen coordinates.