Damping ratio and damping coefficient of two objects connected by spring There are two objects $m_1$ and $m_2$ connected by a spring and a viscous damper (e.g car body and wheel connected by spring).
From wiki: $\zeta ={c \over 2{\sqrt  {mk}}}$
$ζ$ is damping ratio,
$c$ is damping coefficient,
$k$ is spring constant
In the wiki example there is only one mass connected to ground. But here I have two masses $m_1$ and $m_2$. Can I use an equivalent $m$ for $m_1$ and $m_2$? If yes, how to calculate it from $m_1$ and $m_2$?
Edit:
I need to calculate $\zeta$ to find wheter $\zeta = 1$ (critical damping) or $\zeta \neq 1$ (under or over damping)
Sorry, here is the wiki link about damping:
http://en.wikipedia.org/wiki/Damping#Example:_mass.E2.80.93spring.E2.80.93damper
And here is the sketch:

 A: Yes, one can calculate an equivalent mass. Take the following coordinates, that describe the position of the particles' centers with respect to the center of mass of the set, $G$. In the sketch that follows I have arbitrarily chosen $m_1 < m_2$, and so $G$ falls closer to $m_2$:
            

Since $G$ will remain fixed in the absence of external forces, it makes sense to use the coordinates above. The equations of motion of the two particles in these coordinates read
$$m_1\frac{\mathrm{d}^2 x_1}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0 \\
m_2\frac{\mathrm{d}^2 x_2}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0$$
where $l = x_1 + x_2$ (the distance between centers). Multiplying the first equation by $m_1 / (m_1 + m_2)$ and the second one by $m_2 / (m_1 + m_2)$ and adding them we obtain
$$\frac{m_1 m_2}{m_1 + m_2}\frac{\mathrm{d}^2 l}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0$$
which is of the form of the original equation (the one from Wikipedia) if one takes $$m := \frac{m_1 m_2}{m_1 + m_2}$$
Thus, the equivalent mass is the harmonic mean of the masses involved.
