# Two carts connected by spring on frictionless track

I have the following homework problem:

Consider two carts of equal mass m on a horizontal, frictionless track. The carts are connected by a single spring of force constant k, but are otherwise free to move freely along the track. (a) Write down the Lagrangian and find the normal frequencies of the system. Show that one of the normal frequencies is zero. (b) Find and describe the motion in the normal mode whose frequency is nonzero. (c) Do the same for the mode with zero frequency.

As a note, I am just on part (b) now.

For setup, I represented the carts with the location of the edges of the carts that are connected to the spring with the coordinates $x_1$ and $x_2$ and gave the spring an equilibrium length $L$.

This leads to the following terms in the Lagrangian:

$T = \frac{m}{2} \left( \dot{x_1}^2 + \dot{x_2}^2 \right)$

$U = \frac{k}{2} \left( L + x_1 - x_2 \right)^2$

This leads me to the following equations of motion:

$m\ddot{x_1} = k (L + x_1 - x_2)$

$m\ddot{x_2} = -k(L + x_1 - x_2)$

This forms a matrix differential equation

$M \ddot{\vec{x}} = -K \vec{x} + \vec{L}$

with M = $m \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right)$, K = $k \left( \begin{matrix} -1 & 1 \\ 1 & -1 \end{matrix} \right)$, and $\vec{L} = kL \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$.

To solve this, I assume a form of the solution $\vec{x} = \vec{a} e^{i \omega t}$ where $x_i, a_i \in \mathbb{C}$. Getting rid of $\vec{L}$ yields the eigenvalue problem

$\det\left( K - \omega^2M\right) = 0$

When I solve this, I get the solutions

$\omega^2 = 0, -\frac{2k}{m}$

The first solution seems to indicate 0 as a normal mode, which I completely expected. The second solution results in a complex frequency that will have an exponential decay as the carts start to move. This is a bit of a surprise, especially considering that the problem mentions solving for the nonzero normal modes. However, my intuition tells me that this could possibly be partially correct since when the motion of the spring starts an oscillation, the spring should quickly try to return to equilibrium since there's nothing else going on with the carts. But this seems to indicate motion that is oscillatory but exponentially decreasing (of the form $e^{-x}\sin{x}$).

Either way, I feel like I could have done something potentially major wrong. Is there something obvious that I have missed? Did it cause an issue when I threw away $\vec{L}$ and didn't go all the way to finding a particular solution? I didn't think it would be an issue as it wasn't involved in the eigenvalue problem.

Update: After thinking a bit more about it, I believe this may actually be correct at least in the real case, as the spring will tend to return to equilibrium and there's nothing pulling the cars out. However, considering the setup given in this question, I would think that the spring would oscillate perfectly without dampening, resulting in a frequency that may be something like the form of $\omega = \sqrt{\frac{k}{2m}}$. This is similar to the answer that I arrived at if the negative is thrown away (or if a solution of the form $e^{\omega t - i \delta}$ is guessed rather than a solution of the form $e^{i \omega t - \delta}$.

After checking over my math, I realized that the nonzero part results in $\omega^2 = -2k/m$. Taking either of the cases that I looked at in the update just before this (exponential guess or error in signs), we arrive with a normal mode of $\sqrt{2k/m}$, which sounds somewhat reasonable. Anyone have any comments on that?

• Now this is the kind of homework related questions that is good for Physics SE. – dmckee Oct 23 '14 at 2:11
• I'd definitely bet there's a wrong sign somewhere, probably in $K$. An exponentially decaying solution is not an option, it violates too many laws. – QuantumBrick Oct 23 '14 at 4:14
• I don't see any conceptual error...You sure you are not missing a line from the question? because the exponential decay in your solution seems a little out of context....... – GRrocks Oct 23 '14 at 4:33
• Why don't you try assuming a solution of the form $ae^{i(kx-{\omega}t)}$?? maybe a bit more math, but slightly less subtle.. – GRrocks Oct 23 '14 at 4:41
• @GRrocks Are you saying assume solutions $x_1 = f(x, t)$ and $x_2 = f(x, t)$? If so, is this a thing that comes up often? I have always thought of $x_1$ and $x_2$ depending only on time. Also, by violating laws, I suppose you mean energy and momentum conservation? – danielunderwood Oct 23 '14 at 13:27

There's a sign error in your equations of motion. The Lagrangian of the system will be $$L=T-U= \frac{m}{2} \left( \dot{x_1}^2 + \dot{x_2}^2 \right)-\frac{k}{2} \left( L + x_1 - x_2 \right)^2$$ So the equation of motion for $x_1$ is: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1}-\frac{\partial L}{\partial x_1}=0 \\ m\ddot{x}_1+k( L + x_1 - x_2 )=0 \\ m\ddot{x}_1=-k( L + x_1 - x_2 )$$ and for $x_2$: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_2}-\frac{\partial L}{\partial x_2}=0 \\ m\ddot{x}_2-k( L + x_1 - x_2 )=0\\ m\ddot{x}_2=k( L + x_1 - x_2 )$$ So, your matrix $K$ should be $$K=k\pmatrix{1 &-1\\-1 & 1}$$ and the normal frequencies will be the expected $\omega_1=0$ and $\omega_2=\sqrt{\dfrac{2k}{m}}$.
• That was it. For some reason, I did $L = T + U$. I thought that I may have had a sign mistake, but couldn't figure out where it was. Thanks! – danielunderwood Oct 23 '14 at 13:25