# Spring/Mass System with Unequal Masses

1. Oct 17, 2011

### MichalXC

1. The problem statement, all variables and given/known data

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 $l$.

2. Relevant equations

I want to solve this using purely Newtonian methods.

3. 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.

Last edited: Oct 17, 2011
2. Oct 17, 2011

### mathfeel

The equation become SHM in the COM frame. In your notation, just make the substitution: $r_{1CM} = - r_{2CM}$

3. Oct 17, 2011

### MichalXC

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 $m_2$...

4. Oct 17, 2011

### vela

Staff Emeritus
What are the differential equations you get?

5. Oct 17, 2011

### MichalXC

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!)

6. Oct 18, 2011

### vela

Staff Emeritus
Two possible approaches:

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