Spring/Mass System with Unequal Masses

  • Thread starter Thread starter MichalXC
  • Start date Start date
  • Tags Tags
    System
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 4K views
MichalXC
Messages
39
Reaction score
0

Homework Statement



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].

Homework Equations



I want to solve this using purely Newtonian methods.

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?

Thanks.
 
Last edited:
Physics news on Phys.org
The equation become SHM in the COM frame. In your notation, just make the substitution: [itex]r_{1CM} = - r_{2CM}[/itex]
 
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]...
 
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!)
 
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.