How could this damped oscillator ever go to infinity? Or negative infinity for that matter?

Question

This is an ODE problem,but I cannot visualize why it can go to infinity or negative infinity.

Consider

$$x'' -6x' + 8x = 0$$

Where $x''$ is acceleration, $-6x'$ is the damping effect and $8x$ is the spring effect.

If I write it back into $physics$ form, I get $x'' = 6x' - 8x$. This just means the spring wants to pull the mass back and the resistive force is not strong enough.

Now the original question was, "Using the mass–spring analogy, predict the behavior as $t \to \infty$ of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem."

Solving using Maple 15, I got (with Initial conditions)

!

I don't understand how infinity could be at play for this. How could you even have negative infinity? The spring wouldn't stretch that long and wouldn't the "max" point of the position be where the spring is attached (if it does go to negative infinity)

Answer 1

The reason for this result is the sign in you damping term.

For a damped harmonic oscillator you need to have a resistive force on the mass point at $x$. That means if $x=0$ is the equilibrium position the damping term will be proportional to the velocity with an negative constant $F_{\text{damp}} = -ax',a>0$. I.e. the total force on the mass point is the damping term and the spring (Hooke's Laq) term $F_{\text{spring}}=-kx$. The total equation of motion then is:

$$F=ma=-kx-ax'$$ or $$mx''+ax'+mx = 0, a,m>0$$

Your result then is going to infinite because your system is not actually damped but anti-damped. It means there is a force that accelerates your mass point away from the equilibrium position the more the faster it becomes, which will clearly lead to large values very fast.

Answer 2

First, the fact that your graph goes to negative infinity would indeed mean that the spring will break, or at least that it's going to stop behaving like a spring. Basically, consider that you have obtained this solution using a model of a spring in which the force it exerts is proportional to its displacement from equilibrium. Once you reach the point at which that model is not accurate, the solution you obtain from it is no longer accurate. So in practice, what would happen is that the mass would move away from equilibrium until the spring got stretched enough that it no longer behaved linearly. At that point, some other effect would come into play, which would rein in the motion of the mass.

However, it's a moot point in this case because your equation is wrong. Check the signs of the forces and see if you can understand why you have the wrong sign on the drag force. So the preceding paragraph applies not to a mass on a spring, but to some other hypothetical system in which there is an "anti-drag" force.

Your intuition is actually correct for a mass-spring system. With a damping force much greater than the restoring force (more precisely, if $b > \sqrt{4km}$), the mass will never fully return to the equilibrium point, though it will approach it asymptotically. This is called an overdamped oscillator.