# Mass-spring-damper system with damping eigenvalues and eigenvectors

I have a question about finding the eigenvalues and eigenvectors of the fixed point at equilibrium for this system.

The equation given is the follows: $\ddot{x}=\frac{-c}{m}\dot x - \frac{K}{m} x$

where $m$ is the mass, $k$ the stiffness and $c$ the damping.

I have transformed into a state space which looks something like this,

$x_1=x$ and $x_2=\dot{x}$

$$\begin{bmatrix} \dot{x_1}\\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} 0 & 1\\ \frac{-K}{m} &\frac{-c}{m} \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix}$$

So right now I'm stuck on getting the eigenvalue from this... I did some computation and got something like $\lambda^2+\frac{c}{m}\lambda+\frac{K}{m}$ but now I'm stuck. I tried doing the quadratic formula from here and got something like this,

$\frac{\frac{-c}{m}\pm \sqrt{(\frac{c}{m})^2-\frac{4k}{m}}}{2}$

If anyone can help me out, with possibly getting the eigenvalues and vectors from this, it would be greatly appreciated.

Thank you!

## 1 Answer

You are practically done! Now to get the eigenvalues you have to guess the sign of the term in the square root: if it's negative you'll have imaginary eigenvalues (oscillations), if it's positive you'll have real eigenvalues (when the damping $c$ is too big the system goes to the equilibrium without oscillating). Obviously if you had the values of the parameters you could check the sign directly...

To get the eigenvectors you can follow any standard procedure, just google "how to find eigenvectors"!

• I know how to find the eigenvectors but what I keep getting out of the calculation is very messy terms, and nothing is canceling out. In other words, I would like to see how someone is solving this, because I am stuck and get any solutions out. – lnbmoco Oct 23 '17 at 23:52