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)

http://en.wikipedia.org/wiki/Damping#Example:_mass.E2.80.93spring.E2.80.93damper

And here is the sketch:

• Feb 6, 2023 at 17:43

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 $$m := \frac{m_1 m_2}{m_1 + m_2}$$