Coupled Oscillators - matrix methods

What follows is parallel to section 11.2 of Taylor's textbook. However, I'm re-arranging his formalism so as to allow you to use the matrix functions in Sagemath or Octave more easily to solve coupled problems.

Some linear algebra...

Two column vectors are equal if each of their components are equal...just like our more familiar vectors. For example $$\begincv 1 \\ 7\endcv = \begincv 1 \\ 7\endcv $$ $$\begincv 1 \\ 7\endcv \neq \begincv 7 \\ 1\endcv $$ $$\begincv 1 \\ 7\endcv = \begincv 4-3 \\ 4+3\endcv $$

And we can use familiar vector notation, e.g. for two dimensional vectors: $$\myv a \equiv \begincv a_x \\a_y\endcv.$$ And if two vectors are equal: $$\myv a = \myv b \Rightarrow \begincv a_x \\a_y\endcv = \begincv b_x \\b_y\endcv.$$

Coupled oscillators

Our two (coupled) equations of motion were... $$\begin{align} m\ddot{x}_1 =& -(k_1+k_2)x_1 + k_2x_2. \nonumber\end{align}$$ $$\begin{align} m\ddot{x}_2 =&k_2x_1-(k_1+k_2)x_2. \nonumber\end{align}$$ Dividing both equations by $m$: $$\begin{align} \ddot{x}_1 =& -\frac{k_1+k_2}m x_1 + \frac{k_2}m x_2. \label{EOM1}\end{align}$$ $$\begin{align} \ddot{x}_2 =&\frac{k_2}m x_1-\frac{k_1+k_2}mx_2. \label{EOM2}\end{align}$$

We can write our 2 equations of motion, Eq \ref{EOM1} and Eq \ref{EOM2} as an equality between two column vectors: $$\begin{align} \begincv \ddot x_1\\ \ddot x_2 \endcv =& \begincv -\frac{k_1+k_2}m x_1 + \frac{k_2}m x_2 \\ \frac{k_2}m x_1-\frac{k_2+k_1}m x_2 \endcv \label{2c} \end{align}$$

More linear algebra: multiplication and matrices

Scalar multiplication works just like it does with regular vectors: $$5\cdot\begincv \frac25 \\ 2\endcv = \begincv 2\\10\endcv $$ You can multiply a 2 element row vector by a 2 element column vector. It works like the dot product: The dot product of two vectors is a single scalar number: $$\begincv 1 & 2\endcv\cdot\begincv 3\\5\endcv= 1\cdot 3+2\cdot5=13.$$

A new thing is that you can multiply a $2\times 2$ matrix by a 2-element column vector in a dot product'ish way. The dot product of the top (bottom) row of the matrix with the column vector is the top (bottom) element of a new column vector $$\begincv 1 & 2\\ 10 & 20\endcv\cdot\begincv 3\\5\endcv= \begincv 1\cdot3+2\cdot 5\\ 10\cdot3+20\cdot 5 \endcv= \begincv 13\\130\endcv.$$

[In case you need a refresher on or intro to matrix multiplication...]

The right hand side of our column vector Eq $\ref{c2}$ can be written as the product of a matrix with a column vector: $$\begin{align} \begincv -\frac{k_1+k_2}m x_1 + \frac{k_2}m x_2 \\ \frac{k_2}m x_1-\frac{k_2+k_1}m x_2 \endcv =& \begincv -\frac{k_1+k_2}m & \frac{k_2}m \\ \frac{k_2}m & -\frac{k_2+k_1}m \endcv \begincv x_1\\x_2 \endcv = \mym{W}\myv x \end{align}$$ where $\myv{x}$ is a two-element column vector $\myv{x} \equiv \begincv x_1\\x_2 \endcv , $

and $\mym W$ is a $2 \times 2$ matrix: $\mym{W} \equiv \begincv -\frac{k_1+k_2}m & +\frac{k_2}m \\ +\frac{k_2}m & -\frac{k_2+k_1}m \endcv $.

Look for pure oscillation solutions

The video suggests that there are some solutions such that $x_1(t)$ and $x_2(t)$ are moving as simple harmonic oscillators, and both have the *same* frequency.

That is... $$x_1(t) = A_1\cos(\omega t+\delta_1);\ \ x_2(t) = A_2\cos(\omega t+\delta_2),$$

Now, following Taylor, we'll think of $x_1$ as the real part of the complex function $$z_1(t)=A_1e^{i(\omega t+\delta_1)}=(A_1e^{i\delta_1})e^{i\omega t}\equiv a_1e^{i\omega t},$$ where $A_1$ is a real scalar constant and $a_1=A_1e^{i\delta}$ is an imaginary constant, defined such that $|a_1|=A_1$. Similarly, we'll define $z_2(t)=a_2e^{i\omega t}$. And this implies: $$\ddot z_1 = -\omega^2z_1;\ \ \ddot z_2 = -\omega^2z_2 $$ or $$ \begincv \ddot z_1 \\ \ddot z_2\endcv = -\omega^2\begincv z_1 \\z_2\endcv\\ $$ Plugging this in to our equation of motion, and using our matrix $\mym W$ we're looking to solve this (matrix) equation: $$\begineq \mym{W}\myv z =& -\omega^2\myv z\\ \mym{W} \begincv a_1e^{i\omega t}\\a_2e^{i\omega t}\endcv =&-\omega^2\begincv a_1e^{i\omega t}\\a_2e^{i\omega t}\endcv\\ e^{i\omega t} \mym W \begincv a_1\\a_2\endcv =&-\omega^2e^{i\omega t}\begincv a_1\\a_2\endcv\\ \mym W \myv a =&-\omega^2\myv a\\ \endeq $$

This has the form of an eigenvalue equation, $\mym M\myv v=\lambda\myv v$.

There is a trivial, but uninteresting solution $\myv a=\begincv 0\\0\endcv$.

We'll use a Very Important Result from linear algebra to find the more interesting solution: There is one or more non-trivial solutions if the determinant of this matrix vanishes: $$\det\left(\mym{M}-\lambda\mathbb{1}\right)=0.$$ $\mathbb{1}$ is the identity matrix, which for $2\times 2$ matrices is: $$\mathbb{1}\equiv \begincv 1 & 0\\0 &1\endcv.$$ This equation is called the Characteristic Equation. The useful thing about it is that we can solve it for the scalar $\lambda$ without having to worry about the unknown $\myv v$! Once we have $\lambda$, we'll use that to find $\myv v$.

The values of $\lambda$ which solve the characteristic equation are called eigenvalues of the matrix, or in our context of coupled oscillators, our scalar is $-\omega^2$ and we call $\omega$ a normal frequency. Let's do it!

Solving eigenvalue problems in CoCalc

Once you have the equations of motion arranged in matrix form, the calculation of the normal frequencies and normal modes can be carried out in CoCalc. Using the Sagemath kernel, 4 lines of codes are needed to calculate all of those frequencies and amplitude ratios:

The values returned by eigenvector_right() are the solutions to $\mym W \myv a=\lambda\myv a$, but our problem was $\mym W \myv a = -\omega^2\myv a$.

• The first eigenvalue is $\lambda=-(k_1+2k_2)/m$. So $\omega=\sqrt{-\lambda}=\sqrt{k_1+2k_2}$.
• The eigenvector for that first eigenvalue is $\begincv 1\\-1\endcv$. That is, $a_1$ and $a_2$ are in the ratio 1 to -1.
• The multiplicity of the first eigenvalue is 1.

Multiplicity?

It can happen that there are multiple eigenvectors with the same eigenvalue.

For example, if the multiplicity were 2, then 2 eigenvectors are returned, which together define a plane. Any vector in that plane (that is, any linear combination of the two eigenvectors) is a solution to the eigenvalue equation.

When the identity matrix, $\mathbb{1}$, multiplies any vector, you get the same vector. So all vectors are eigenvectors of the identity matrix.

Therefore, any multiple of the identity matrix in two dimensions should have a multiplicity of 2, and sure enough...

Problem in class

• Find the solution (CoCalc)
• How to verify? (and left vs right...)
• How to imagine?