# Small oscillations, strange springs

1. Mar 4, 2012

### fluidistic

1. The problem statement, all variables and given/known data
Consider 2 masses linked via 3 springs in this way |----m----m----| where the | denotes fixed walls and the ---- the springs.
The length between the walls is 2L and the natural length of each spring is b=L/3.
When we move a mass from its equilibrium position, each spring generate a potential of the form $V=V_0e^{c(x-b)^2}$.
1)Demonstrate that the equilibrium positions are L/3 and (2/3)L.
2)Find the normal frequencies and normal modes of the system.

2. Relevant equations
L=T-V.

3. The attempt at a solution
I'm having some troubles. When x=b, in other words, when the springs aren't streched, they still have a non zero potential energy stored, $V_0$. I know that this is the minimal potential energy stored but still... shouldn't the masses constantly move? Hmm, I guess not.
1)So is the argument "the potential energy is minimal when the first mass is at x=b=L/3 and the second mass at $x=2b=(2/3)L$ so these are the equilibrium positions" valid?
2)I chose my 2 generalized coordinates as being x_1 and x_2 where x_1 is the distance the first mass streches the first spring, and x_2 the distance the second mass streches the 3rd spring.
My expression for the potential energy is then $V(x_1,x_2)=V_0e^{c(x_1^2+x_2^2+x_2-x_1)}$.
Usually one would approximate V(x_1,x_2) by a function, say $U(x_1,x_2)$ equal to $\frac{k}{2}x^2$. Or in my case by a second order Taylor's expansion of a function of 2 variables. But my big problem is that $V_0 \neq 0$.
So even if I had only 1 spring here with 1 mass, I wouldn't know how to approximate the potential energy function.
Can someone help me here?

2. Mar 4, 2012

### M Quack

You have 3 springs. Each spring has a potential energy. The total potential energy is the sum of these 3 energies. The PE of the leftmost spring depends only on x1, of the rightmost only on x2, and the middle on x1-x2...

3. Mar 4, 2012

### fluidistic

Yes I know this, as you can see my potential energy is a sum of 3 terms. In fact I made an error and it should be $V(x_1,x_2)=V_0[e^{cx_1^2}+e^{cx_2^2}+e^{c(x_2-x_1^2)}]$. When the system is at equilibrium, $x_1=x_2=0$, but unlike in common springs, here $V(0,0)=3V_0 \neq 0$. And I cannot just set the lower potential energy value possible to 0 because, unlike with common springs, the potential energy of each springs is directly proportional to the lowest possible value of their potential energy.

4. Mar 4, 2012

### M Quack

Who said that the minimum in potential energy has to be 0? The minimum for each exponential function is 1, so 3 V0 is the absolute minimum you can get.

However, I still don't agree with your expression for the potential energy. Compare your first term with the formula given...

5. Mar 4, 2012

### Liquidxlax

pretty sure your potential should be

Voecx12 + Voecx22 + Voec(x2-x1)2

where x1 x2 are spring deviations relative to either m1 or m2

Last edited: Mar 4, 2012
6. Mar 4, 2012

### fluidistic

It looks like the problem was badly copied. $V=V_0e^{c(x-b)^2}$ seems to be the potential energy of the first spring when stretched a distance x from its equilibrium point, $x_0=b$.
The important thing to notice is that for each spring, if we stretch it a distance x, then $V=V_0e^{cx^2}$. (only a guess though, but seems reasonable)

This is exactly what I get and wrote in post #3 :)

Using a Taylor expansion for the potential, I reach $V(x_1,x_2)\approx 2V_0c(x_1^2+x_2^2-x_1x_2)+V(0,0)$. Usually one takes $V(0,0)=0$ because since it's a constant it shouldn't modify the equation of motions. However here, as you (M Quack) pointed out, $V(0,0)=3V_0$. So if I take it to be 0, it means $V_0=0$ and there's absolutely no motion.
If I still take $V(0,0)=0$ but $V_0 \neq 0$ (by some magic) then $V(x_1,x_2)\approx 2V_0c(x_1^2+x_2^2-x_1x_2)$.
In form of matrices, $V= \begin{pmatrix} 2V_0c & -V_0c \\ -V_0c & 2V_0c \end{pmatrix}$ and $T= \begin{pmatrix} m/2 & 0 \\ 0 & m/2 \end{pmatrix}$.
So that $|V-\omega ^2 T|=0$ gave me $\omega _1 = \sqrt {\frac{2V_0c}{m}}$ and $\omega _2 = \sqrt {\frac{6V_0c}{m}}$.
The normal modes: $\vec \eta _1=c_1\begin{pmatrix}1\\-1\end{pmatrix}\cos (\omega _1t+\varphi _1)$ and $\vec \eta _2=c_2\begin{pmatrix}1\\ 1\end{pmatrix}\cos (\omega _2t+\varphi _2)$.
I don't really have a lot of confidence in all this.

7. Mar 4, 2012

### OldEngr63

The mode vector - mode frequency grouping appears reversed. The mode vector col(1,1) should go with the lower mode frequency while the mode vector col(1,-1) should go with the higher mode frequency. Go back through your algebra and watch your signs very carefully.

8. Mar 4, 2012

### fluidistic

Thank you very much for this comment and for pointing this out. I've recheked the algebra twice now and doesn't see any mistake. Hmm...