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Finding Normal Modes of Oscillation with matrix representations

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Two equal masses (m) are constrained to move without friction, one on the positive x axis and one on the positive y axis. They are attached to two identical springs (force constant k) whose other ends are attached to the origin. In addition, the two masses are connected to each other by a third spring of force constant k'. The springs are chosen so that the system is in equilibrium with all three springs relaxed (length equal to unstretched length), What are the normal frequencies? Find and describe the normal modes. Consider only small displacements from equilibrium.


    2. Relevant equations
    Lagrangian Formalism: L = T (Kinetic Energy) - U (Potential Energy)
    Taking T = (1/2)(summation across all j, k) M(j,k)*qdot(j)*qdot(k)
    And U = (1/2)(summation across all j, k) K(j,k)*q(j)*q(k)

    Where qdot is dq/dt, j and k represent subscripts, M(j,k) represents the (j,k) value in a jxk matrix and likewise for K(j,k). After this point there are some additional equations to find eigenvalue solutions (which represent the normal modes), but I'm not having trouble with that part so I'll leave them out for now.

    3. The attempt at a solution
    The problem I'm having is the set-up. The closest I've gotten to a reasonable solution took x and y as the general coordinates, where x represented the distance past equilibrium length for the x-mass, and likewise for the y-mass. This gave me the following equation:

    T = (1/2)m*xdot^2+(1/2)m*ydot^2

    Then, to find the distance the third spring (with constant k') is stretched, I used:
    L represents the equilibrium distance for the two axial springs.

    L'(at equilibrium)^2 = 2L^2
    L'(stretched)^2 = (L+x)^2 + (L+y)^2
    =2L^2 +2xL + 2yL + x^2 + y^2

    So (delta)L^2 = 2xL + 2yL + x^2 + y^2

    Thus, the potential energy of the system is:

    U = mgL + mgy + (k/2)(x^2+y^2) + (k'/2)(2xL + 2yL + x^2 + y^2)

    The first two terms represent the gravitational potential, with the first being a constant that can be thrown out of the Lagrangian.

    This U needs to be rewritten as in the above formula to define the K matrix, but the formula I've got demands that each term contain either a cross-product between the general coordinates or a square of a single general coordinate. If what I'm doing is right, then how do I fit the 2xL and 2yL terms into my matrix, and if not, then what am I doing wrong?

    It's occurred to me gravity might not be part of the problem as well - so that everything is symmetrical - but since the issue I'm having doesn't stem from that I haven't thought about it much further.

    Thanks for any help you can give!
     
  2. jcsd
  3. Nov 15, 2012 #2

    haruspex

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    I read the question as being in the horizontal plane.
    Because of the linear terms, don't you need the vector to be (x y 1), and the matrix 3x3?
     
  4. Nov 15, 2012 #3
    As far as I can tell, the potential energy I have written is simply wrong. I just don't know why. Throwing out gravity makes it simpler, but the issue is still that I can't use the typical step of simply pulling out matrix K using the definition I gave. A 3x3 matrix wouldn't help either, since I need to equate it to the 2x2 matrix M (which is easily attainable from T) for the next step, which demands that:

    det(K-(w^2)M)=0

    A 3x3 matrix only occurs when three masses (and thus three equations of motion) are present, as far as I know.

    In short, I need to make the first order (of general coordinates) factors go away. The book I'm using (Classical Mechanics by John Taylor) doesn't have any instances in which the potential energy has any first order terms, so I must have mucked up the set-up for the problem. Does anyone have a hint on which direction to go in?
     
  5. Nov 16, 2012 #4

    haruspex

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    In the simple case of vertical motion under gravity there'd be a linear term. How's that dealt with?
     
  6. Nov 16, 2012 #5

    haruspex

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    Another thought...
    U =(k/2)(x2+y2) + (k'/2)(2xL + 2yL + x2 + y2) is of the form A(x2+y2)+2AB(x+y), so can't you complete the square by change of variable: x' = x+B, y' = y+B, U'=U+AB2?
     
  7. Nov 17, 2012 #6

    ehild

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    The contribution of the third spring to the potential energy is 0.5k'(ΔL)^2.
    You used 0.5k'Δ(L2) which is wrong.

    ehild
     
  8. Nov 17, 2012 #7

    haruspex

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    Doh! Thanks ehild, I missed that completely.
     
  9. Nov 17, 2012 #8

    ehild

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    Aware of Big Brother watching you.:tongue2:

    ehild
     
  10. Nov 17, 2012 #9
    Derp! Thanks. Will try this again once I get a minute.
     
  11. Nov 17, 2012 #10
    L'(unstretched) = root(2)*L
    L'(stretched) = root((L+x)^2+(L+y)^2)

    So delta L' = root((L+x)^2+(L+y)^2) - root(2)*L

    Squaring delta L' is still going to leave me with a handful of linear terms. Here's the potential I got (skipped a few steps since this is difficult to type out, and leaving out gravity):

    U= (K/2)(x^2 + y^2) +(K'/2)[(L+x)^2 + (L+y)^2 + 2L^2 - 2L*root(2*(L+x)^2 +2*(L+x)^2)]

    Going all the way to the Lagrangian gave me as an equation of motion for x:

    m*(dotdotx) = kx - k'(L+x) + (k'L)(L+x)/(root(2(L+x)^2+2(L+y)^2))

    The y EOM will be the same. I'm not sure if it's possible to relate this in matrix form M(dotdotx) = -K(x), where M and K are 2x2 matrices and (dotdotx), (x) are vectors for [dotdotx, dotdoty] and [x, y], respectively.

    So... is there another problem with the set-up? Do I need to change my general coordinates? Or maybe, is there a way for me to bypass this ridiculous matrix formulation? This problem doesn't seem that hard, so I feel like I'm missing something simple here. The matrix formulation works really well in some simple cases, but it seems to fall a bit short when you have anything complicated.

    Thanks for all the help, by the way!

    PS: This is just to find the normal modes of oscillation. The Lagrangian equations of motion having linear terms in them doesn't pose a problem when considering the case of vertical motion due to gravity, but if you were to find the normal modes of that motion it would - because they don't exist in that case.
     
    Last edited: Nov 17, 2012
  12. Nov 21, 2012 #11

    ehild

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    Replace the square-root-term with its approximation linear in x and y. (x and y are small)

    ehild
     
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