Coupled Oscillators
What happens when you have a number masses coupled to each other with springs? Motivation:
- What happens when you have a number of different "atoms" coupled to each other with "bonds"?--This is a model for thinking about molecules.
- This is also a model of a particle coupled to its "environment". (In the simplest possible model, the "environment" consists of just one other particle :->
Finding an exact solution for the positions of the two masses, $x_1(t)$ and $x_2(t)$, will involve:
- Finding the equations of motion. But yikes, these equations will be coupled differential equations!
- We'll write the equations of motion in terms of matrices and column vectors.
- We'll try to find "pure" oscillatory solutions:
- All the masses oscillating with the same frequency
- Each mass has its own constant (not time dependent) amplitude.
- Solving an "Eigenvalue problem" for the matrix equation to find the matrix' "eigenvectors" = normal modes.
- We'll find that we can express any solution in terms of a sum (superposition) of the normal modes. In this sense, the normal modes are something like the individual fourier components of a fourier series.
2 masses, 3 springs: Equations of motion
Consider the symmetrical arrangement below of 2 frictionless carts of of the same mass, $m$, connected to each other and surrounding walls with 3 springs in a symmetrical arrangement with two different spring constants.
When the carts are at rest, the horizontal coordinates of their CMs are $x_1=0$ and $x_2=0$.
[We can say that the springs are also at their equilibrium lengths when $x_1=x_2=0$. But (Problem 11.1) it turns out this is not a necessary assumption.]
Now, I'm following Matthew Schwarz' development
Left cart: Away from equilibrium, it looks like the spring on the left would be exerting a force on the left mass of $-k_1x_1$. That is, if $x_1\gt 0$ the spring is stretched out and pulling the cart back (negative) to the left.Spring 2's force, which depends on the positions of both particles, is $k_2(x_2-x_1)$: $$\begin{align} m\ddot{x}_1 =& -k_1x_1 + k_2(x_2-x_1) \nonumber\\ =& -(k_1+k_2)x_1 + k_2x_2. \label{EOM1}\end{align}$$
Right cart: By the same kind of analysis, if the spring at right is compressed, it's exerting a force $-k_1x_2$ to the left, and the middle spring force depends on the difference difference of the coordinates: $$\begin{align} m\ddot{x}_2 =& -k_2(x_2-x_1) - k_1x_2 \nonumber\\ =&k_2x_1-(k_1+k_2)x_2. \label{EOM2}\end{align}$$
2 solutions or 2 "normal" modes
If we add Eq $\ref{EOM1}$ and Eq $\ref{EOM2}$ together: $$m(\ddot x_1+\ddot x_2)= -k_1(x_1+x_2)$$ If we define $y(t)\equiv x_1(t)+x_2(t)$, we can readily see that $\ddot{y} = \ddot{x}_1+\ddot{x}_2$ and we can re-arrange the equation above as $$\ddot y = -\frac{k_1}m y$$ Our spring equation! Writing the solution as a cosine function with a phase shift: $$y(t)=x_1(t)+x_2(t)=A_s\cos(\omega_st+\delta_s);\ \text{where } \omega_s=\sqrt{ \frac{k_1}m}.$$
A second solution is needed if we consider Eq $\ref{EOM1}$ - Eq $\ref{EOM2}$: $$m(\ddot x_1-\ddot x_2)= (-k_1-2k_2)(x_1-x_2).$$ The solution is $$x_1(t)-x_2(t)=A_f\cos(\omega_ft+\delta_f);\ \text{where } \omega_f=\sqrt{ \frac{k_1+2k_2}m}.$$ The reason for the subscripts is that the second solution has a faster angular frequency, $\omega_f$, than the slower first solution.
Since we were solving 2 differential equations, each of 2nd order, the general solution should have $2\times 2=4$ adjustable constants, which are $A_s,\ \delta_s,\ A_f,$ and $\delta_f$. So, all possible solutions should be linear combinations of the fast and slow solutions. We can add and subtract these two "modes" to find these solutions for $x_1(t)$ and $x_2(t)$: $$x_1(t)=\frac 12\Big(A_s\cos(\omega_st+\delta_s) +A_f\cos(\omega_ft+\delta_f)\Big)$$ $$x_2(t)=\frac 12\Big(A_s\cos(\omega_st+\delta_s) -A_f\cos(\omega_ft+\delta_f)\Big)$$
If we release both carts from rest, that is $\dot x_1(0)=0=\dot x_2(0)$, you can find that this implies that $\delta_s=0=\delta_f$. Now, pulling both carts to the right by the same amount, i.e. $x_1(0)=x_2(0)$, and then releasing them from rest with 0 initial speed implies that $A_f=0$. So both the carts will oscillate back and forth in tandem with angular frequency $\omega_s$. This is the lower frequency normal mode. It is symmetric.
The two particles swing back and forth with a constant separation between them.
Conversely releasing both carts from rest, but with $x_1(0)=-x_2(0)$ will imply that $A_s=0$, and both carts oscillate with the higher frequency, such that they alternately approach and retreat from each other. This is the anti-symmetric normal mode.
This is the second normal mode.
The two particles swing exactly opposite each other at a higher frequency $\omega_f > \omega_s$.
But for an arbitrary solution, e.g. for $A_s=1$, $A_f=0.2873$, $\delta_s=0$, $\delta_f=0.455$ (pictured below), neither cart appears to be moving with a unique frequency or period:
$x_1(t)$