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What is the definition of the equilibrium separation in the force of a spring equation

  1. Oct 30, 2014 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    Two masses slide freely in a horizontal frictionless track and are connected by a Spring whose force constant is k. Fid the frequency of oscillatory motion for this system.

    2. Relevant equations
    My professor posted the solution but I am having trouble understanding everything that he did.
    according to the solution, the first step is defining the Force equation as
    F = -k(x2 - x1 - deltaXe) where deltaXe is the equilibrium separation. x1 is the distance to the center of mass of m1 from some point to the left of m1. x2 is the distance from that point to the center of m2. so that x2-x1 equals the distance between the center of m1 and m2.

    3. The attempt at a solution
    the distance between the atoms at which the force on each atom is zero. Is the force repulsive (atoms are pushed apart) or attractive (atoms are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation
    is this the distance between the two masses when the spring is neither compressed nor stretched? Wouldn't that just be x2-x1? so the the Force would be 0?
     
    Last edited: Oct 30, 2014
  2. jcsd
  3. Oct 30, 2014 #2

    grandpa2390

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    and then later on in the solution he takes says that dXe is equal to x2e - x1e (x2e = the distance of m2 from the center of the spring during equilibrium and likewise with x1e) shouldn't dXe = x2e + x1e ?
     
  4. Oct 30, 2014 #3

    BvU

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    Doesn't look like a relevant equation to me. Don't you have a few things that look like

    ##m_1\ddot x_1 = - F_s##

    ##m_2\ddot x_2 = F_s##

    ## F_s = -k (x_2 - x_1 - \Delta x_{eq})##
    so that we can do some mathematics on them ?

    Because in that case we could easily derive

    ##m_1\ddot x_1 + m_2\ddot x_2 =0##

    and concentrate on

    ##m_1 \ddot x_1 - m_2 \ddot x_2 + 2k( x_2 - x_1 - \Delta x_{eq})=0## ?

    And perhaps the masses happen to be equal ? Saves work and yields a pleasant answer.

    Then: In your attempt there is the sudden appearance of atoms that weren't there before. Please read what you wrote as if you were a helper trying to make sense of your post. I am one and I can't.
     
  5. Oct 30, 2014 #4

    grandpa2390

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    that third equation is the one. I figured out what it was. the origin was defined in the middle of the spring. So that x1 is negative. so it does turn out right.
     
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