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Tricky Lagrangian

  1. Aug 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A system consists of 3 identical masses (A,B & C) in a line, connected by 2 springs of spring constant k. Motion is restricted to 1 dimension. at t=0 the masses are at rest. Mass A is the subjected to a driving force given by:

    F=F0*cos(omega*t)

    Calculate the motion of C

    2. Relevant equations
    L=T-V, Euler lagrange equation.


    3. The attempt at a solution
    I figured that the easiest way to do this is to write a lagrangian and solve for the equation of motion for the masses.

    I denoted the initial positions of the masses x1, x2 and x3 respectively.
    the kinetic energy term for the lagrangian then becomes

    T= 1.5*m*((x1dot)^2 + (x2dot)^2 + (x3dot)^2)

    For the potential, i have stated that the distance between the masses at equilibrium is equal to J. Hence the potential due to the springs is:

    V= k(x1^2 + 2*x2^2 - 3*x2*x1 - x2*x3 + x1*x3 + J(x1+x3-2*x2)

    I am unsure if this is correct and i have no idea how to incorporate the driving force into the equation. Any ideas?
     
  2. jcsd
  3. Aug 18, 2009 #2

    Avodyne

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    Science Advisor

    Your kinetic term should have a coefficient of 0.5, not 1.5. If we choose to measure each position from its own equilibrium point, then we can set J=0. Then the potential from the springs should be V = 0.5 k [(x1-x2)^2 + (x2-x3)^2]. You can take the force into account by adding -F(t)x1 to V, because a force is minus the gradient of the potential.
     
  4. Aug 18, 2009 #3
    i was uncertain whether i could ascribe an equilibrium position to each mass as it is determined by its position relative to the adjacent masses rather than a fixed point in space. Hence it is constantly changing. that is why i thought i had to introduce the value J. Is this wrong?
     
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