Coupled Oscillator: Coupled Mass-Spring System with Input: Difference between revisions

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\Rightarrow\ \ddot{x}_1 = \frac{-k_1}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2 - g + \frac{F}{m_1}
\Rightarrow\ \ddot{x}_1 = \frac{-k_1}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2 - g + \frac{F}{m_1}
</math>
</math>


==State Space Equation==
==State Space Equation==

Revision as of 20:26, 18 November 2009

Setup State Space Equation

Problem Statement

Find an input function such that the lower mass, m1, is stationary in the steady state. Find the equation of motion for the upper mass, m2.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

Problem Setup
Problem Setup


Initial Conditions and Values

m1=140kg

k1=108000Nm

m2=80kg

k2=80000Nm


Let the initial conditions be zero for the time being.


Force Equations

FBD for m2
FBD for m2
FBD for m1
FBD for m1

Sum of the forces in the x direction yields

For m1

+Fx2=m2x¨2

m2x¨2=Fs2m2g

Since Fs=kx

x¨2=k2m2x2g

And for m1

+Fx1=m1x¨1

m1x¨1=Fs1+Fm1gFs2

Where F=F(t) is the input force

x¨1=k1m1x1+k2m1x2g+Fm1


State Space Equation

The general form of the state equation is

x˙_(t)=A^x_(t)+C^u_(t)

Where M^ denotes a matrix and v_ denotes a vector.

Let x1, x˙1, x2, and x2˙ be the state variables, then

[x˙1x¨1x˙2x¨2]= [0100k1m10k2m10000100k2m20][x1x˙1x2x˙2]+[000011m10000001000][gF00]


Solve Using Laplace Transform Method