Monday 27 May 2019

homework and exercises - 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 $120^{\circ}$.


2d mass spring system



Answer



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.


When you add more masses to the system, things get interesting.


What manner of driving force are you planning on applying?


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...