I am trying to model this system but am having difficulties in setting up the equations of motion. This is a simplified model of a motorcycle suspension system, where P is the ground in the image below. In addition, the variable $u$ is the input to the system, which is the displacement from the ground (like a motorcycle going over a bump).

enter image description here

Consider mass $m_2$ first. Keeping $m_1$ fixed and displacing $m_2$ a bit yields the equation of motion for $m_2$ as $$ m_2 \ddot{y} = -k_2 (y-x) -b_2 (\dot{y} - \dot{x}) $$

And similarly, for $m_1$, keeping $m_2$ and $u$ fixed and displacing $m_1$ a little bit, the equation of motion comes out to be $$ m_1 \ddot{x} = -k_1 (x-u) -b_1(\dot{x} - \dot{u}) - k_2(x-y) - b_2(\dot{x}-\dot{y}) $$

The first two term on the RHS come from the coupling of $x$ with $u$ while the second and third come from the coupling of $m_1$ and $m_2$, or equivalently from Newton's third law. However, the textbook solution only has the first two terms, and not the interaction between $m_1$ and $m_2$ for the EOM of mass 1. I am extremely confused as to why this is the case. Is it because they are vertical and not horizontal? How would the EOM change if the scenario was all horizontal?

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    $\begingroup$ Your EOM are correct, the terms for the coupling between m1 and m2 are definitely needed. $\endgroup$
    – am304
    Feb 23 '18 at 9:22
  • $\begingroup$ two experts in this field are karnopp and margolis, see their papers- NN $\endgroup$ Oct 1 '18 at 4:23
  • $\begingroup$ Is there anything saying that mass 2 is a lot smaller than mass 1 or that the spring constant of k2 is really high? If either of those are true, you might be able to neglect the contribution of the top mass. $\endgroup$
    – John
    Feb 5 '21 at 18:28

You're equations of motion are correct. Your book is either wrong, or is making the assumption that $m_1 \gg m_2$ in which case the terms contributing from the interaction with $m_2$ would be negligible.

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    $\begingroup$ $$m_2$$ is the motorcycle. $$m_1$$ the unsprung mass. No way in which the unsprang mass is larger. Leaves only one conclusion: The book must be wrong. But without knowing the context, it's difficult. Maybe the equation is annotated to make clear, it is only part of the actual equation? $\endgroup$
    – Ingo
    Dec 14 '19 at 18:37

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