# Oscillations: mass in the center of an octahedron -- eigenvalues?

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1. Aug 23, 2016

### leialee

You have an infinitesimally small mass in the center of octahedron. Mass is connected with 6 different springs (k_1, k_2, ... k_6) to corners of octahedron.
Equilibrium position is in the center, you don't take into account gravity, only springs.
Find normal modes and frequencies.

Relevant equations
$m\ddot R=\sum_{n=1}^{6} \vec F_i$
$R=(x,y,z)$is a vector of location of the mass and $A$ is matrix.
My idea is to transform differential equations that I will get for each coordinate into this form:
$\ddot R+AR=0$

View attachment 105010
3. The attempt at a solution
When you move the mass away from equilibrium position, every spring works on it with its own force, problem is 3D so I used vectors.
$\vec F_i=\frac{\vec r_i}{|\vec r_i|}u_i k_i$, where I used Hook's law: $F=-kx$, and in using this law I keep in mind that $x,y,z$ are small.
$u_i=b-|\vec r_i|$, where $\vec r_i = \vec R - \vec R_i$.
$R_i$ is a vector from center to corner for each spring (picture)
View attachment 105009
For every sping i defined its vector $R_i$
$\vec R_1=b(1,0,0)\\ \vec R_2=b(0,1,0)\\ \vec R_3=b(-1,0,0)\\ \vec R_4=b(0,-1,0)\\ \vec R_5=b(0,0,1)\\ \vec R_6=b(0,0,-1)$
Where $b$ is a lenght of un-expanded spring, $b=a\frac {\sqrt{2} } {2}$ and $a$ is a lenght of a side of octahedron. I marked $r^2=x^2+y^2+z^2$.
So i get:
$\vec r_1= (x,y,z)-(b,0,0)=(x-b,y,z)\\ \vec r_2=(x,y-b,z)\\ \vec r_3=(x+b,y,z)\\ \vec r_4=(x,y+b,z)\\ \vec r_5=(x,y,z-b)\\ \vec r_6=(x,y,z+b)$

and for the lenghts of thse vectors:
$|\vec r_1|= \sqrt{(x-b)^2 + y^2 + z^2}=\sqrt{r^2+b^2-2xb}\\ |\vec r_2|=\sqrt{x^2 + (y-b)^2 + z^2}=\sqrt{r^2+b^2-2yb}\\ |\vec r_3|=\sqrt{(x+b)^2 + y^2 + z^2}=\sqrt{r^2+b^2+2xb}\\ |\vec r_4|=\sqrt{x^2 + (y+b)^2 + z^2}=\sqrt{r^2+b^2+2yb}\\ |\vec r_5|=\sqrt{x^2 + y^2 + (z-b)^2}=\sqrt{r^2+b^2-2zb}\\ |\vec r_6|=\sqrt{x^2 + y^2 + (z+b)^2}=\sqrt{r^2+b^2+2zb}$

My equation is then:
$m\ddot R=\sum_{n=1}^{6} \vec F_i=\sum_{n=1}^{6} k_i u_i \frac{\vec r_i}{|\vec r_i|}$, and i mark $u_i \frac{\vec r_i}{|\vec r_i|}=\vec r_i (\frac{b}{|\vec r_i|}-1)$

I started solving for each coordinate, and equation for x is:
$$m\ddot x=k_1(\frac {b}{\sqrt{r^2+b^2-2xb}}-1)(x-b)+k_2(\frac {b}{\sqrt{r^2+b^2-2yb}}-1)x+k_3(\frac {b}{\sqrt{r^2+b^2+2xb}}-1)x+k_4(\frac {b}{\sqrt{r^2+b^2+2yb}}-1)x+k_5(\frac {b}{\sqrt{r^2+b^2-2zb}}-1)x+k_6(\frac {b}{\sqrt{r^2+b^2+2zb}}-1)x=\\ -x(\sum_{n=1}^{6} k_i)+ k_1 b - k_3 b+ b^2(\frac {k_3}{\sqrt{r^2+b^2+2xb}}-\frac {k_1}{\sqrt{r^2+b^2-2xb}})+ \\bx(\frac {k_1}{\sqrt{r^2+b^2-2xb}}+\frac {k_2}{\sqrt{r^2+b^2+2xb}}+\frac {k_3}{\sqrt{r^2+b^2-2yb}}+\frac {k_4}{\sqrt{r^2+b^2+2yb}}+\frac {k_5}{\sqrt{r^2+b^2-2zb}}+\frac {k_6}{\sqrt{r^2+b^2+2zb}})$$ (1)

And then similar equation for $y$ and $z$.
And here is where i have a problem. As far as i know i should get linear equatios so i can put together matrix A but no matter how i try to approximate, i get mixed parts of equations.

I went like this: because $x,y,z$are small i neglected all $r^2$ and then developed roots with Taylor approximation to linear order.
$\frac{1}{\sqrt{1\pm x}}=1\mp \frac{1}{2}x$
Which i thought would go nicely and everything odd will fall out, even constant $k_1 b - k_3 b$ went away.
But I get such equations:
$m\ddot x= -x(k_3+k_1)+\frac{x}{b}(k_1 x + k_2 y-k_3 x -k_4 y +k_5 z -k_6 z)$ and similar for y, z.
This all translates into a matrix which isnt constant but has $x,y,z$ dependencies.

So my question would be, how to develop equation 1 differently so i only get linear dependencies, meaning:
$\ddot x= a_1x+a_2y+a_3z\\ \ddot y=b_1x+b_2y+b_3z\\ \ddot z=c_1x+c_2y+c_3z$ so I can then get this system: $\ddot R+AR=0$

...or is my whole approach wrong? In that case...any suggestions, advice?
We are also allowed to use numerical approaches, but given i'm looking for normal modes i think matrix is a way to go.
Anything would be helpful at this point, because i'm on the verge of just neglecting mixed parts and getting diagonal matrix from the equations alone.

2. Aug 23, 2016

### haruspex

Not sure you can assume this is the same for each spring. If they have different constants, the same relaxed length, yet equilibrium is in the centre than that says there is a relationship between the constants. So maybe they have different relaxed lengths, and the central position is where each spring exerts the same force.

3. Aug 25, 2016

### leialee

True, that is one of my problems - as you don't take into account gravity, are springs pre-stressed or not? if they are relaxed and same lenghts equilibrium is still in the center, but if they are pre-stressed then equlibrium (which the task says is in the center) gives you relationship between $k_1$ and $k_3$, $k_2$ and $k_4$, $k_5$ and $k_6$.
So, arm? Not sure anymore?

4. Aug 25, 2016

### leialee

Actually the task's instruction is scpecific: all springs have the same lenght and are not pre-stressed.

5. Aug 25, 2016

### haruspex

Ok. I realised later I was wrong to say it implied a relationship between the constants.

6. Aug 26, 2016

### haruspex

Since the forces equations look messy, I tried looking at potential energy instead.
I considered a small perturbation length s at angle θ to the X axis. I represent the corners of the octagon as being at distance R from the origin and at angles θI to the X axis.
Approximating as far as the s2 terms, when the fog clears I get $\frac{s^2}2\Sigma k_i\cos^2(\theta_i-\theta)$

7. Aug 26, 2016

### TSny

Yes, I think you are right to drop the nonlinear terms. So, you get a fairly trivial solution.

8. Sep 3, 2016

### leialee

Thank you all, i solved this later on - i did drop nonlinear terms and that was it.