I am having troubles deriving the 2nd order differential equation for the system below, where $r=y-s$. According to my lecture notes the differential equation is

$$ M\frac{d^2r}{dt^2}+b\frac{dr}{dt}+kr=-M\frac{d^2s}{dt^2}=-Ma \\ \ddot r+2\zeta \omega_0 \dot r+\omega_0² r=-a, $$

whereas $ \omega_0 = \sqrt{\frac{k}{M}} $ and $ \zeta = \sqrt{\frac{b}{2M\omega_0}} $.

My understanding: So I know that the force exerted by a spring follows $ F_F=-kr $ and the force by a damper $ F_D=-b\dot r $. The resulting force then equals $F_a = F_F + F_D$ or $F_a - F_F - F_D = 0$. This can also be seen in the formula from the lecture notes, but the $-M\frac{d^2s}{dt^2}$ on the right hand side confuses me a bit. Why does the absolute acceleration equal the differential equation on the left?

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    $\begingroup$ As per our usage of the term, the homework-and-exercises fits the type of problem. See the info tab in that link for details. $\endgroup$
    – Kyle Kanos
    Commented Mar 10, 2017 at 13:23


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