# How can I interpret the normal modes of this mechanical system?

How can I interpret the normal modes of this mechanical system? The equations of motion for the system are as follows: $$\left[\begin{array}{ccc} m_{1}\\ & m_{2}\\ & & 0 \end{array}\right]\left\{ \begin{array}{c} \ddot{u}_{1}\\ \ddot{u}_{2}\\ \ddot{\alpha} \end{array}\right\} +\left[\begin{array}{ccc} k_{1}+k_{2} & 0 & -ak_{2}\\ 0 & k_{3}+k_{4} & -bk_{3}\\ -ak_{2} & -bk_{3} & a^{2}k_{2}+b^{2}k_{3} \end{array}\right]\left\{ \begin{array}{c} u_{1}\\ u_{2}\\ \alpha \end{array}\right\} =\left\{ \begin{array}{c} 0\\ 0\\ 0 \end{array}\right\}$$

To simplify the analysis, it is assumed that (omitting units) $$m_{1}=m_{2}=1$$, $$k_{1}=k_{2}=k_{3}=k_{4}=1$$, and $$a=b=1$$. Therefore, the characteristic equation of the system is: $$\left(2-\omega^{2}\right)\left(1-\omega^{2}\right)=0$$ Consequently, there are two frequencies associated with the normal modes, $$\omega_{1}=1$$ and $$\omega_{2}=\sqrt{2}$$. As I understand it, in this case, there should be three frequencies associated with the system's normal modes. What happens to the third frequency, $$\omega_{3}$$?

• all four answers below allude to the underlying mathematical problem without naming it: singular perturbation. The subject is vast and very interesting; in physics/engineering it shows up, among others, in idealizations such as this (inertia free rigid rod on a fulcrum) and, e.g., in questions such as what happens when two ideal batteries of different voltages are connected ++ and --, etc. Commented Jul 14 at 15:04

The system you wrote is a system of DAEs (differential-algebraic equations) since the third equation is a algebraic equation representing an algebraic constraints. As the rod is massless, its "equation of motion" for the rotation

$$I \ddot{\theta} = \sum_k M_k^{ext}$$

"degenerates" in the algebraic equation representing the equilibrium of moments

$$0 = \sum_k M_k^{ext} \ .$$

Once you're a bit familiar with these systems, you can think at an algebraic equation - in this problem representing rigid body with no mass - as dynamical equations with infinitely fast dynamics.

Limit $$I \rightarrow 0$$. If you do so you should get two modes that are very similar to the "exact" modes, and one mode mainly involving the rotation of the rod and approximately no motion of the masses. As a "check", here the mode frequencies and amplitude (I'm too lazy to build animations) with data you provide and $$I = 0.1$$

$$\Omega^2 = \{ 21.0499, \ 2. , \ .9501 \} \qquad , \qquad V = \left\{ \begin{bmatrix} -5.2\cdot10^{-2} \\-5.2\cdot10^{-2} \\ 1.\end{bmatrix}, \begin{bmatrix} 1. \\ -1. \\ 0. \end{bmatrix}, \begin{bmatrix} -9.5 \cdot 10^{-1} \\ - 9.5\cdot10^{-1} \\ -1.\end{bmatrix} \right\}$$

and $$I = 0.01$$

$$\Omega^2 = \left\{ 201.0050, \ 2. , \ .9950 \right\} \qquad , \qquad V = \left\{\begin{bmatrix} -5.\cdot10^{-3} \\-5.\cdot10^{-3} \\ 1. \end{bmatrix}, \begin{bmatrix} 1. \\ -1. \\ 0. \end{bmatrix}, \begin{bmatrix} -9.9 \cdot 10^{-1} \\ - 9.9 \cdot10^{-1} \\ -1. \end{bmatrix} \right\}$$

Taking a look at the modes, you can realize the "meaning" of the three modes:

1. high-frequency mode mainly involving the rotation of the rod. This "local mode" is the "lost mode" when the rod has no mass. As inertia decreases, the contribution of the motion of the cart decreases as well (from $$10^{-2}$$ to $$10^{-3}$$)
2. second "anti-symmetric" mode involving the "counterphase" oscillation of the carts, with no rotation of the rod for symmetric systems
3. third "low frequency mode", with the carts moving in the same directions (along with the extreme points of the rod)

From DAEs to ODEs. Sometimes, it can be useful to exploit algebraic constraints and reduce the system of DAEs to a system of ODEs. As an example, here you could use the algebraic equation to find

$$\alpha = \frac{a k_2}{a^2 k_2 + b^2 k_3} u_1 + \frac{b k_3}{a^2 k_2 + b^2 k_3} u_2 \ ,$$

and replace it the other two equations, to get the system of ODE with two 2nd-order equations. This system of ODEs - representing an undamped mechanical system - has 2 pairs of complex conjugate eigenvalues, as expected.

I have not attempted to do the algebra, but I presume that you have found that the coeffeicient of $$(\omega^2)^3$$ in your characteristic equation is zero. To understand what this means consider the simpler case of a quadratic equation $$ax^2+b x+c=0$$ so $$x= \frac{- b\pm \sqrt{b^2-4ac}}{2a}.$$ As $$a\to 0$$ you find that the limits of the above expressions for $$x$$ are $$-c/b$$ and $$\infty$$. The same thing happens in your case: If you gave the rod some non-zero moment of inertia $$I$$ and then took $$I\to 0$$ you would find that one of the frequencies becomes infinite. This means that the massless rod instantly adjusts its position so that the net torque on the rod is zero. This condition is exactly what you assumed in deriving your equation for the given massless rod.

In this example all the linkage does is convert a displacement of $$u_3$$ at the top to a displacement of $$-\frac ba u_3$$ at the bottom instantaneously so you should not expect there to be any associated natural frequency.
A similar concept is that of the massless string which is used to change the point of application and direction of a force.
As is pointed out in other posts it is best to assign a finite moment of inertia to the rod and at an appropriate time in the analysis see what happens as the moment of inertia tends to zero.

Ok, I'm not doing the math beyond counting degrees-of-freedom (DoF), but here's my thought process:

$$3 x 3$$ matrix: 3 dimensions should have 3 eigenvalues for 3 DoF.

But you have 2 masses, so it's a 2 dimensional problem.

Oh wait, you include the angle of this massless rod that can oscillate without moving m1 or m2, and it stores potential energy with zero kinetic energy (ideal springs)--so it should have infinite frequency: there better be a problem in the math.

Meanwhile the other 2 DoFs should be the usual: COM moving (masses have a fixed separation), and COM stationary with masses moving in opposite directions.

If we label the DOF's as $$u$$ for all of the DOF's with inertia ($$u_{1}$$ and $$u_{2}$$ in your case), and $$u_{0}$$ those with no associated inertia ($$\alpha$$ in your case), the EOM can be written in partitioned form as: $$\left[\begin{array}{cc} M & 0\\ 0 & 0 \end{array}\right]\left\{ \begin{array}{c} \ddot{u}\\ \ddot{u_{0}} \end{array}\right\} +\left[\begin{array}{cc} K_{uu} & K_{uu_{0}}\\ K_{u_{0}u} & K_{u_{0}u_{0}} \end{array}\right]\left\{ \begin{array}{c} u\\ u_{0} \end{array}\right\} =\left\{ \begin{array}{c} 0\\ 0 \end{array}\right\}$$ where $$M$$ and $$K_{uu}$$ are the mass and stiffness matrix associated with the $$u$$ DOF's, and $$K_{uu_{0}}=K_{u_{0}u}^{T}$$ , then $$\begin{array}{ccc} M\ddot{u}+K_{uu}u+K_{uu_{0}}u_{0} & = & 0\\ K_{u_{0}u}u+K_{u_{0}u_{0}}u_{0} & = & 0 \end{array}$$ From the second equation: $$u_{0}=-K_{u_{0}u_{0}}^{-1}K_{u_{0}u}u$$, thus substituting in the first equation leads to: $$M\ddot{u}+\left(K_{uu}-K_{uu_{0}}K_{u_{0}u_{0}}^{-1}K_{u_{0}u}\right)u=0$$ calling $$K^{*}=K_{uu}-K_{uu_{0}}K_{u_{0}u_{0}}^{-1}K_{u_{0}u}$$, we obtain: $$M\ddot{u}+K^{*}u=0$$ Which is the EOM of the system written in the DOF's with associated inertia (the $$u_{0}$$ has been eliminated or “condensed”). This latter equation leads to the modes shapes and frequencies of the system.

$$\def \b {\mathbf}$$ The EOM's are:

$$\b M\, \begin{bmatrix} \ddot{u}_1 \\ \ddot{u}_2 \\ 0 \\ \end{bmatrix}+\b Q\, \begin{bmatrix} {u}_1 \\ {u}_2 \\ \alpha \\ \end{bmatrix}= \begin{bmatrix} 0\\ 0 \\ 0 \\ \end{bmatrix}\tag 1$$

form the third equation you obtain that

$$2\alpha-u_1-u_2=0$$ form here: $$\begin{bmatrix} {u}_1 \\ {u}_2 \\ \alpha \\ \end{bmatrix}= \underbrace{\left[ \begin {array}{cc} 1&0\\ 0&1 \\ 1/2&1/2\end {array} \right]}_{\b T} \begin{bmatrix} {u}_1 \\ {u}_2 \\ \end{bmatrix}$$

so actually the EOM's with the generalized coordinate $$~u_1~,u_2~$$ are

$$\underbrace{\b T^T\, \b M\,\b T}_{2\times 2} \begin{bmatrix} \ddot{u}_1 \\ \ddot{u}_2 \\ \end{bmatrix}+ \underbrace{\b T^T\,\b Q\,\b T}_{2\times 2} \begin{bmatrix} {u}_1 \\ {u}_2 \\ \end{bmatrix}= \begin{bmatrix} 0\\ 0 \\ \end{bmatrix}\tag 2$$

the eigenvalues are now $$~\omega^2_1=2~,\omega^2_2=1~$$ , this is the correct interpretation of the eigenvalues