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Extension of spring Lagrangian

  1. Feb 1, 2017 #1
    1. The problem statement, all variables and given/known data
    An ideal spring of relaxed length l and spring constant k is attached to two blocks, A and B of mass M and m respectively. A velocity u is imparted to block B. Find the length of the spring when B comes to rest.

    2. Relevant equations
    [tex] ∆K + ∆U = 0 [/tex]
    [tex] U = \frac {kx^2}{2} [/tex]
    [tex] ∆p = 0 [/tex]
    [tex] L = K - U [/tex]

    3. The attempt at a solution

    I used energy conservation to find the answer. I got:

    [tex] (mu)^2( \frac 1m - \frac1M) = kx^2 [/tex]

    Then divide by k and take the square root.

    I then tried with the Lagrangian. My coordinates are x and X, where X is the position of B, and x is the position of A. u is the velocity of A and U is the velocity B. And because the solutions will not change if I multiply the Lagrangian, I eliminated the factor of half from the energies. Thus my Lagrangian is:

    [tex] mu^2 + MU^2 - k(x - X - l)^2 = L [/tex]

    I used the Euler Lagrange equations with respect to x and X. However when I used them I simply end up with Hooke's law. But Hooke's law relates the acceleration to the extension of the spring and I do not have the acceleration. Thus I can't solve for the extension. Plus the solution should depend on the final and initial velocities but it does not.
    How could I solve this using the Lagrangian?
    Thank you for answering
     
  2. jcsd
  3. Feb 1, 2017 #2
    I set up a new Lagrangian that is in terms of the displacement of the spring. Then I used Hooke's laws from the Lagrangian, and found an equation for the displacement's acceleration. Its Hooke's law, but with the reduced mass, so I solved it, used the velocity of the displacement to find the time, then plugged that in back in the equation of the displacement. I know it is the right answer because the energy answer matches this one, but is there a simpler way of doing this?
    Thank you
     
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