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

Click For Summary
SUMMARY

This discussion focuses on the analysis of third-order differential equations in mass-spring-damper systems, specifically addressing the addition of the third derivative, x''', and its implications for damping constants. The user seeks to understand the equivalent damping constant for x''' and its real-world interpretation, proposing that it should be mass multiplied by time (kg·s). The conversation also touches on the characteristic equations for second and third-order systems, highlighting the role of the damping ratio (ζ) and natural frequency (ωn) in system behavior.

PREREQUISITES
  • Understanding of differential equations, particularly in mechanical systems.
  • Familiarity with mass-spring-damper system dynamics.
  • Knowledge of Laplace transforms and their application in system analysis.
  • Basic concepts of damping ratios and natural frequencies in oscillatory systems.
NEXT STEPS
  • Research the concept of "jerk" in mechanical systems and its implications in dynamics.
  • Study the derivation and application of characteristic equations for higher-order systems.
  • Explore the role of damping ratios in stability analysis of third and fourth-order systems.
  • Learn about the physical interpretation of coefficients in differential equations related to mechanical systems.
USEFUL FOR

Mechanical engineering students, vibration analysts, and professionals working with dynamic systems who seek to deepen their understanding of higher-order differential equations and their applications in real-world scenarios.

Dnandrea
Messages
1
Reaction score
0
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!
 
Engineering news on Phys.org
Dnandrea said:
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
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. )
 

Attachments

Similar threads

  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K