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how to find max velocity in a spring-mass system?

  1. Jul 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Two masses connected with a spring with contant k. The string streched by l . Find the max velocity of mass m!
    M2 ___spring___ M1

    M2--stretched by x2--____spring____--x1--M1
    l = x1+ x2
    2. Relevant equations

    F = k.l
    Mass1.x1 = mass2.x2
    (x= displacement?)
    a=w^2 x
    v = wx

    Ep + Ek1 = Ep + Ek2
    Ep = 1/2kx^2
    Vmax at equilibrium = ##\sqrt{\frac{A^2k}{m}}##

    3. The attempt at a solution
    kl = m.a
    a = k.l/m
    While l = x1 + x2
    After i get a, i can find w, and can find v

    My question is, why can't i use Vmax at equilibrium = ##\sqrt{\frac{A^2k}{m}}## ? The A = x1
    m1x1 = m2x2
    X1 = m2.(l-x1)/m1
    X1=m2.l /(m1+m2)
    Substitute x1 to A. But i get different answer.
     
  2. jcsd
  3. Jul 11, 2017 #2

    TSny

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    Is k the effective spring constant for m1's motion? Hint: Write Hooke's law in terms of x1 rather than L.
     
    Last edited: Jul 11, 2017
  4. Jul 12, 2017 #3
    Ah, i get it..
    The Hooke's law = k.x1
    F=m.a
    m1.a = kx1
    K= m1.a/x1
    Then substitute k to equation.
    So, when we use total constant the displacement is the total L. While we use only partly displacement the k constant different one?
     
  5. Jul 12, 2017 #4

    TSny

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    Yes. The force felt by each mass is F = kL where L is the total amount of stretch of the spring from equilibrium. Using some of your equations in your first post, you should be able to express L in terms of just x1. Then you can identify the effective spring constant k1 for mass m1.
     
  6. Jul 12, 2017 #5
    If the system can move freely then isn't the total momentum always zero?
     
  7. Jul 12, 2017 #6

    TSny

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    Yes.
     
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