Vibrational Modes and Imaginary Frequencies of a Three Spring System

Question

This question is an extension of the one I posted a few days ago: Rigid Body and Two-Spring System and the Lagrangian.

I am attempting to find the vibrational modes and their frequencies of an airplane-suspension system by hand. The model is fairly simple, it is a plate supported by three springs with a COM at some location.

$k_1$ while both the back suspensions has spring constant $k_2$." />

The first picture is a top view of the system. The red dot represents the center of mass, and the black boxes represent the suspensions. The front suspension has spring constant $k_1$ while both the back suspensions has spring constant $k_2$. $d_1$ represents the distance from the front suspension to the COM while $d_2$ represents the horizontal distance from the rear suspension. $w$ represents the distance from either rear suspension to the longitudinal axis of the COM.

The second picture is a side view of the system. Notice that since the back two suspensions are aligned, only one rear suspension is visible. $x$ is the coordinate for the height of the center of mass above the ground. Let $x=0$ be the point where the springs touch the ground. Also notice, $θ$ is the angle of the pitch of the suspension.

The third picture is a front view of the system. Here $φ$ represents the roll angle of the aircraft.

Just from thinking about it, I would imagine there to be 3 modes: a translational mode in which the COM goes up and down, a pitching mode in which the COM stays put but $θ$ changes and the plane pitches up and down, and a roll mode in which the COM again stays put but $φ$ changes and the plane rolls along the longitudinal axis.

Here's what I have so far.

Kinetic Energies

Translational Kinetic Energy of the COM: $\frac{1}{2}m(\frac{dx}{dt})^2$

Rotational Kinetic Energy of Pitch about the COM: $\frac{1}{2}I_θ(\frac{dθ}{dt})^2$

Rotational Kinetic Energy of Roll about the COM: $\frac{1}{2}I_φ(\frac{dφ}{dt})^2$

Potential Energies GPE: $mgx$

Front Spring: $\frac{1}{2}k_1(x+d_1\tan(\theta)-L)^2$

the $x+d_1\tan(\theta)-L$ term comes from the fact that if $L$ is the equilibrium length of the spring, the spring is compressed by both the COM's position, $x$ and the pitching compression $d_1\tan(\theta)$

Back Springs:

$\frac{1}{2}k_2(x-d_2\tan(\theta)+w\tan(\phi)-L)^2$

AND

$\frac{1}{2}k_2(x-d_2\tan(\theta)-w\tan(\phi)-L)^2$

For the back springs, it is -$d_2\tan(\theta)$ because when $\theta$ is positive, the back springs become compressed. Similarly, the $w\tan(\phi)$ term in each represents the fact that the back springs are also compressed by the rolling of the aircraft.

Lagrangian and Equations of Motion

Putting this into a nice Lagrangian, the equations of motion become:

$m\frac{d^2x}{dt^2} = (-2k_2-k_1)x+(2d_2k_2-d_1k_1)\theta + 2Lk_2 + Lk_1 -gm$

$I_\theta\frac{d^2\theta}{dt^2} = (-2d_2^2k_2-d_1^2k_1)\theta + (2d_2k_2-d_1k_1)x - 2Ld_2k_2 + Ld_1k_1$

$I_\phi\frac{d^2\phi}{dt^2} = -2k_2w^2\phi$

Note: I used the small angle approximation to make things linear.

Determining Frequencies

Now, from what I remember, I will need to do:

$\det(\textbf{K}-\omega^2\textbf{M}) = 0$ where $\textbf{K}$ is the stiffness matrix and $\textbf{M}$ is the "mass" matrix.

I notice immediately, that the current equations of motion have a few constants in them: $−2Ld_2k_2+Ld_1k_1 - gm$ and $2Lk_2+Lk_1$. From what I understand, I can drop these when trying to find the frequencies, as they are constant forces and do not influence the natural frequencies.

Therefore:

$\begin{pmatrix}m&0&0\\\ 0&I_\theta&0\\\ 0&0&I_\phi\end{pmatrix} \begin{pmatrix} \frac{d^2x}{dt^2}\\\ \frac{d^2\theta}{dt^2} \\\ \frac{d^2\phi}{dt^2}\end{pmatrix} = \begin{pmatrix}(-2k_2-k_1)&(2d_2k_2-d_1k_1)&0 \\\ (2d_2k_2-d_1k_1)&(-2d_2^2k_2-d_1^2k_1)&0 \\\ 0&0&-2k_2w^2 \end{pmatrix} \begin{pmatrix}x\\\ \theta \\\ \phi \end{pmatrix}$

giving

$ \textbf{K} = \begin{pmatrix}(2k_2+k_1)&(-2d_2k_2+d_1k_1)&0 \\\ (-2d_2k_2+d_1k_1)&(2d_2^2k_2+d_1^2k_1)&0 \\\ 0&0&2k_2w^2 \end{pmatrix}$

Taking $\det(\textbf{K}-\omega^2\textbf{M}) = 0$ yields a polynomial equation of the form $x^6 + ax^4 + bx^2 + c = 0$ which indeed is analytically solvable, but whose solutions are quite nasty and won't post here. There are only two that are strictly real. The other four have the potential to be complex if the above constants and coefficients are specific values. However, it was my understanding that no matter what the values are for this characteristic equation, the roots should be real given that frequencies are real-valued.

So, should I be concerned with this? Otherwise, is my reasoning for deriving the frequencies correct? I am concerned because the model implies that the roll mode is not coupled with the either two modes while the pitching and dip modes are coupled together. I find this hard to believe.

Answer 1

$\def \b {\mathbf}$

this is the EOM's

$$\b M\,\ddot{\b{q}}+\b K\,\b q=\b 0\tag 1$$

the ansatz for the solution $~\b q(t)~$

$$\b q(t)=\b c\,e^{i\,\omega\,t}$$

thus from equation (1) you obtain

$$\det\left(-\b\omega^2\,\b M+\b K\right)=0$$

the eigenvalues $~\b\omega^2~$ must be all real and positive, the frequency $~\omega=\pm \sqrt{\b\omega^2}$

---

$$\b q=\begin{bmatrix} x \\ \phi_y \\ \phi_z \\ \end{bmatrix}$$

$$\b M=\begin{bmatrix} m & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \\ \end{bmatrix} $$

$$\b K= \left[ \begin {array}{ccc} k_{{1}}+2\,k_{{2}}&-k_{{1}}d_{{1}}+2\,k_{{ 2}}d_{{2}}&0\\ -k_{{1}}d_{{1}}+2\,k_{{2}}d_{{2}}&2\, k_{{2}}{d_{{2}}}^{2}+k_{{1}}{d_{{1}}}^{2}&0\\ 0&0&2 \,k_{{2}}{w}^{2}\end {array} \right] $$