# Why is the damping force larger than the restoring force in this model?

I'm reading the book "Differential equations with boundary-value problems" by D.Zill and I cannot understand the following:

In the Free Damped Motion Model,

we have a flexible spring suspended vertically from a rigid support. We attach an object of certain mass $$m$$ to the spring so this is stretched $$s$$ units (in this situation, the object will be in an equilibrium position). Assume that there is a resisting force due to the surrounding medium (as a viscous medium, for example). If $$x(t)$$ is the position of the object after $$t$$ time units, we have the ODE $$mx''=mg-\underset{Hooke\; Law}{\underbrace{k(x+s)}}-\underset{damping\; force}{\underbrace{\beta x'}},$$ where $$g$$ is the gravity acceleration , $$k$$ is the constant given by the Hooke Law (restoring force) and $$\beta>0$$ is the damping -constant given by the damping force.

The ODE has as solutions for its auxiliary equation: $$-\lambda\pm \sqrt{\lambda^2-\omega^2}$$, where $$\lambda=\beta/(2m), \omega^2=k/m$$.

We have three cases to analyze, but it is one that I cannot get:

If $$\lambda^2-\omega^2>0$$, then * the system is said to be overdamped because the damping coefficient b is large when compared to the spring constant k.*

Why does the author state that $$b$$ is large when compared to $$k$$? I cannot get why he affirms this (which I guess he does in order to conclude that the damping force is large compared to the restoring force). I know that the graph of the solutions helps to visualize the model but I want to understand this model just by looking at the equation and its variable involved (basically, a qualitative analysis of the problem).

Starting with the inequality $$\lambda^2-\omega^2>0$$, plug in the definitions of $$\lambda$$ and $$\omega$$ in terms of $$\beta$$, $$k$$, and $$m$$. Then simplify the resulting inequality to get all the $$\beta$$'s on one side and all the $$k$$'s on the other.
• Can you give more details? Because I get $\beta^2>4mk$ and I know how to conclude that the damping force is large compare to the restoring force. – Math Guy Jan 2 at 20:12
• @MathGuy It doesn't say the damping force is large compared to the restoring force but rather that the damping coefficient is large compared to the spring constant. And that's just an intuitive, non-quantitative summary of the information given precisely by $\beta^2>4mk$. – Andreas Blass Jan 3 at 18:16
If $$λ^2−ω^2>0$$, then obviously $$β>2\sqrt{mk}$$, so that the qualitative relation of damping constant and spring constant is as stated.
• how do you get $\beta>2k$ ? am I missing something? I get $\beta^2>4mk$, how do you conclude your statement? – Math Guy Jan 2 at 20:11
• You are right, I somehow missed the square of $ω$. – Lutz Lehmann Jan 2 at 20:54