Spring/Mass System with Unequal Masses

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Homework Statement



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

Homework Equations



I want to solve this using purely Newtonian methods.

The Attempt at a Solution



The position of m_1 in the center of mass frame is given by:

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

Likewise, the position of m_2 in the CM frame is:

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

I can write down Newton's equations for each mass using for Hooke's law r_{2_{CM}} - r_{1_{CM}} - l 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?

Thanks.
 
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The equation become SHM in the COM frame. In your notation, just make the substitution: r_{1CM} = - r_{2CM}
 
Is that substitution justifiable even though the two masses are unequal? Certainly the distance from the center of mass to m_1 need not equal the distance from the center of mass to m_2...
 
What are the differential equations you get?
 
The force on mass one will be:

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

And on mass two:

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

(Please correct me if this is wrong!)
 
Two possible approaches:

1. You should be able to get an equation of the form
\frac{m_1m_2}{m_1+m_2} \ddot{r} = -krwith 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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