Question

I'm studying harmonic motion by understanding it through potential energy. But there seems to be so many things going on, and I was hoping somebody could help me get things sorted out.

I'll post more details in a moment. But I'm wondering about the secular equation. What's the point, and how did we get it?

Answer 1

My textbook and professor used potential energy and the Lagrangian to get the secular equation. On the internet, I've only found information on finding the equations of motions using the forces.

The three general steps we use to get to this equation is
1. Find the Lagrangian of the system
2. Find the equations of motion from the Lagrangian
3. Write the components out

I'm good with (1) and (2). That was last chapter.
Then (3) is like creating a linear system of equations from this, with the columns each being a generalized coordinate. I get that, too.

I don't know why we take the determinant and set it to zero.
It's something with eigenvectors, eigtnvalues, eigenstuff, eigenfrequencies...
I know the fundamental relations of eigenvalues and eigenvectors for a given matrix.
\(A\vec x=\lambda\vec x\)
so
\({\rm det}(\lambda I-A)\vec x=0\)

But I don't understand what I'm actually doing...

Answer 2

There are so many matrices we're learning about... And a glossed over "normal coordinate system." And I know we do some things to find the normal modes, which are when the displacements of the masses or particles are either equal to each other or opposite to each other. As I understand, these are symmetric and antisymmetric, respectively.

Answer 3

So...
\(A\) is this matrix of equations of motion. Each row is an equation of motion and each column divides the terms by the particles in question.

This, multiplied by an eigenvector, is equivalent to a that eigenvector times the scalar eigenvalue. Is this eigenvalue an eigenfrequency?

I have no idea what's going on with some of this stuff!

Answer 4

Hm..
\(A\) would not the the matrix of equations of motion. That matrix of equations of motion would be \(A\vec x\). If that is true, \(\vec x\) is simply \(\begin{bmatrix} x_1\\x_2\\ x_3\end{bmatrix}=\vec x\)

So now I wonder what the eigenvalue would signify.

Answer 5

\(\begin{bmatrix}\\\\\\
eqns.& of& motion\\
without && coordinates\\\\\\
\end{bmatrix}
\vec x\\
=\begin{bmatrix}\\\\\\
eqns.& of&\ motion\\\\
\end{bmatrix}\\
=\lambda \vec x\\
=\lambda \begin{bmatrix}\\\\
coordinates\\\\
\end{bmatrix}\)

Answer 6

I am not familiar with the Lagrangian method. What I can explain is what happens when you solve the problem using Newton's equations.

The acceleration of every oscillator will obey a coupled differential equation.
When you inject solutions of identical angular frequency ω and (different) amplitudes Ai, you will get a linear algebraic system.

For this system to have a non zero (i.e. all Ai = 0) solution, the determinant must be 0.
The eigenvalues will lead you to all the ω² possible. These lead to the angular frequencies the normal modes of the oscillating system.

Any motion is a linear combination of these modes.

Answer 7

Okay, thanks! A lot of that was helpful! :)

I've read a little about using that method to determine \(\omega^2\), and it is neat. But I'll have to use the Lagrangian.

Wikipedia suggests that the eigenfrequencies are the normal frequencies. So, the \(\omega^2\) values will be the eigenvalues in that problem. All because we look for non-zero amplitudes. And the frequencies can then be derived by \(\sqrt{\left|\omega^2\right|}\), or just \(\sqrt{\omega^2}\) if \(\omega \ge0\). And if \(\omega=0\) we have \(A=0\), so we wouldn't consider any \(\omega=0\), and it probably wouldn't come from the linear algebraic system you mentioned.

Thank you for your input, @Vincent-Lyon.Fr ! I have to go now. But thanks for looking!

Answer 8

Looking and helping :)

Answer 9

You're welcome.