Consider that this is a simple mass spring damper system:
$$ \ m \frac{d^{2}x}{dt} = F – b\frac{dx}{dt} – kx\ $$
What I allready know is the force $ F $ and the mass $ m $. Then I can find the spring stiffness k by sett all the derivatives to zero.
Then I can compute the damper $ b $ throught this formula:
$$ \ b = \zeta 2 \sqrt{k m}\ $$
According to Wikipedia, the damping ratio $ \zeta = 0.7 \ $ is a good number.
And this really works too! I have test it by my own ODE simulations, and it works OK! Perfect!
But the question is:
If I have two or more ODE:s with diffrent spring stiffness and dampness like this:
$$ \ m_{2}\frac{d^{2}z}{dt} = F – C_{2}(\frac{dz}{dt} – \frac{dy}{dt}) – k_{2}(z – y) \\ m_{1}\frac{d^{2}y}{dt} = C_{2}(\frac{dz}{dt} – \frac{dy}{dt}) + k_{2}(z – y) – C_1\frac{dy}{dt} – k_1 y\ $$
I can compute the stiffness $ k_1 , k_2 $ if I know the masses $ m_1 , m_2 $ and the force $ F $ , and also I need to set some position values for $ y , z $
Very easy! But how do I compute the damper $ C_1 , C_2 $ if I determine that $ \zeta = 0.7 $ ?
Best Answer
I am going to tackle a special case of the general problem stated in the question. I hope this would leads to some insight into how to handle the general problem.
Consider the 2-DOF system, but with each mass equal to $m_1=m_2 =\frac{m}{2}$ (so the total mass is $m$), each stiffness $k_1=k_2 = 2 k$ (so the combined stiffness is $k$) and each damping is $c_1=c_2 =2 c$ for the same reason.
To solve the problem, I formulated the equations of motion in terms of the absolute displacements $x_1$ and $x_2$, but then I switched the DOF of the system to relative displacements $ q_1 = x_1 $ and $q_2 = x_2-x_1$.
As a system of equations this is
$$ \pmatrix{\ddot{q}_1 \\ \ddot{q}_2 } + \left[ \matrix{ \frac{4 k}{m} & -\frac{4 k}{m} \\ -\frac{4 k}{m} & \frac{8 k}{m} } \right] \pmatrix{q_1 \\ q_2} + \left[ \matrix{ \frac{4 c}{m} & -\frac{4 c}{m} \\ -\frac{4 c}{m} & \frac{8 c}{m} } \right] \pmatrix{\dot{q}_1 \\ \dot{q}_2}= \pmatrix{0\\0} $$
This system has an "in-phase" solution of
$$\pmatrix{q_1 \\ q_2 } = \pmatrix{Q_1 \\ Q_2} \exp(-\beta t) \sin(\omega t) $$
Since this is a 2-DOF system it has two natural frequencies
$$ \omega_n^2 = \begin{cases} \frac{2 k( 3-\sqrt{5})}{m} & \mbox{mode A} \\ \frac{2 k( 3+\sqrt{5})}{m} & \mbox{mode B} \end{cases} $$
These are found from the eigenvalues of the 2×2 matrix multiplying $\pmatrix{q_1 & q_2}$ in the equations of motion.
The values of $\beta$ and $\omega$ that solve the system are $$\begin{aligned} \beta & = \frac{c\, \omega_n^2}{2 k} \\ \omega^2 & = \omega_n^2 \left(1- \frac{c^2\, \omega_n^2}{4 k^2} \right) \end{aligned} $$
Using the values of $\beta$ and $\omega$ that solve the system of equations, one can parametrise them as $$ \begin{aligned} \beta & = \omega_n \zeta \\ \omega & = \omega_n \sqrt{1-\zeta^2} \end{aligned}$$ which is exactly what you do in the 1-DOF system. This means that the stiffness $k$ and damping $c$ needed for a response with natural frequency $\omega_n$ and damping ratio $\zeta$ are
$$\boxed{ \begin{aligned} k & = \frac{m \, \omega_n^2}{2 (3 \pm \sqrt{5})} \\ c & = \frac{m \, \omega_n \zeta}{(3 \pm \sqrt{5})} \end{aligned} }$$
Also note that to target a specific half-life of $t_H$ where $\exp(-\beta t_H) = \frac{1}{2}$ you need a damping ratio of $\zeta = \frac{ \ln(2)}{\omega_n\, t_H}$.
Finally, for completeness I am going to mention that the response amplitudes are arbitrary, but must be related to each with $Q_2 = -Q_1 \frac{1\pm\sqrt{5}}{2}$.