Normal modes and system's energy

[Tosh5457]

Hi, why does the energy of the system equals the sum of the energy of the modes? The book I'm reading only states it, it doesn't prove it.

 

[marcusl]

The motion/excitation/configuration of a system, whatever it is (you don't say), can always be expressed as a sum of normal modes. An example: lift a guitar string at one point so the string displacement is triangular, and let go. Subsequent motion is extremely complicated, but it can be expressed as a sum of sinusoidal displacements at the fundamental and harmonic frequencies that are relatively easy to predict from a Fourier decomposition of the initial triangular excitation. Since the normal mode expansion describes the "actual" displacement, the sum of its energies equals the total energy of the system.

 

[AlephZero]

The matiematical reason is that the mode shapes are orthogonal. If $x_i$ amd $x_j$ are two different modes ($i \ne j$), then $x_i^TMx_j = 0$ and $x_i^TKx_j = 0$ where $M$ and $K$ are the system mass and stiffness matrices.

You can express any motion of the system as a linear combination of all the modes, i.e.
$$x = \sum_i a_i x_i.$$ So the total strain energy of the system is
$$x^T K x/2 = (\sum_i a_ix_i)^T K (\sum_j a_j x_j)/2
= \sum_i\sum_j (a_ia_jx_i^T K x_j)/2 = \sum_i (a^2_i x_i^T K x_i)/2$$because the only non-zero terms are when $i = j$. The same is true for the kinetic energy.

The math proof that the modes are orthogonal requires quite a bit of linear algebra, and may be just assumed, or demonstrated by a numerical example, in a first course in dynamics. For practical engineering work, knowing the result is true is a lot more important than knowing how to prove it!

 

[AlephZero]

The math proof that the modes are orthogonal requires quite a bit of linear algebra...

It's easy to show this for the special case of two modes with different frequencies, if you assume the mass and stiffness matrices are symmetric. (None of those assumptions are necessary, but the proof without them is much harder).

For the two modes we have $Mx_i + \omega_i^2Kx_i = 0$ and $Mx_j + \omega_j^2Kx_j = 0$.

Multiply the first equation by $x_j^T$ and the second by $x_i^T$:
$x_j^TMx_i + \omega_i^2x_j^TKx_i = 0$ and $x_i^TMx_j + \omega_j^2x_i^TKx_j = 0$.
If M and K are symmetric, $x_j^TMx_i = x_i^TMx_j$ and $x_j^TKx_i = x_i^TKx_j$.
So by subtracting the two equations we get $(\omega_i^2 - \omega_j^2)x_i^TKx_j =0$.

If the two frequencies are different, this means that $x_i^TKx_j = 0$, and back substituting, $x_i^TMx_j = 0$.