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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:

enter image description here

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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.

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    $\begingroup$ You've set up the problem correctly but made an error when combining the equations. m1*d2x1/dt2 + m2*d2x2/dt2 is not (m1+m2)d2l/dt2. It is d2(m1x1+m2x2)/dt2, which isn't particular useful. Instead, multiply the first eq by m2/(m1+m2) and the 2nd eq by m1/(m1+m2) and add them together. You'll get m1m2/(m1+m2)*d2l/dt2+c*dl/dt+kl = 0. So the equivalent mass is m1m2/(m1+m2). This also makes sense because it approaches m1 as m2 approaches infinity and approaches m2 as m1 approaches infinity. $\endgroup$ Commented Feb 5, 2023 at 20:58
  • $\begingroup$ @DeanBrettle: You are absolutely right; I have modified the answer accordingly. I remember at the time being a bit surprised by the result, but it came out so easily as an average that I did not give it a second thought. I should have! $\endgroup$ Commented Feb 6, 2023 at 10:33

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