# simple physical system involving convergence to equilibrium at decreasing speed

I want to teach an economic model to my students (Solow, 1956), using a simple "device", or physical system that I can take to my lectures. The system must have these three properties:

1. There is one equilibrium, which is stable (e.g. the bottom inside an empty sphere).
2. Variable (say, $x$) converges over time to it (e.g. a ball inside the sphere with friction on its inner surface will end up at the equilibrium).
3. The speed of convergence is directly related to the distance between $x$ to that equilibrium. In other words, the speed decreases the closer $x$ is to the equilibrium. Unfortunately, the example of a ball inside a sphere does not fulfill this condition, because the ball oscilates around the equilibrium before reaching it.

No particular speed of convergence is needed (in fact, in the Solow model it takes infinite time to reach the equilibrium). The key is point 3, that the movement towards it depends directly on its distance.

I'm looking for a system as simple as the ball inside a sphere (which I could easily replicate with basic tools) but where the three conditions hold. I just can't think of anything like this.

• Could you be precise about your concept of "inversely related"? If you just want the speed of approach to equilibrium to decrease proportionately to the distance from equilibrium, then exponential relaxation, i.e. $\mathrm{d}x/\mathrm{d}t = -\gamma (x - x_{\mathrm{eq}})$ does the trick. This behaviour describes an enormous number of simple physical systems. – Mark Mitchison Jan 22 '15 at 16:24
• I mean I want an example of something I can physically (i.e. manually, materialistically) have, experiment with, touch, like my example of a ball and sphere (ping-pong ball and a salad bowl). – luchonacho Jan 22 '15 at 22:20

Assume that a block of mass $m$ initially at $x=0$ is given an initial velocity $v_0$ (by you pushing it and then letting go). When the block is in motion, friction is the only force acting on it. The friction force is proportional to the velocity of the block, with coefficient of proportionality $\gamma$. Then Newton's law $F = ma$ tells you that $$m \ddot{x} = -\gamma \dot{x},$$ where dots denote derivatives with respect to time, and $m$ is the mass of the block. This can be integrated to give $$\dot{x} = v_0 - \frac{\gamma}{m} x.$$ If you define the equilibrium position $x_{\mathrm{eq}} = mv_0/\gamma$, the equation can be rewritten $$\dot{x} = -\frac{\gamma}{m}(x-x_{\mathrm{eq}}). \qquad \qquad (1)$$ This equation tells you that as the distance to the equilibrium point $(x-x_{\mathrm{eq}})$ gets smaller, the block slows down ($\dot{x}$ decreases) until the block reaches the equilibrium point and stops. The actual solution to the equation is $$x(t) = x_{\mathrm{eq}}(1-e^{-\gamma t/m} ).$$