# Solving Newton-Euler ODE as eigenvalue problem

I'm trying to make sense of some lecture slides which skip a lot of steps so I'm just deriving all the equations etc. myself.

I have rigid body system with the following FBDs

I am investigating its harmonic motion. With Newton-Euler I have derived the following state-space representation of the equations.

\begin{align} \label{eq:system} \begin{bmatrix} m & 0\\ 0 & I \end{bmatrix} \begin{bmatrix} \ddot{x}\\ \ddot{\phi} \end{bmatrix}+2k \begin{bmatrix} 2 & b-a\\ b-a & a^2+b^2 \end{bmatrix} \begin{bmatrix} x\\ \phi \end{bmatrix} = 0 \end{align} I have assumed the following forms for the solutions \begin{align} \label{eq:sol} \begin{split} x &= x_0 \cdot e^{i \omega t}\\ \phi &= \phi_0 \cdot e^{i \omega t} \end{split} \end{align} and \begin{align} \label{eq:dd_sol} \begin{split} \ddot{x} &= x_0 \left(-\omega^2 \cdot e^{i \omega t}\right)\\ \ddot{\phi} &= \phi_0 \left(-\omega^2 \cdot e^{i \omega t}\right). \end{split} \end{align}

which, by substitution, yield \begin{align} \label{eq:system2} \underbrace{ \left( - \omega^2 \begin{bmatrix} m & 0\\ 0 & I \end{bmatrix}+2k \begin{bmatrix} 2 & b-a\\ b-a & a^2+b^2 \end{bmatrix} \right)}_{\text{Y}} \underbrace{ \begin{bmatrix} x_0\\ \phi_0 \end{bmatrix}}_{\text{Z}} \cdot e^{i \omega t} = 0 \end{align}

This is where it gets funky. In the slides the exponential term is gone. I'm not quite sure why. The second funky part is where we want to solve the eigenvalue problem $(A-\lambda I)\mathbf{x} = 0$ and $A$ is then defined as \begin{align} A = 2k \begin{bmatrix} \frac{1}{m} & 0\\ 0 & \frac{1}{I} \end{bmatrix} \begin{bmatrix} 2 & b-a \\ b-a & a^2+b^2 \end{bmatrix} \end{align} with $\lambda = \omega^2$.

It seems that the terms between the brackets, Y, have been set equal to each other. Why? What I can infer from this is that Y is a non-trivial solution whereas, Z, has a trivial solution of 0.

Basically, I do not understand why the exponential term disappears and then how $A$ is defined to solve the eigenvalue problem.

• 1) You have equality of form $YZ \cdot e^{i \omega t} \equiv 0$, where $Y$,$Z$ are constants and this equality holds for all $t$. Since $e^{i \omega t}$ is never zero, you can divide by it and obtain just a system of linear equations. 2) This is a sort of trick. Matrix $Y$ looks very much like matrix for searching eigenvalues of some matrix -- that's because you have $\omega^2 \times {\rm something \; diagonal}$. If it were identity matrix, it would be truly eigenvalue problem. So we just correct the matrix by multiplying $(x_0, \phi_0)$ with inverse of this diagonal matrix. – Evgeny Jan 15 '17 at 15:29
• @Evgeny the first part makes perfect sense! However, you lost me on the second part with the correction... EDIT: Nevermind I see it now! – Ortix92 Jan 15 '17 at 20:05
• Well, It's great then if there are no questions :) btw, do you know how these amazing pictures were drawn? – Evgeny Jan 15 '17 at 22:59
• @Evgeny I drew them myself with a Wacom tablet :) – Ortix92 Jan 16 '17 at 10:59
• It's an off-topic, but these images are fantastic! They look like handwritten and computer typed at the same time :) – Evgeny Jan 17 '17 at 7:39