# Damping Oscillation

1. Oct 5, 2011

### coki2000

Hello all,

I want to find position function of a spring which is on a frictional surface so there is a friction force like k.m.g. Its differential equation is like that

$\ddot{x}+\omega ^2x=-kmg$

But I can't solve this. Can you help me to solve that equation? And also I don't understand why most of classical mechanics books say that friction is proportional to velocity. I was said that friction is proportional to velocity at only under water or another liquid thing. Can you also explain this?

2. Oct 5, 2011

### DiracRules

To solve that ODE(=ordinary differential equation), you first need to solve the homogeneous ODE, and then find a particular solution. In fact, $x(t)=x_{homog.}+x_{particular}$ is the solution of the equation.

1) $\ddot{x}+\omega^2x=0$ (homogeneous) -> $x(t)=Ae^{i\omega t}+Be^{-i\omega t}$
2) You need to find a particular solution to that equation, that is, you must find a solution $\bar{x} \rightarrow \ddot{\bar{x}}+\omega^2\bar{x}=-kmg$. Usually you start guessing (there are some techniques, but I won't tell them here), but here is simple: if you choose $\bar{x}=x_{particular}=-\frac{kmg}{\omega}$ the equation is satisfied.

So, the solution of the ODE $\ddot{x}+\omega^2x=-kmg$ is $x(t)=Ae^{i\omega t}+Be^{-i\omega t}-\frac{kmg}{\omega}$

Now, you expected to see a damping. The thing is that the ODE you are trying to solve has not the damping term :D
The damping term occurs when friction is proportional to speed. This is why in textbooks, when they talk about the damping, they say that friction is proportional to speed. In fact, if you put
$\ddot{x}+\omega^2x=-k\dot{x}\Rightarrow\ddot{x}+k\dot{x}+\omega^2x=0$
the solution is
$x(t)=Ae^{-\frac{\sqrt{k^2-4\omega^2}+k}{2}t}+Be^{\frac{\sqrt{k^2-4\omega^2}-k}{2}t}$ that can be written as $x(t)=e^{-\frac{k}{2}}\left(Ae^{-\frac{\sqrt{k^2-4\omega^2}}{2}t}+Be^{\frac{\sqrt{k^2-4\omega^2}}{2}t}\right)$

If $k^2<4\omega^2$, then x(t) is a damped periodic oscillation. You can see it if you put $4\omega^2-k^2=\gamma^2$, it can be written as $x(t)=e^{-\frac{k}{2}}\left(Ae^{-i\frac{\gamma}{2}t}+Be^{i\frac{\gamma}{2}t}\right)$. It is damped because the exponential term $e^{-\frac{k}{2}t}$ decreases with time.

3. Oct 5, 2011

### olivermsun

Think of "damping" caused not by friction (like the spring mass system sliding on a rough surface) but by something like a shock absorber.

4. Oct 5, 2011

Well for starters, your equation isn't quite right. You shouldn't have mass in it because your units are no longer correct. If you start from a force balance, you then get your equation and divide through by mass to arrive at the one you tried to arrive at. Then you are missing an important part of your friction term. As it stands, your friction term is constant, but in reality it depends on the direction of movement. In reality, your equation should look more like this:

$$\ddot{x} + kg\;\mathrm{sgn}(\dot{x}) + \omega^2 x = 0$$
or
$$\ddot{x} + kg\frac{\dot{x}}{|\dot{x}|} + \omega^2 x = 0$$

I honestly am not quite sure how one would go about solving this analytically. Using a numerical method such as Runge-Kutta would be the best bet. I just did it in Matlab to make sure my equation was sane and it comes out right. It is just an oscillator whose amplitude decreases linearly to zero, which is what I found.

Last edited: Oct 5, 2011
5. Oct 5, 2011

### AlephZero

The analytic solution is just a sequence of "half sine waves", with a different origin for the ones with positive and negative velocities.

While the velocity is one direction (either positive or negative), this is just a mass-on-a-spring with a constant applied force. Exactly the same as a vertical spring with a mass when you include the weight of the mass in the equation of motion.

The period of oscillation is not affected by the friction force, so you could write one complicated formula to represent the whole motion if you really wanted to, but in practice there's not much point doing that. But the analytic solution for each part of the motion IS useful if you want to look how the amplitude decays, the rate or energy dissipation, etc.

Unlike damping proportional to velocity, the motion does not decay exponentially "for ever", it only continues for a finite number of cycles until the static friction is enough to hold the mass stationary.

FWIW an accurate numerical solution of this is not as simple as it might appear. If you include the friction force in the model in a naive way, you will find the output from the model does not come to a stop, but goes into an apparently "chaotic" oscillation whose properties depend on the size of the time step you use, and not on the stiffness and mass of the structure, With a "clever" adaptive time stepping integration method, the results can look quite pretty - but completely wrong.

6. Oct 5, 2011

Sure you could set up a piece-wise analytic solution that way, but I didn't think that is what the OP had in mind. I may have just fallen victim to assumptions, but I thought he/she was looking for an explicit form of $x=f(t)$.

Assuming you mean the absolute value is constant, I agree. I don't see how it is the same as a mass on a spring since the force always opposes the motion in this case. For a mass on a spring that is hanging, the added force is always down so it just becomes a non-homogeneous term on the EoM. For a horizontal spring, the only way that weight matters is with friction, which is exactly what is being discussed here. Maybe I made a mistake in my reasoning here, in which case I am certainly open to being corrected.

I agree.

Indeed. The Runge-Kutta solution I got to this showed this exact thing for all the initial conditions and physical constant values that I tried (needed a 30-min break from work earlier today, haha).

I recognize that this is always a possibility in any dynamical system, but I don't see that being a problem with this equation. I haven't done a stability analysis of it or anything, but just in a physical sense, you always have a constant force opposing the motion and no input force, so the amplitude should monotonically decrease regardless of initial conditions or the physical constants. The system seems to be dissipative to me based solely on inspection, so I feel like in this particular case, the system is globally stable.

I know what your saying though. That is, after all, related to how Ed Lorenz discovered strange attractors in the first place.