# What is the Equation of Motion for a Harmonic Motion Damped by Constant Friction?

In normal physics class, we usually learn about damped harmonic motion where the damping force is directly proportional to the velocity.

$${\bf F} = -b {\bf v}$$

This would give us the equation

$$-b{\bf v} - k{\bf x} = m{\bf a}$$

which we can solve to find the equation of motion.

However, the simplest damping force is friction. For a mass attached to a spring on a table, the kinetic friction only depends on the normal force and the coefficient of friction. In other words, it is a constant.

Our equation would then be

$$-b\hat{\bf v} - k{\bf x} = m{\bf a}$$

How would we solve this differential equation?

Without loss of generality, this answer will work in 1D, but it is easily generalised to 2D and 3D. Then your equation of motion is $$m\ddot{x}(t)=-kx-b\mathrm{sgn}(\dot{x}(t))$$, where $$\mathrm{sgn}(x)=\frac{x}{|x|}$$ is the sign function. This equation is piecewiese, meaning that the sign of $$b$$ when $$\dot{x}>0$$ is negative and when $$\dot{x}<0$$ is positive.

You have an analysis of the system here: https://aapt.scitation.org/doi/10.1119/1.1976111

How does one solve this? I don't think there is any analytical solution, so one solves it numerically. Here's the solution using MATLAB with $$m$$=1 kg, $$b$$=0.1 N and $$k$$=1 N/m:

This would need to be solved piecewise between each set of turning points, since the frictional force changes sign discontinuously at the turning points.

Specifically, if the object is released from rest at $$x_0 > 0$$, then friction will act in the positive direction initially. We will then have by the work-energy principle ($$\Delta T = \int F dx$$) $$\frac{1}{2} m \dot{x}^2 = - \frac{1}{2} k x^2 + \frac{1}{2} k x_0^2 + b(x - x_0) \\ \frac{1}{2} m \dot{x}^2 + \frac{1}{2} k \left(x - \frac{b}{k} \right)^2 = \frac{1}{2} k \left(x_0 - \frac{b}{k} \right)^2$$ where I have completed the square on both sides. We can further substitute $$y \equiv x - b/k$$ to rewrite this as $$\frac{1}{2}m \dot{y}^2 + \frac{1}{2} ky^2 = \frac{1}{2} k y_0^2.$$ In other words, while the particle is moving in the negative direction, its motion is that of a harmonic oscillator centered at $$y = 0$$, or $$x = b/k$$. By similar logic, while the particle is moving in the positive direction, its motion is that of a harmonic oscillator centered at $$x = -b/k$$. The full solution for $$x(t)$$ would be a piecewise function of sinusoids alternately centered at $$\pm b/k$$, with successive amplitudes chosen so that $$x(t)$$ is continuous.

Finally, for one of these half-cycles, we will get to the point where the particle ends a a cycle where the force of friction is greater in magnitude than the restoring force from the spring (i.e. $$|x| < b/k$$.) The particle would presumably then be "stuck" there for all subsequent $$t$$. Moreover, the force of static friction is usually taken to be greater than the force of kinetic friction, so this "sticking point" might well occur at a greater value of $$x$$ than this.

• This seems remarkably simple, to be honest — so much so that I wonder if I've made a mistake. Comments are welcome. Nov 9 '21 at 1:12