Small oscillations, strange springs

[fluidistic]

Homework Statement

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 [itex]V=V_0e^{c(x-b)^2}[/itex].
1)Demonstrate that the equilibrium positions are L/3 and (2/3)L.
2)Find the normal frequencies and normal modes of the system.

Homework Equations

L=T-V.

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, [itex]V_0[/itex]. 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 [itex]x=2b=(2/3)L[/itex] 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 [itex]V(x_1,x_2)=V_0e^{c(x_1^2+x_2^2+x_2-x_1)}[/itex].
Usually one would approximate V(x_1,x_2) by a function, say [itex]U(x_1,x_2)[/itex] equal to [itex]\frac{k}{2}x^2[/itex]. Or in my case by a second order Taylor's expansion of a function of 2 variables. But my big problem is that [itex]V_0 \neq 0[/itex].
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?

 

[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...

 

[fluidistic]

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...

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 [itex]V(x_1,x_2)=V_0[e^{cx_1^2}+e^{cx_2^2}+e^{c(x_2-x_1^2)}][/itex]. When the system is at equilibrium, [itex]x_1=x_2=0[/itex], but unlike in common springs, here [itex]V(0,0)=3V_0 \neq 0[/itex]. 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.

 

[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...

 

[Liquidxlax]

pretty sure your potential should be

V_oe^cx₁2 + V_oe^cx₂2 + V_oe^{c(x₂-x₁)2}

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

 

[fluidistic]

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...

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

pretty sure your potential should be

V_oe^cx₁2 + V_oe^cx₂2 + V_oe^{c(x₂-x₁)2}

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

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

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

 

[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.

 

[fluidistic]

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.

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...