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Homework Help: Three masses two strings system: lagrange and eigenvalues

  1. May 16, 2013 #1
    1. The problem statement, all variables and given/known data
    We have a three mass two strings system with:
    [itex]m_1 [/itex] string M string [itex]m_2[/itex]

    The end masses are not attached to anything but the springs, the system is at rest, and k is equal for both strings and [itex]m_1[/itex] and [itex]m_2[/itex] are equal. The distance between to [itex]m_1[/itex] and [itex]m_2[/itex], on both sides of M, is b.

    It appears to be the same problem as in this thread:

    But we want to find the Lagrange equation, use the Euler-Lagrange equation to find the equations for movement and then finally the eigenvalues.

    2. Relevant equations

    [itex]\frac{\partial L}{\partial x} = \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}[/itex]

    3. The attempt at a solution

    V=[itex]\frac{k}{2}((x_2 -x_1)-b)^2 + \frac{k}{2}((x_3-x_2)-b)^2[/itex]

    L=T-V=[itex]\frac{m\dot{x}^{2}_{1}}{2}+\frac{M\dot{x}^{2}_{2}}{2}+\frac{m\dot{x}^{2}_{3}}{2}+\frac{k}{2}(b- (x_2 -x_1))^2 + \frac{k}{2}(b - (x_3-x_2))^2[/itex]

    Using the Euler-Lagrange we then got:

    [itex]m_1 \ddot x_1 +k ((x_2 - x_1)-b)=0[/itex]

    [itex]m_2 \ddot x_2 + k (x_3 - x_1)=0[/itex]

    [itex]m_3 \ddot x_3 +k (b - (x_3 - x_2))=0[/itex]

    And then we're pretty much stuck with what to do to find the eigenvalues, especially with b, but an idea was:

    Using [itex]\omega^{2}=\frac{k}{m}[/itex]:

    [itex]\ddot x_1 = -\omega^{2}x_2+\omega^{2}x_1+\omega^{2}b[/itex]

    [itex]\ddot x_2 = -\omega^{2}x_3 +\omega^{2}x_1[/itex]

    [itex]\ddot x_3 = -\omega^{2}b + \omega^{2}x_3 -\omega^{2}x_2[/itex]

    Then we assumed [itex]x=Ae^{iwt}[/itex], so [itex]\ddot x=-\omega^{2}Ae^{iwt}=-\omega^{2}x[/itex]

    Then [itex]λ=-\omega^{2}[/itex]


    Calculating [itex]\left|A+ \omega^{2}I\right|=0[/itex] gave us [itex]10 \omega^{6}b+3 \omega^{4}=0[/itex] which is... how about no.

    Any hint on what to do with b? Can we just exclude it or how should we think? All and any help/hints would be highly appreciated!
  2. jcsd
  3. May 16, 2013 #2


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    Hello, kejal. Welcome to PF!

    Note that in a normal mode, each mass oscillates about its equilibrium position. Things will be easier if you let ##x_1## denote the displacement of ##m_1## from its equilibrium position. Similarly for the other masses. You will then find that ##b## will not appear in the Lagrangian.
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