Let's say $m_1$ is attached to $m_3$ via a spring of constant $k_1$ and $m_3$ is attached to $m_2$ via a spring of constant $k_2$. Just to simplify the problem we can make $m_1=m_2=m_3$ and $k_1=k_2$.

Now, using Lagrangian formalism, the equations of motion are:

\begin{equation} \begin{split} \ddot{x}_1 & =-\omega_0^2 (x_1-x_3) \\ \ddot{x}_2 & =-\omega_0^2 (x_2-x_3)\\ \ddot{x}_3 & =\omega_0^2 (x_1+x_2-2x_3)\\ \end{split} \end{equation}

If we suppose solutions of the form $x_i=A_ie^{\alpha t}$, we get the solutions to two coupled harmonic oscillators. So maybe that's not the correct answer.

Is it absolutely necessary to create a enormous matrix to reduce it to first order differential equations? Is it possible to make any physical assumptions to facilitate the problem?

  • $\begingroup$ $3\times 3$ matrices aren't large in my opinion. And linear differential equations are so easy to handle will matrices. In assuming ideal springs, frictionless surfaces, etc. I'm not sure what more simplification you could find. It's not needed though. $\endgroup$ – Aaron Stevens Jun 13 at 3:12
  • $\begingroup$ @AaronStevens If you reduce the system to first order differential equations the result would be a $9 \times 9$ matrix. $\endgroup$ – IchVerloren Jun 13 at 3:15
  • $\begingroup$ Oh ok I misread it. Why do you need to make everything first order? $\endgroup$ – Aaron Stevens Jun 13 at 4:02
  • $\begingroup$ You can use $x_1+x_2-2x_3=(x_1-x_3)+(x_2-x_3)$ to reduce this to a pair of second order equations. $\endgroup$ – StephenG Jun 13 at 4:04

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