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

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)

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Comment to the question(v1): The so-called "damping effect term" is not damping at all because it has the wrong sign. It should be called a "positive-feedback-term" instead. –  Qmechanic Feb 7 '12 at 1:20
Make calculations with $-6|x'|$ instead. –  Vladimir Kalitvianski Feb 7 '12 at 10:33
If you flip the graph around the y-axis, so that negative times are positive, this is the correct motion. –  Ron Maimon Mar 2 '12 at 15:45

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.

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But the spring force, restoring force, 8x also draws it back, so ti can't grow that fast? –  sidht Feb 7 '12 at 1:51
well but the spring force is proportional to the acceleration, while the velocity is $\int_0^t a dt'$ so that term dominates. –  luksen Feb 7 '12 at 13:11
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.