# I Find the practical resonance frequencies in a system of linear differential equations

#### Kostas1335

Summary
I would like to know what is the equation upon which I can use to determine the practical resonance frequencies in a system of second order, linear differential equations.
Hi all,

I would like to know what is the equation upon which I can use to determine the practical resonance frequencies in a system of second order, linear differential equations.

First some definitions: What I mean by practical resonance frequencies, is the frequencies that a second order linear system of differential equations with damping, exhibits in its solution. In other words the practical resonance frequency approaches the natural frequency, if the damping on the system approaches zero.

Now with that clear, let's look a the math:

We have the following differential equation with its coefficients for mass $\mathbf{M}$ and stiffness $\mathbf{K}$ being $n\times n$ matrices:
$$\mathbf{M} \ddot{x} + \mathbf{K} x = \mathbf{F_0}$$
From the above, we know that the system's natural frequency for every mass element in the mass matrix is the square root of the generalised eigenvalues:
$$-\mathbf{M}^{-1} \mathbf{K} x= \omega_n^2 x$$
where it is visible that the eigenvalues $\lambda = \omega_n^2$. Consequently we solve to find $\lambda$ in the equation above, and take its square root to determine the natural frequencies.

My question lies on how does the above eigenvalue problem gets expressed in order to find the practical frequency this time, for the case where damping is taken into account. So the system of equations would now be:
$$\mathbf{M} \ddot{x} + \mathbf{C} \dot{x} + \mathbf{K} x = \mathbf{F_0}$$
I know that in the case were we are not dealing with a system of equations, but with coefficients with one value only, we can find the practical frequency as:
$$\omega_r = \sqrt{ \frac{k}{m} - \frac{c^2}{2m^2} }$$
However I cannot seem to be able to understand how to generalise the above for the case for when I have a system of equations.

Kind Regards,

Kostas.

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