# Spring-mass system with two springs and three masses [closed]

I'm trying to solve a system of springs and masses that is confusing me. First, the balls are all lined up linearly. Secondly, the ball in the middle has a smaller mass $m$ while the first and last balls have a larger mass of $M$. The larger balls are each connected to the middle ball with a spring with a spring constant of $k$. They are assumed to move in the right direction and there is no additional external force.

     x_1      x_2       x_3
*--------*---------*           ------> x
M    k   m    k    M


I'm trying to solve this system using the eigenvalue concept, but I'm having trouble. I've dealt with a two mass spring system, but never a three mass, so I'm a bit confused.

From what I gather, I just set the $\ddot{x}$ to $F/m$ and set up the matrix, much like for a two mass system. But I don't know how to do that with the third mass. Like for example, for $\ddot{x}_1$, I think that we have

$$0 -(k/m)(x_1-x_2) -(k/M)(x_2-x_3)$$

and for $\ddot{x}_2$

$$0 -(k/m)(x_2-x_1) -(k/M)(x_2-x_3)$$

and for $\ddot{x}_3$

$$0+0-(k/M)(x_3-x_2)$$

but I really don't know if I'm right. I highly doubt it. So if some kind soul could tell me what I'm doing wrong/right and point me in the next direction, I would be forever grateful.

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## closed as off-topic by jinawee, David Z♦May 29 '14 at 14:19

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First hint to the question(v4): Re-check where the springs and masses enters the coupled diff. eqs. (=Newton's 2nd law) . – Qmechanic Nov 4 '12 at 11:37
A sketch of the system would really help making the question clearer. – Bernhard Nov 4 '12 at 11:43
A system of ordinary differential equations can be written as a matrix [ordinary] differential equation whose solution is obtained looking for the eigenvalues of the associated matrix, see en.wikipedia.org/wiki/Matrix_differential_equation – DaniH Nov 4 '12 at 15:45

OK. You must use the Newton's second law $F=ma$ for each mass. But you also need to specify your system of reference. For example, lets consider our origin in the rest position of the mass $M$ in the left. Respect that system of reference we can write the forces acting upon each mass. The first mass undergoes two forces due to the first spring and to the movements of the two masses 1 and 2. That's why we can write $$F_1\equiv M\frac{d^2x_1}{dt^2}= kx_1+k(x_2-l)$$ Where $x_1$ and $x_2$ means the position of the first $M$ and second $m$ mass with respect our origin. Here $l$ represents the rest length of the springs. Notice that if $x_1=0$ and $x_2=l$, then there is no force upon the left mass $M$. I understand that the problem can be difficult, but in order to understand and clarify the ideas you should make a diagram showing the forces acting upon each mass when vary the positions of each of the three masses.

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The object on the left at $x_1$ is only connected to the small mass so the 'equation of motion' should be,

$$M \, \ddot{x}_1 = k(x_2 - x_1) = F,$$

The sign of the force experienced by the mass at $x_1$ must be positive because the spring is pulling the mass along the $x$-axis to the right.

You are mixing the masses up as well. The other equations of motion should be,

$$m \,\ddot{x}_2 = -k(x_2 - x_1) + k(x_3 - x_2)\\ M \,\ddot{x}_3 = -k(x_3 - x_2).$$

The next thing you should do is to try and divide the masses over on the right side of the equations and write them as a matrix equation which in general can be written as $$\ddot{\mathbf{x}} = \mathbf{A}\mathbf{x}$$ where $\mathbf{A}$ is the system-matrix you should derive from the equations above. The eigenvalues of the system-matrix corresponds to the eigenmode frequencies squared as you can see for yourself by assuming oscillatory solutions of the form $\mathbf{x}(t) = \mathbf{x}_0\cos(\omega t)$ since the second derivative above yield two factors of $\omega$. But this is your job!

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It cannot work that way! The linear combination $Mx_1+mx_2+Mx_3$ has to correspond to the zero eigenvalue, as it is, up to the factor $1/(2M+m)$, the position of the center of mass which freely moves. For that eigenvalue the solution has not the form $\cos(\omega t)$ but instead $At+B$... – Valter Moretti May 29 '14 at 8:36