# Modeling a 2-dimensional mass spring system

First of all, I am unfortunately not an expert in physics, so please be indulge with me. I am trying to model a $$2$$-dimensional mass-spring system with $$1$$ mass and $$3$$ springs to solve a dynamics problem in frequency domain. I've been looking for a solution for a similar problem but I couldn't find anything useful. Are these classical newton equation of motion mass-spring systems limited to $$1D$$?

The mass $$m$$ is connected to $$3$$ springs $$k_1, k_2, k_3$$, which are fixed at their endpoints, rotations are possible. The springs are assumed linear and can be simplified $$k_1=k_2=k_3$$. In the equilibrium state, the angle between the springs is $$120^{\circ}$$. • In the case of 4(or 2) perpendicular springs, the motion of the mass can be approximated by Lissajous curves.
– Ali
Feb 28, 2014 at 1:44
• @HansBlix: I'm pretty sure the two-dimensional potential surface associated with the system is not separable as a quadratic form, so the question becomes, are you considering small or large displacements from equilibrium? If they are small then the Hessian of the potential can be used to determine the two eigenfrequencies. If they are large, it's more complicated. Feb 28, 2014 at 2:00
• @HansBlix: Also, when you say "in frequency domain", does that imply that the mass is being forced by some source? Or is it just freely vibrating? Feb 28, 2014 at 3:38
• In the first step Feb 28, 2014 at 11:17
• I want to investigate perpendicular springs later, so thanks Ali! DumpsterDoofus: I will consider small displacements from equilibrium. So I will have a look into the Hessian potential to determine the eigenfrequencies. For the beginning I just want to know the eigenfrequencies, also from systems where more masses are interconnected in the same pattern as above. Later on there should be put a force on (homogeneous wave and wave spectra). Feb 28, 2014 at 11:25

The simplest setup is for small displacements. Suppose the spring rest lengths are $L_1,L_2,L_3$, the mass has mass $m$, the springs have constants $k_1,k_2,k_3$, the angle is 120 degrees between attachments, and the attachment points are set up so that at rest, the springs are all unstretched.
The potential becomes $$U(\mathbf{r})=\sum_{j=1}^3\frac{k_j}{2}(|\mathbf{r}-\mathbf{u}_j|-L_j)^2$$ where $$\mathbf{u}_j=\{L_j\cos(2\pi j/3),L_j\sin(2\pi j/3)\}.$$ It's easy to verify that $$\nabla U(\mathbf{0})=\mathbf{0}$$ which means that the system is at equilibrium when the mass sits at the origin.
Defining $$H=\nabla\nabla U(\mathbf{0})=\left( \begin{array}{cc} \frac{1}{4} \left(k_1+k_2+4 k_3\right) & \frac{1}{4} \sqrt{3} \left(k_2-k_1\right) \\ \frac{1}{4} \sqrt{3} \left(k_2-k_1\right) & \frac{3}{4} \left(k_1+k_2\right) \\ \end{array} \right)$$ we obtain eigenvalues $$\lambda_\pm=\frac{1}{2} \left(k_1+k_2+k_3\pm\sqrt{k_1^2-k_2 k_1-k_3 k_1+k_2^2+k_3^2-k_2 k_3}\right)$$ and the ordinary vibrating frequencies become $$\omega_\pm=\frac{1}{2\pi}\sqrt{\frac{\lambda_\pm}{2m}}.$$ Notice that the lengths are irrelevant, and that in the case $k_1=k_2=k_3$ the frequencies become identical, thus becoming like a 2D spherical oscillator.