Solving for the resonant frequencies of a mass-spring system

Question

Given 3 masses ($M_1, M_2, M_3$) connected linearly with 2 springs ($K_1,K_2$), let $X_1,X_2,X_3$ be the displacement of the masses relative to their rest positions. Using the Lagrangian I can write down the following equations of motion:

$$M_1\ddot{X_1}+K_1X_1-K_1X_2=0$$

$$M_2\ddot{X_2}-K_1X_1+(K_1+K_2)X_2-K_2X_3=0$$

$$M_3\ddot{X_3}-K_2X_2+K_2X_3=0$$

I then rewrite this in matrix form using the following matrices:

$$M:=\begin{bmatrix}M_1 & 0 & 0\\0 & M_2 & 0 \\ 0 & 0 & M_3\end{bmatrix} $$

$$K:=\begin{bmatrix}K_1 & -K_1 & 0\\-K_1 & M_1+M_2 & -K_2 \\ 0 & -K_2 & K_2\end{bmatrix}$$

$$M\ddot{\vec{X}}+ K\vec X = 0$$

The site says to use the following equation: However I believe it is wrong.

$$K=\lambda M $$

Solving for the eigenvaleus or $\lambda$:

$$ M_1= M_2 =M_3=4Kg$$

$$ K_1 = 3 \frac Nm, K_2 = 7 \frac Nm $$

$$ \lambda = \begin{bmatrix}.75 & -.75 & 0\\-.75 & 2.5 & -1.75 \\ 0 & -1.75 & 1.75\end{bmatrix}$$

So the resonant frequencies are: .75, 2.5, and 1.75 $\frac{m}{s^2}$? I don't think so $\frac{m}{s^2}$ = acceleration!

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John from Physics.SE answered this questions, but moderators there deleted his beautiful answer. He substituted $X(t)=Ae^{\omega stuff}$ into the K matrix and then solved for $\omega$, or the resonant frequencies.

Answer 1

You have some confusion about units. You wind up with the generalized eigenvalue problem

$$(K-\omega^2 M)X=0$$

where $\omega^2$ is the generalized eigenvalue. This is NOT the problem $K=\omega^2M$; this would assume that $\omega^2$ is a generalized eigenvalue of full multiplicity. Also, for this to make sense, $K$ and $\omega^2 M$ must have the same units. Since $M$ has units of mass and $K$ has units of force/length=mass/time^2, that implies $\omega^2$ has units of time^{-2}, as expected.

One can convert this generalized problem into an ordinary eigenvalue problem by multiplying through on the left by $M^{-1}$, since $M$ is always invertible. The result is the ordinary eigenvalue problem

$$M^{-1}KX=\omega^2X$$

for the matrix $M^{-1}K$.

Answer 2

How would you justify the equation: $K = \lambda M$?

I would make the ansatz: $\vec X(t) = \vec A e^{-i\omega t}$ then you get the algebraic equation: $(K - \omega^2M)\vec A = 0$. In order for this to have a non trivial solution for $\vec A$ we require that $\det (K - \omega^2M) = 0$. Thus:

$$\begin{vmatrix}K_1 -\omega^2 M & -K_1 & 0\\-K_1 & K_1+K_2-\omega^2 M & -K_2 \\ 0 & -K_2 & K_2-\omega^2 M\end{vmatrix} \overset{!}{=} 0 $$

$$\begin{vmatrix} 3 -4\omega^2 & -3 & 0\\-3 & 10-4\omega^2 & -7 \\ 0 & -7 & 7-4\omega^2 \end{vmatrix} \overset{!}{=} 0 $$

Which has solutions: $\omega_1 = 0,\ \omega_2 = \frac{\sqrt{10 +\sqrt{37}}}{2}, \ \omega_2 = \frac{\sqrt{10 -\sqrt{37}}}{2}$. Which are the eigenfrequencies of the system. And in absence of drag, they are equal to the resonant frequencies.