# Damped (harmonic) oscillation with constant friction

I have the following problem: Suppose a harmonic oscillator with constant (in measure) friction, which is of course opposite to the velocity. We ultimately want to know how the amplitude decreases and where the body stops.

Without much thought we can write: $mx'' +kx = -T$ . However the friction changes direction every $P/4$. How do we tackle this problem? It looks like every quarter of the period P we have a different differential equation : $$mx'' +kx = -T : (eq. 1)$$ and $$mx'' +kx = T :(eq. 2)$$depending on the direction of $\vec T$.

I solved the homogeneous equation $mx''+kx=0$ and found the solution to be: $x_h=c_1 cos(\sqrt{k \over m}t) + c_2sin(\sqrt{k \over m}t)$ . So to solve say eq. 1 I need to find one particular solution of it, which is $x_p ={-T \over k}$ (and $+T \over k$ for the 2nd equation). So the general solution(s?) is/are : $$x=c_1 cos(\sqrt{k \over m}t) + c_2sin(\sqrt{k \over m}t) \pm {T \over k}$$

So what do we do after this?How do we know how its amplitude changes?How do we see if (and where/when) the body stops moving? Also is there a nicer (more rigorous??) way of approching the problem as far as the differential equations are concerned? Thanks in advance :)

• Seems to me the direction of friction changes every $\frac{T}{2}$. Also, you're using the same symbol ($T$) twice, that's a bit confusing. – Gert Jan 21 '17 at 18:40
• Indeed, every half a period, changed the symbol to P. I gave it some thought, I think we have to give some initial conditions to the problem, say start from x=-A with v=0, solve the corresponding equation, then once the direction of the friction changes (x=+A, v=0)we solve the second one, etc. – Dimitris Jan 21 '17 at 18:46
• Yes but I didn't think that was what you were looking for. – Gert Jan 21 '17 at 18:49
• Note that you can summarize the equation as $m\ddot x + T\, \mathrm{sign}(\dot x) + kx = 0$. This equation is not linear, so you will not have success trying to find a homogeneous / particular solution etc. The only way I can think of is what you just wrote yourself in the last comment. – Noiralef Jan 21 '17 at 18:49
• Your solution has to have some kind of casework in it. To get the solution for each stretch of time $T/2$, note that adding the $T$ term is equivalent to changing $x$ to $x \pm T/k$, so it's just like a frictionless oscillator with a displaced center. This should also tell you how much the amplitude changes every cycle. – knzhou Jan 21 '17 at 18:51

Without much thought we can write: $mx'' +kx = -T$.

The system is the mass $m$ and on it there are two external force, the force due to the spring of spring constant $k$ and the frictional force opposing motion $T$.

Assume that the system and the initial conditions are defined as shown below.
At $t=0\, \; x=A+\frac T k$ the initial extension of the spring and the velocity $v$ of the mass is zero.

A little later in time the displacement of the mass is $x$ so the force on the mass due to the spring is in the negative x-direction and the force $T$ is in the positive x-direction opposing the upward (negative x) motion.
Hopefully the reason for drawing the x-axis down and other annotations will become apparent later.

When travelling back to the origin the equation of motion is $mx'' =-kx + T$.
This would cause a change of sign to the solution for $x$ given by the OP.

What worried me was the step function nature of the solution but that can be explained as follows.

As drawn this is a diagram of the motion for half a period of a spring-mass system in a constant gravitation field with $T=mg$.

When one solves that sort of problem it reduces down to simple harmonic motion about the equilibrium position of the spring-mass system $x_{\rm o}= = \frac{mg}{k}$ where $x_{\rm o}$ is the static extension ie the displacement of one end when it was unextended.

So the current example the motion will be undamped simple harmonic about a position $x = +\frac T k$.

The system has lost mechanical energy $= \frac 12 k (+A+\frac T k)^2 - \frac 12 k (-A+\frac T k)^2= 2TA$ as one would expect from work done equals force $T$ times distance $2A$.
Unlike the spring-mass system in a gravitational field that energy is lost to the this system.

What happens next depends on the relative values of $\frac T k$ and $A$ as shown in the diagram below.

If the spring is compressed and the force due to the spring is greater than the frictional force the motion will continue when the mass stops for the first time and a similar analysis can be made with the direction of force $T$ reversed.