# Equation of motion for a Mass-Spring-Damper-system, one mass 2 DOFS

1. May 21, 2013

### Lelak

1. The problem statement, all variables and given/known data:

Find the equation of motion for the system below (see the attached files)
https://www.physicsforums.com/attachment.php?attachmentid=58905&stc=1&d=1369155073
Solve the problem with the state vector approach. Choose realistic values on k1,k2,c1,M and F

2. Relevant equations

See the attatched file.
https://www.physicsforums.com/attachment.php?attachmentid=58906&stc=1&d=1369155073

3. The attempt at a solution

I am having trouble finding the right equation of motion for this system since it has two degrees of freedom and only one mass. For a "normal", where there is a mass at every node, I have not had a problem so far.

I would appreciate any help or tips how to handle this problem! Thank you very much!

Last edited: May 21, 2013
2. May 21, 2013

### rude man

Why do you say this is a 2DF problem? It isn't.

3. May 21, 2013

### rude man

One way you might want to do this is to put an imaginary second mass m2 where the springs connect to each other, solve for x1 and x2, then let m2 approach zero.

4. May 21, 2013

### Lelak

I tried doing that but then the eig function in matlab will not solve the eigen values and the eigenvectors. I also checked with one of the teachers, he did not seem to think that was the right way to go.

5. May 21, 2013

### Lelak

Okay, how would you set up the equation of motion then?

I figured since there are two springs, one with damping and one without, it would be of interest to know the displacement in the node between the two springs. It would also be intereseting to find the displacement of the mass. That would give me two translations and that why I though there was 2dofs.

How would you approach this problem?

6. May 21, 2013

### rude man

I would put a 2nd mass m at the junction of the springs. Then, when you get the result for x1 (the position of your real mass M), let m << M. I know this has to work.

So x1 is the position of real mass M when the system is quiescent (no motion of either mass), and x2 is the position of the second mass m. Then write F = mx'' for both masses, solving the system of ODE's and getting x1 and x2. Then just let m << M.

It is a 1DF system since there is motion in one direction only.

I imagine one can do without a second mass but that would involve knowledge of how to handle springs and dampers in series & parallel which I don't possess.

7. May 21, 2013

### rude man

OK, there's another way: realizing that force is the same at every node from the left wall to the left end of the mass M, that gives us
k1 x1 + c dx1/dt = k2 x2 where

x1 = stretch of spring 1,
x2 = stretch of spring 2.

Let x = 0 be at the left wall;
L1 = relaxed length of spring 1
L2 = relaxed length of spring 2.

then position of mass M is at x = L1 + L2 + x1 + x2.
So now what is/are the equation(s) of motion for M? Hint: use the fact at the top of my post.

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