Differential equation of frictional force

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1. May 12, 2015

Dazed&Confused

A question from a classical mechanics past paper described a particle of mass
$m$ that had a pair of horizontal identical springs of spring constant $k$ attached on either side and that the mass is free to move horizontally. The mass is also placed on a table that gives rise to an additional frictional force; the coefficient of sliding friction between the mass and the table is $\mu$.

You have to 'verify' that the following is a solution of the equation of motion:

$$x_B(t) = B \sin( \omega_B t) + C( \cos( \omega_B t) - 1).$$

In my opinion this is wrong. The solution clearly should decay in time. It seems that the differential equation they wanted was of the form

$$x''(t) = -2kx(t) - \mu mg.$$

I think it should be

$$x''(t) = -2kx(t) - \mu mg \text{Sign}(x'(t)),$$

where $\text{Sign}(x)$ is defined to be 1 if $x > 0$ and -1 if $x < 0.$

Is there something wrong with my thinking? Keep in mind that Mathematica's numerical solution does show a particle approximately exhibiting SHM motion, but also decaying with time, however it does not ultimately stop moving at the equilibrium position.

2. May 12, 2015

Orodruin

Staff Emeritus
You definitely need a sign in the friction term or it only describes a constant force acting in a determined direction. This is equivalent to a gravitational force on a vertical spring with a mass attached and does nothing but translate the solution to be oscillations around the new equilibrium position.

Why do you think it would stop moving at the equilibrium position? (I assume you here mean the position where the spring is at its lowest potential energy.)

3. May 12, 2015

Dazed&Confused

Well if it stopped at any other position there would be a net force on the particle due to the springs, since I haven't added a static friction term.