So I'm not sure whether this should be posted in the Physics forum or not, but here goes:(adsbygoogle = window.adsbygoogle || []).push({});

I'm a Junior ME student at NC State taking a Vibrations course. We've gone over the general differential equation for mass-damper-spring systems where

M_{eq}⋅x'' + C_{eq}⋅x' + k_{eq}⋅x = F_{extermal}

What I was wondering is, if you had a third order equation, where x''' (jerks, I believe) was added in, what would be the "damping constant" equivalent that would be multiplied by x''' and how could this be interpreted in the real world?

If you follow units, I think it should be mass ⋅ time as the coefficient. For k_{eq}, if you assume N/m, can be canceled out into kg/s^{2}, C_{eq}into kg/s, and M_{eq}into kg. Following that trend, the coefficient would have to be kg⋅s.

Mass ⋅ time I've tried researching and came across something called "action" that was really interesting. I'm not sure whether I'm making this much harder than it needs to be. Any explanation?

Thanks!

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# Third Order Differential Equations (Mass-Spring-Damper-etc)

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