# Setup for a two spring/two mass system (ODEs)

1. Apr 5, 2013

### SithsNGiggles

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

Suppose a cart of mass 2 kg is attached by a spring of constant k = 1 to a cart of mass 3 kg, which is attached to the wall by a spring also of k = 1. Suppose the initial position of the first cart is 1 m in the positive direction from the rest position, and the second mass starts at the rest position. The masses are not moving and are let go. Find the position of the second mass as a function of time.

2. Relevant equations

3. The attempt at a solution

It's been a while since I've taken physics, so I'm a bit lost on the setup for this system. I've attached an image of the scenario. Are my forces (green arrows/text) for each mass correct?

If this is right, then my system of equations is
$\begin{cases} m_1x_1''=-k_1x_1+k_1x_2\\ m_2x_2''=k_1x_1 + (k_2-k_1)x_2 \end{cases}$

Plugging in the given values, it becomes
$\begin{cases} 2x_1''=-x_1+x_2\\ 3x_2''=x_1 \end{cases}$

Or, as a matrix equation,
$\left(\begin{matrix}2&0\\0&3\end{matrix}\right)\vec{x}\;'' = \left(\begin{matrix}-1&1\\1&0\end{matrix}\right)\vec{x}$

Is this all right so far? I'm confident I can solve the system, but not as much about the setup. Thanks

2. Apr 5, 2013

### SithsNGiggles

Forgot to attach in first post

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3. Apr 5, 2013

### vela

Staff Emeritus
The sign of the k2 in your second equation is wrong. To see this, suppose both masses are at the equilibrium positions. Now if x2 increases, both springs will push back in the same direction. The effects of the springs add; they don't subtract.

4. Apr 5, 2013

Is that it?

5. Apr 5, 2013

### vela

Staff Emeritus
Yes.

6. Apr 5, 2013

### SithsNGiggles

Thanks, I can definitely handle the solution from here. But could you walk me through the setup again (if you have the time, of course)? I don't really follow.

I have to get to class now, but I'll be sure to check back if you or anyone else could explain.

7. Apr 5, 2013

### vela

Staff Emeritus
The force on mass 1 due to spring 1 generally has to be either $k_1(x_2-x_1)$ or $-k_1(x_2-x_1)$. You just have to figure out which one gives the right sign. To do this, look at a simple case. Start off with the masses in their equilibrium positions so that $x_1=x_2=0$. Now imagine moving mass 1 to the right, which is the positive direction according to your convention, while holding mass 2 stationary. The spring is compressed, so it's going to push mass 1 to the left or in the negative direction. So which expression gives a negative result when $x_1>0$ and $x_2=0$? That's the one that contributes to the sum of the forces that equals $m_1 x''_1$. Do a similar analysis for mass 2.

Then move on to spring 2. It's connected only to mass 2, so this time, move mass 2 a little to the right and keep mass 1 at $x_1=0$. Then analyze the situation the same way.

8. Apr 8, 2013

### SithsNGiggles

I think I get it now. I'll try out some more problems. Thanks for the break-down!