Infinite coupled oscillators

Consider an infinite system of masses $m$ each distance $a$ apart connected by springs with constant $k$ . We want to find the position of each mass over time.

The equation of motion for block $j$ is

\ddot x_j = \omega_0^2 (x_{j-1} -2x_j +x_{j+1}).

Where $\omega_0^2 = \frac km$ .

Since the system is infinite it has translational symmetry. Substituting $\tilde{j} = j+1$ results in the same physical system. We can write down a symmetry matrix representing this translation:

\begin{align*} S = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 1 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \end{bmatrix}. \end{align*}

There are zeros down the main diagonal and $1$ s down the adjacent diagonal (it doesn’t matter which, but in this case we have chosen it such that $\tilde j = j+1$ .)

Now we guess a solution $x_j = A_j e^{i \omega t}$ . By symmetry we can relate the amplitudes to say $S \mathbf A = \beta \mathbf A$ where

\mathbf A = \begin{pmatrix} \vdots \\ A_{j-1} \\ A_j \\ A_{j+1} \\ \vdots \end{pmatrix}.

This gives us

\begin{align*} A_j &= \beta A_{j+1} \\ A_{j+1} &= \beta A_{j+2} \\ A_j &\propto \beta^j. \end{align*}

We can’t have the amplitudes going to positive or negative or infinity since that wouldn’t make sense physically, so we know that $\lvert \beta \rvert = 1$ so that repeatedly multiplying by $\beta$ keeps the magnitude of the amplitudes constant. This gives us

A_j = e^{i \mathcal K j a}.

We choose $\beta = e^{i \mathcal K a}$ . Here $\mathcal K$ is called the wave number. We choose this form of $\beta$ because it will make our result more useful later (no good justification for this except that it makes $\mathcal K$ physically useful).

We can plug in our initial guess into the equation of motion to find

\begin{align*} -\omega^2 A_j &= \omega_0^2 (A_{j-1} -2A_j + A_{j+1}) \\ \omega^2 &= \omega_0^2 \left(-\frac{A_{j-1}}{A_j} +2 + \frac{A_{j+1}}{A_j}\right) \\ \omega^2 &= \omega_0^2 \left(-\frac{1}{\beta} +2 - \beta \right). \end{align*}

Plugging in $\beta$ we get

\begin{align*} \omega^2 &= \omega_0^2 (-e^{-i \mathcal K a} + 2 - e^{i \mathcal K a}) \\ &= 2 \omega_0^2 (1-\cos (\mathcal K a)). \end{align*}

This relation between $\omega$ , $\omega_0$ , and $\mathcal K$ is called the dispersion relation.

We plug the value for $A_j$ back into our guess to find

\begin{align*} x_j &= A_j e^{i \omega t} \\ &= (c_1 e^{i \mathcal K j a} + c_2 e^{-i \mathcal K j a}) e^{i \omega t} \\ &= C \cos(\mathcal K j a + \alpha) \cos(\omega t + \beta). \end{align*}

This is a normal mode solution for a mass in a system of infinite coupled oscillators. Note that it is the product of two sinusoids, one oscillating with time and one with space ( $ja$ is just an initial $x$ coordinate). The general solution will be a superposition of normal mode solutions, where each $\omega$ and $\mathcal K$ pair defines a normal mode.

Note

We used the example of masses moving horizontally connected by springs, but the solution we found is totally general. Depending on the system you will get a different $\omega_0$ .

Applying the infinite solution to finite systems

Consider a system like shown below, with $n$ masses connected by springs between two stationary walls.

How can we apply the solution for infinite masses to this (finite) system? Imagine this system is just a part of an infinite system, with masses continuing forever in both directions.

Recall our general solution is $x_j = C \cos(\mathcal K j a + \alpha) \cos(\omega t + \phi)$ . Having a wall before block $1$ is equivalent to having a block $0$ where $x_0 = 0$ , as pictured above.

x_0 = C \cos(\alpha) \cos(\omega t + \phi) = 0.

The only way we can make $x_0 = 0$ and still have an interesting solution (i.e. $\omega \ne 0$ ) is if we set $\alpha$ to be a multiple of $\frac\pi 2$ . We choose $\alpha = -\frac \pi 2$ .

We can do the same thing for the right wall. We set $x_{n+1}=0$ . Again the first cosine must be zero, and the only free parameter is $\mathcal K$ .

\begin{align*} x_{n+1} = 0 &= C \cos(\mathcal K (n+1) a - \frac\pi 2) \cos(\omega t + \phi) \\ m \pi &= \mathcal K (n+1) a \\ \mathcal K &= \frac{m \pi}{a(n+1)}, \quad m \in \mathbb Z. \end{align*}

String attached to a ring on a pole

Let’s say now we have a system of masses connected by string constrained to move vertically. One end of the system is connected to a massless ring on a pole such that it exerts no force in the vertical direction on the connected mass. Assume the string has a constant tension $T$ .

In this case our equation of motion (assuming small displacements) becomes $\ddot y_j = \frac{T}{ma} (y_{j-1} - 2y_j +y_{j+1})$ , so $\omega_0^2 = \frac{T}{ma}.$

Since ring is free to move vertically on the pole without friction, we know that the $y$ displacement of the ring will always be equal to the displacement of the block attached to it. As with the system attached to the wall, we will treat this system as a part of an infinite system, where blocks $0$ and $n+1$ have special behavior.

The algebra will be easier in complex form, so we rewrite $y_j = (Ae^{i \mathcal K j} + Be^{-i \mathcal K j}) \cos(\omega t + \phi)$ .

\begin{align*} y_0 = y_1 &\implies& Ae^0 + Be^0 &= Ae^{i \mathcal K a} + Be^{-i\mathcal K a} \\ && A(1-e^{i \mathcal K a}) &= B(e^{-i \mathcal K a} - 1). \tag{1} \\ y_n = y_{n+1} &\implies& Ae^{i \mathcal K n a} + Be^{-i \mathcal Kna} &= Ae^{i \mathcal K(n+1)a} + Be^{-i\mathcal K(n+1)a} \\ && Ae^{i\mathcal Kna}(1-e^{i \mathcal K a}) &= Be^{-i\mathcal Kna}(e^{-i \mathcal K a} - 1). \tag{2} \end{align*}

We divide equation (1) by equation (2) to find

\begin{align*} \frac{(1)}{(2)}:\quad e^{i \mathcal K na} &= e^{-i \mathcal Kna} \\ 1 &= e^{i 2 \mathcal K na} \\ 2 \mathcal K na &= 2 m \pi \\ \mathcal K &= \frac{m\pi}{na}. \end{align*}

Now we return to equation (1):

\begin{align*} A(1-e^{i\mathcal Ka}) &= B(e^{-i\mathcal Ka}-1) \\ \text{Let } Q &= e^{i\mathcal Ka} \\ A(1-Q) &= B(\frac{1}{Q}-1) \\ B &= \frac{A(1-Q)}{\frac1Q-1} \cdot \frac QQ \\ &= AQ = A e^{i \mathcal K a}. \end{align*}

Which gives us our final solution

y_j = A(e^{ij\mathcal Ka} + e^{-i(j-1) \mathcal Ka})\cos(\omega t + \phi)

where $\mathcal K = \frac{m\pi}{na}$ and $\omega^2 = 2 \omega_0^2 (1-\cos(\mathcal K a))$ .