Int he three-mass coupled oscillator problem, we often see it stated that you have three masses, (they can be equal or not, but we'll assume they are equal here) connected by two springs and then another set of springs connecting the masses to the walls on the end. We assume that the spring constants are all the same. (We're assuming one-dimensional motion here as well).

So the equations of motion are usually written thus:

$$m\ddot x_1+kx_1-k(x_2-x_1)=0$$ $$m\ddot x_2+k(x_2-x_1)+k(x_3-x_2)=0$$ $$m\ddot x_3+k(x_3-x_2)+k(-x_3)=0$$

(And lord knows tell me if this is wrong and I missed a concept)

Well, ok, but what if we assume that the three masses are not connected to anything else? That is, they are connected by two springs but there are no walls? Would we just have to pick a reference frame and go from there, declaring one mas "stationary"? For instance, if we connect the first mass to a wall but leave the other end unconnected, would the EoMs look like:

$$m\ddot x_1+kx_1-k(x_2-x_1)=0$$ $$m\ddot x_2+k(x_2-x_1)+k(x_3-x_2)=0$$ $$m\ddot x_3+k(x_1+x_2)+k(x_3)=0$$

Anyhow, I was just curious, because all the problems in this vein I see seem to assume the three masses are attached between two walls.

Thanks for your insights, in advance.


1 Answer 1


Conceptually, the reason they are attached to walls is to remove rigid body motion from consideration. When they are tethered to the wall, there is no translation or rotation about the center of mass of the system.

So, when you break the connections to the walls and they are now free to drift through space, the number of admissible solutions increases.

If you connect one mass to the wall and leave the other end unconnected, then you remove the rigid body modes because it cannot rotate or translate and you're back to a similar, but different, problem as when both are connected to walls.

The equations of motion are easily derived for 3 bodies in both conditions when one draws the free body diagram (if deriving from a Newtonian approach) or through the application of Lagrange's equation or Hamilton's principle, depending on your choice of weapon.

But it's not that it's harder or impossible to have the three masses floating in space, it just adds more admissible modes to the solution which may or may not obscure the intent of the example/problem.

  • $\begingroup$ OK, so when you set up a problem where they AREN't attached to walls, you can say you are ignoring the other motions and focus on the normal modes of the oscillating masses relative to each other along the line of the spring, which would reduce it to a single dimension, yes? (This assumes you aren't looking for the normal modes in rotation and the like) yes? $\endgroup$
    – Jesse
    Commented Dec 14, 2013 at 18:49
  • 1
    $\begingroup$ It's not that you are ignoring them per se -- whatever solution method you choose to use will have to include those modes as potential solutions. But you will get oscillating solutions. So for example, if you did an eigenfunction analysis, you would get a mode with a zero natural frequency which corresponds to pure translation. Then you would get modes with non-zero natural frequencies that correspond to the oscillating modes. $\endgroup$
    – tpg2114
    Commented Dec 14, 2013 at 20:10

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