For a simple system where you have a mass attached to a spring and damper in parallel:

enter image description here

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:

  1. $c^2 <4mk$ (under damping)
  2. $c^2 >4mk$ (over damping)
  3. $c^2 =4mk$ (critical damping)

The damping ratio is then expressed by $\frac{c}{\sqrt{4mk}}$.


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:

enter image description here

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?

enter image description here

$σ = (η_1+η_2)\dot{ϵ} + \frac{η_1η_2(E_1+E_2)}{E_1E_2} - (\frac{η_1}{E_1} + \frac{η_2}{E_2})\dot{σ} - \frac{η_1η_2}{E_1E_2}\ddot{σ}$

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.


1 Answer 1


Re-writing the strain-stress equation $$ 0 = E_1Ɛ + \frac{η(E_1+E_2)}{E_2}\dot{Ɛ} - \frac{η}{E_2}\dot{σ} - σ $$ for displacement/restoring force variables $$ 0 = E_1x + \frac{η(E_1+E_2)}{E_2}\dot{x} + m\frac{η}{E_2}\dddot{x} + m\ddot{x} $$ and applying the Laplace transform, we arrive at the equation $$ 0 = E_1 + \frac{η(E_1+E_2)}{E_2}s + ms^2 + m\frac{η}{E_2}s^3 \tag1 $$

The formula for root $s = \frac{-E_1E_2}{η(E_1+E_2)} $ is only valid when $m=0$, and for root $s = ±\sqrt{-\frac{E_1}{m}}$ is only valid when $η=0$. In a general case of nonzero $m, η$, the equation (1) has three roots. The general formula for cubic equation roots is complicated. To write it down, the Wikipedia article introduces two first-level "determinants" $Δ_0 = b^2-3ac$ and $Δ_1 = 2b^3-9abc+27a^2d$, where $a, b, c, d$ are the cubic equation coefficients: $ax^3+bx^2+cx+d=0$, Then it defines an "upper-level" determinant $C$, built with the "first-level" determinants, $$ C=\sqrt[3]{{Δ_1 ± \sqrt{Δ_1^2 - 4Δ_0^3}}\over{2}} $$ the sign ± is selected to be plus or minus in order to guarantee that $C$ is never zero, and the three roots are $$ x_k = -{1\over{3a}}\left(b + ξ^kC+{{Δ_0}\over{ξ^kC}}\right), \, k \in \{0,1,2\}, \, ξ={{-1+\sqrt{-3}}\over2} \tag2 $$ For a parallel-serial-connected spring-damper-mass system of your question, $a = m\frac{η}{E_2}, b = m, c = \frac{η(E_1+E_2)}{E_2}, d = E_1$.

The equation (1) reduces to a second-order ODE describing the simple damped harmonic oscillator:

  1. for a trivial case of $η=0$, where there is no damping at all, and the spring E2 does not participate in the motion equations;
  2. the other non-dissipative case is $η\to{\infty}$, the springs act in parallel with the total stiffness of $E_1+E_2$;
  3. with $E_2\to\infty$, the bottom spring is "turned off", the system has one spring and one damper in parallel, this system you consider in the beginning of your question;
  4. with $E_1=0$, the upper spring is "turned off", the system has one spring and one damper in series, the system also has overdamped, underdamped and critically damped solutions that can be found with the formulas from your link.

For non-zero, finite values of $η, E_0, E_1, m$, iff the system has degenerated roots for certain combinations of $η, E_0, E_1, m$ values, then one of its linearly independent solutions has the form $t·exp(x_kt)$, the root $x_k$ has a negative real part. This solution may be called the critically damped solution. To find these combinations of $η, E_0, E_1, m$ values, one must search for degenerate solutions of Eq. 1 with multiple roots. The formulas are certainly not that simple as the formulas for a damped harmonic oscillator, but the ODE for the mechanical system of eq. 1, by virtue of an electrical-mechanical analogy, is identical to the ODE of a circuit with a Cauer topology of the generic three-pole unbalanced T-section LC low pass filter, the latter being often discussed on electronics forums.


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