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Small oscillations, strange springs

  1. Mar 4, 2012 #1

    fluidistic

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    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 [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.

    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, [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?
     
  2. jcsd
  3. Mar 4, 2012 #2
    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...
     
  4. Mar 4, 2012 #3

    fluidistic

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    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.
     
  5. Mar 4, 2012 #4
    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...
     
  6. Mar 4, 2012 #5
    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
  7. Mar 4, 2012 #6

    fluidistic

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    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)

    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.
     
  8. Mar 4, 2012 #7

    OldEngr63

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    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.
     
  9. Mar 4, 2012 #8

    fluidistic

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    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...
     
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