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

1. Sep 30, 2015

### Dnandrea

So I'm not sure whether this should be posted in the Physics forum or not, but here goes:

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

Meq⋅x'' + Ceq⋅x' + keq⋅x = Fextermal

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 keq, if you assume N/m, can be canceled out into kg/s2, Ceq into kg/s, and Meq 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!

2. Oct 1, 2015

### Hesch

As for a 2. order system, written in the Laplace domain, the characteristic equation could be expressed:

s2 + 2ζωns + ωn2 = 0

where ωn is the freqency [rad/s] and ζ is the damping ratio of a sinusoidal oscillation. If ζ<1, the roots of the equation will be complex, and as a complex root will have a conjugated root, the roots will have "used" both roots in the polynomium. There is no further to add. ( See fig. 2. )
As for a 3. order system, you could express the characteristic equation:

( s + a )( s2 + 2ζωns + ωn2 ) = 0

so the same ζ is still there, in principle unchanged. Only a factor ( s + a ) has been added, which is just an overlayed exponential function, that has no damping ratio. The interpretation of the ζ is the same. ( See fig. 3. )

The charateristic equation for a 4. order system could be expressed:

( s2 + 2ζ1ωn1s + ωn12 )( s2 + 2ζ2ωn2s + ωn22 ) = 0

thereby having 2 different damping ratios and 2 frequencies. ( See fig. 4. )

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