Background
For a simple system where you have a mass attached to a spring and damper in parallel:
We can calculate the critical damping from the equation of motion:
$mx_{tt} + cx_t + kx = 0$
$ms^2 + cs + k = 0$
$s= \frac{-c ± \sqrt{c^2-4mk}}{2m}$
There are then three conditions:
- $c^2 <4mk$ (under damping)
- $c^2 >4mk$ (over damping)
- $c^2 =4mk$ (critical damping)
The damping ratio is then expressed by $\frac{c}{\sqrt{4mk}}$.
Question
I'm wondering if or how this can be extended to more complex systems.
Let's say you add a second spring to the parallel system:
This is a viscoelastic model where the constitutive relationship is in terms of stress (σ) and strain (Ɛ):
$σ = E_1Ɛ + \frac{η(E_1+E_2)}{E_2}\dot{Ɛ} - \frac{η}{E_2}\dot{σ}$
Is it possible to calculate the critical damping or damping ratio in the same manner? If so how would it work?
What about with two springs and two dampers like this?
$σ = (η_1+η_2)\dot{ϵ} + \frac{η_1η_2(E_1+E_2)}{E_1E_2} - (\frac{η_1}{E_1} + \frac{η_2}{E_2})\dot{σ} - \frac{η_1η_2}{E_1E_2}\ddot{σ}$
Is it possible to do the same and if so how?
My Guess
My guess is strain is equivalent to $x$ and stress is equal to force, so we can reword the 3 element system:
$0 = E_1Ɛ + \frac{η(E_1+E_2)}{E_2}\dot{Ɛ} - \frac{η}{E_2}\dot{σ} - σ$
$0 = E_1x + \frac{η(E_1+E_2)}{E_2}\dot{x} - m\frac{η}{E_2}\dddot{x} - m\ddot{x}$
$0 = E_1 + \frac{η(E_1+E_2)}{E_2}s - ms^2 - m\frac{η}{E_2}s^3$
Which according to Wolfram gives the following roots:
$s = \frac{-E_1E_2}{η(E_1+E_2)}$
$s = ±\sqrt{\frac{E_1}{m}}$
But that is not useful for solving any critical damping scenario that I can figure out.
Is there some other approach or what did I screw up?
Any help or guidance is appreciated. Thanks.