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Spring/Mass System with Unequal Masses

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data

    I want to find the equations of motion of two masses [itex]m_1[/itex] and [itex]m_2[/itex] attached to each other by a spring on a smooth surface assuming [itex]m_2[/itex] is given an instantaneous velocity [itex]v_0[/itex] at time zero. Call the unstretched length of the spring [itex]l[/itex].

    2. Relevant equations

    I want to solve this using purely Newtonian methods.

    3. The attempt at a solution

    The position of [itex]m_1[/itex] in the center of mass frame is given by:

    [tex] r_{1_{CM}} = r_1 - R_{CM} = \frac {m_2 (r_1 - r_2)}{m_1+m_2} [/tex]

    Likewise, the position of [itex]m_2[/itex] in the CM frame is:

    [tex] r_{2_{CM}} = r_2 - R_{CM} = \frac {m_1 (r_2 - r_1)}{m_1+m_2} [/tex]

    I can write down Newton's equations for each mass using for Hooke's law [itex]r_{2_{CM}} - r_{1_{CM}} - l[/itex] as the displacement of the length of the spring from its equilibrium position.

    At this point, I get two differential equations that I do not know how to solve. (Not SHM.) Can anybody help me?

    Last edited: Oct 17, 2011
  2. jcsd
  3. Oct 17, 2011 #2
    The equation become SHM in the COM frame. In your notation, just make the substitution: [itex]r_{1CM} = - r_{2CM}[/itex]
  4. Oct 17, 2011 #3
    Is that substitution justifiable even though the two masses are unequal? Certainly the distance from the center of mass to [itex]m_1[/itex] need not equal the distance from the center of mass to [itex]m_2[/itex]...
  5. Oct 17, 2011 #4


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    What are the differential equations you get?
  6. Oct 17, 2011 #5
    The force on mass one will be:

    [tex] F_1 = m_1 r''_{1_{CM}} = - k (r_{2_{CM}} - r_{1_{CM}} - l) [/tex]

    And on mass two:

    [tex] F_2 = m_2 r''_{2_{CM}} = + k (r_{2_{CM}} - r_{1_{CM}} - l) [/tex]

    (Please correct me if this is wrong!)
  7. Oct 18, 2011 #6


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    Two possible approaches:

    1. You should be able to get an equation of the form
    [tex]\frac{m_1m_2}{m_1+m_2} \ddot{r} = -kr[/tex]with an appropriate definition of r.

    2. You could write your equations as single matrix equation and then diagonalize the matrix. That'll decouple the equations for you.

    I know the first approach is definitely doable because it's a standard result in classical mechanics. The second one might work. I haven't worked it out, so there could be complications I'm not aware of.
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