Coupled oscillators and stability of equilibrium points

Question

My question is about parts (e) and (f). I have found the matrix to equation of motion to be $\frac{d}{dt}\begin{bmatrix} x_1 \\ x_2 \\ p_1 \\ p_2\end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & \frac{1}{2} \\ -400 & 500 & 0 & 0 \\ 500 & -\frac{1900}{3} & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ p_1 \\ p_2\end{bmatrix}$.

For part (e), the eigenvalues, computed from Mathematica, to be $\lambda_1, \lambda_2 = -2.97523 \times 10^{-16} \pm 26.727i$ and $\lambda_3, \lambda_4 = -1.68896 \times 10^{-15} \pm 1.52747i$. I will denote their corresponding eigenvectors as $v_1, v_2, v_3, v_4$. Let $A = \begin{bmatrix} v_1 & v_2 & v_3 & v_4\end{bmatrix}$.

Is it true then that since all the real parts of the eigenvalues are negative, the system will always tend to the equilibrium $x_1 = x_2 = p_1 = p_2 = 0$, no matter the initial set of conditions for $x_1, x_2, p_1, p_2$?

Even so, I take it that we want the system to tend to the equilibrium at 0 quickly, which would mean that in the general solution to the equation of motion $\begin{bmatrix} x_1 \\ x_2 \\ p_1 \\ p_2\end{bmatrix} = A\begin{bmatrix} e^{\lambda_{1}t} & 0 & 0 & 0 \\ 0 & e^{\lambda_{2}t} & 0 & 0 \\ 0 & 0 & e^{\lambda_{3}t} & 0 \\ 0 & 0 & 0 & e^{\lambda_{4}t} \end{bmatrix}A^{-1}\begin{bmatrix} x_{1, 0} \\ x_{2, 0} \\ p_{1, 0} \\ p_{2, 0}\end{bmatrix} = c_{1}e^{\lambda_{1}t}v_1 + c_{2}e^{\lambda_{2}t}v_2 + c_{3}e^{\lambda_{3}t}v_3 + c_{4}e^{\lambda_{4}t}v_4$, where $\begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4\end{bmatrix} = A^{-1}\begin{bmatrix} x_{1, 0} \\ x_{2, 0} \\ p_{1, 0} \\ p_{2, 0}\end{bmatrix}$, I need to choose $c_1, c_2, c_3, c_4$ (and therefore, $x_{1, 0}, x_{2, 0}, p_{1, 0}, p_{2, 0}$) such that I can somehow make the complex part of $e^{\lambda_{n}t$, which is responsible for oscillation, have smaller magnitude and the real part more negative. Is this reasoning correct? How do I do this?

As for part (f), I have written my matrix $M$ as $\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & \frac{1}{2} \\ 100-k & k & 0 & 0 \\ k & -\frac{400}{3} - k & 0 & 0 \end{bmatrix}$. I think that for the system to be stable, the real parts of all the eigenvalues need to be less than or equal to zero. As such, I need to find the smallest k such that all real parts of the eigenvalues of $M$ are less than or equal to zero. Is it possible to do this manually or do I have to use numerical software like Mathematica?

Sorry for the long post, and any help or hints or corrections to my reasoning would be appreciated. Thanks!

Answer 1

Without damping, eigenvalues are pure imaginary, representing periodic oscillations of the free response.

The real part of the eigenvalues you get from Mathematica are zero, except for numerical errors (it's likely they're slightly negative because of the algorithms used in the numerical computation of eigenvalues, that needs "stable eigenvalues" to converge).

Point (e). Thus, the answer to point (e) is that there is no such a condition.

Point (f). For the answer to point (f), you're right and you can easily do it by hand, either exploiting the block structure of the matrix $M$, or evaluating eigenvalues of the $2^{nd}$ order system $$\tilde{M} \ddot{z} + \tilde{K} z = 0$$, being $\tilde{M}$ and $\tilde{K}$ mass and stiffness 2x2 matrices of the system. You just need to solve the generalized eigenvalue problem

$$[s^2 \tilde{M} + \tilde{K}]\hat{z} = 0 \ .$$

You can evaluate the eigenvalues by hand, evaluating the determinant of the matrix, setting it to zero to find the eigenvalues$^2$ as a function of $k$ and check the condition to avoid unstable eigenvalues.

You should look for a bifurcation from a pair of pure imaginary eigenvalues$^2$, to real eigenvalues.

Answer 2

starting with the linearized EOM's

$$m_1\,l_1^2\,\ddot\phi_1+m_1\,g\,l_1\phi_1-k\,(x_2-x_1)=0\\ m_2\,l_2^2\,\ddot\phi_2+m_2\,g\,l_2\phi_2+k\,(x_2-x_1)=0$$ where $$x_1=l_1\,\phi_1\quad,x_2=l_2\,\phi_2$$

from here you obtain

$$\mathbf{\dot{y}}=\mathbf M\,\mathbf y$$

where $$ y_1=\phi_1~,y_2=\phi_2~,y_3=\dot\phi_1 ~,y_4=\dot\phi_2$$

$$ \begin{align*} &\mathbf{M}= \left[ \begin {array}{cccc} 0&0&1&0\\0&0&0&1 \\-{\frac {m_{{1}}g+k}{l_{{1}}m_{{1}}}}&{\frac {kl_ {{2}}}{m_{{1}}{l_{{1}}}^{2}}}&0&0\\{\frac {kl_{{1}} }{m_{{2}}{l_{{2}}}^{2}}}&-{\frac {m_{{2}}g+k}{l_{{2}}m_{{2}}}}&0&0 \end {array} \right] \end{align*} $$

the eigenvalues of $~\mathbf M~$ are

$$\mathbf \lambda= \left[ \begin {array}{c} - 8.658454123\,i\\ 8.658454123\,i\\ - 82.20927265\,i \\ 82.20927265\,i\end {array} \right] $$

thus, the real parts is zero and the imaginary parts are conjugate complex as it should be for this system.

I need to find the smallest k such that all real parts of the eigenvalues of M are less than or equal to zero

the real parts is always zero

I used this data

$$[m_{{1}}= 1.0,m_{{2}}= 2.0,l_{{1}}= 0.10,l_{{2}}= 0.15 ,k= 500.0,g= 10.0] $$