I was given the attached 3 degree of freedom spring system with the purpose of analyzing it. system

I came up with the following equation of motion equation

and then I ran Matlab to calculate the corresponding natural frequencies and mode shapes using eigenvalues and eigenvectors; I was asked to see what happens when value of stiffness $k_{12}$ is changed. This is the plot of the value $k_{12}$ against the natural frequencies. plot

The problem is that I do not know WHY values of natural frequencies are insensitive at low values of $k_{12}$ and why both the 1st and 2nd natural frequencies are insensitive to changes in $k_{12}$ when values are large (first two level off and the third one seems to go to infinity).

I assume it has something to do with the equation for force due to a spring between 2 masses but I cannot figure it out. That is why I ask for your help - thanks in advance.

  • $\begingroup$ One thing to note is that Matlab sorts the eigenvalues which means the green curve on the left becomes the red curve on the right. $\endgroup$ Apr 10, 2021 at 19:19

1 Answer 1


It's easy to see without doing any math, but just by looking at the picture.

Let's consider first the case of low $k_{12}$. In this case, $m_1$ and $m_2$ basically don't notice $k_{12}$ because it is so weak that it is drowned out by the other springs. So the low $k_{12}$ case basically gives the same value as the $k_{12}=0$ case for all three frequencies (you can check this).

As an exercise, let's think about what would happen if you removed $k_{20}$ and $k_{23}$ (i.e., set them equal to zero). Now $m_2$ can see $k_{12}$ because there are no other springs drowning out its effect, and you should get one frequency changing as $k_{12}$ goes to zero. It is only one because two of the frequencies will just be for the symmetric and antisymmetric modes of $m_1$ and $m_3$, which don't really care about $m_2$. The third one will be slower.

Now lets like about the high $k_{12}$ limit. Here $m_2$ sees only $k_{12}$, and since $k_{12}$ is so big, $m_1$ and $m_2$ are basically rigidly attached. Thus two modes will be the symmetric and antisymmetric modes of $m_3$ and $m_1+m_2$ (you can check this, you have to add $k_{10}$ and $k_{20}$ as well as $k_{13}$ and $k_{23}$ to get the effective spring constants), and the third mode will be a quick oscillation of $m_2$ relative to $m_1$. You can check this too, the frequency ought to be $\sqrt{k_{12}/\mu}$, where $\mu$ is the reduced mass for $m_1$ and $m_2$: $\frac{1}{\mu} = \frac{1}{m_1}+\frac{1}{m_2}$


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