Solving a coupled ODE

Question

I am trying to solve a couple mechanical system and have ended up with the following equations: $$\underbrace{\begin{pmatrix} M & 0\\ 0 & m \end{pmatrix}}_{\mathbf{M}} \ddot{\mathbf{q}} + \underbrace{\begin{pmatrix} k_s & -k_s\\ -k_s & k_s+k_t \end{pmatrix}}_{\mathbf{K}} \mathbf{q}=0 $$ and I have been given that $\mathbf{q}$ is of the form: $$\mathbf{q}=\mathbf{v}_j\sin\omega_jt$$ I need to show how the system matrix $\mathbf{M}^{-1}\mathbf{K}$ can be used to determine $\mathbf{v}_j,\omega_j$ but I am not quite sure. What I have done so far is: $$\mathbf{M\ddot{q}}+\mathbf{Kq=0}\Rightarrow \mathbf{I\ddot{q}}+\mathbf{M}^{-1}\mathbf{K}\mathbf{q}=0$$ then from this I can make the substitution: $\mathbf{q}=\boldsymbol{\mu}e^{\lambda t}$ which I think would leave me with: $$\mathbf{I}\lambda^2+\mathbf{M}^{-1}\mathbf{K}=0$$ where $\lambda$ is the eigenvalues and $\boldsymbol{\mu}$ is the eigenvector. I am just not sure how to go from here and make the link to the $\sin$ functions, I would assume that $\lambda$ will come out to be imaginary but I do not see how.

Answer 1

From the given ansatz, we have $$\ddot{\boldsymbol{q}} = -\omega_j^2 \boldsymbol{q}$$

Using the hint given, $$I (-\omega_j^2) \boldsymbol{q} + M^{-1} K \boldsymbol{q} = \boldsymbol{0}$$

or $$(-\omega_j^2 I + M^{-1} K) \boldsymbol{q} = \boldsymbol{0}$$

Usually the trivial solution $\boldsymbol{q} = \boldsymbol{0}$ is not interesting. A necessary and sufficient condition to have non-trivial solution is

$$\det(-\omega_j^2 I + M^{-1} K) = 0$$

After you found the eigenvectors for the above matrix, you can take the appropriate linear combination to satisfy any initial conditions given.

Additional comments to uncouple the ODE in general:

If we assume that the masses in $M$ are positive, then the quantities $\sqrt{M}$ and $M^{-1}$ are well-defined. Consider the matrix $M^{-1/2}K M^{-1/2}$. It is symmetric because $K$ and $M$ are symmetric. Thus it is able to be orthogonally diagonalized. So we can write

$$M^{-1/2}K M^{-1/2} = S \Lambda S^T$$ where $\Lambda$ is diagonal and $S^T = S^{-1}$. Then observe that

$$M^{-1/2} S \Lambda S^T M^{1/2} = M^{-1} K$$

and

$$M^{-1/2} S S^T M^{1/2} = I$$

So plugging into your original equation, we get

$$\left(M^{-1/2} S S^T M^{1/2} \right)\ddot{\boldsymbol{q}} + \left(M^{-1/2} S \Lambda S^T M^{1/2} \right)\boldsymbol{q} = \boldsymbol{0}$$

which can be simplified by multiplying $S^T M^{1/2}$ on the left to get

$$S^T M^{1/2} \ddot{\boldsymbol{q}} + \Lambda S^T M^{1/2} \boldsymbol{q} = \boldsymbol{0}$$

Define new coordinates

$$\boldsymbol{u} = \Lambda S^T M^{1/2} \boldsymbol{q}$$

$\Lambda$, $S$ and $M$ all have no time-dependence. Therefore,

$$\ddot{\boldsymbol{u}} = \Lambda S^T M^{1/2} \ddot{\boldsymbol{q}}$$

and the final equation is

$$\ddot{\boldsymbol{u}} + \Lambda \boldsymbol{u} = \boldsymbol{0}$$

which you will undoubtedly recognize as simple harmonic motion equations that are uncoupled. (This is why the ansatz is given in that form.) The diagonal entries of $\Lambda$ are just the eigenvalues of $M^{-1} K$.

Answer 2

If you find that the eigenvalues of $M^{-1}K$ are negative, then $\omega$ is purely imaginary.