# Potential Energy of a Two Spring, Two Mass System

1. May 7, 2008

### logic smogic

1. The problem statement, all variables and given/known data
Find the total potential energy described by a system consisting of a mass hanging by a spring, connected to a second mass also hanging by a spring. Assume that the masses are the same, and the springs are identical (in spring constant and length).

2. Relevant equations
$$V_{spring} = \frac{1}{2} k y^{2}$$
$$V_{gravity} = -mgy$$

3. The attempt at a solution
This is one-dimensional, so I'm working only with "y". I'm calling "up" the positive y-direction, and I'm labeling the top mass "1" and the bottom mass "2".

The initial position of mass 1 will be $y_{01}$ and the initial position of mass 2 will be $y_{02}$.

I'm assuming that (is this right?):

$$V_{total} = V_{spring,1} + V_{spring,2} + V_{gravity,1} + V_{gravity,2}$$

So shouldn't the answer just be:

$$V_{total} = \frac{1}{2} k (y_{1} - y_{01})^{2} + \frac{1}{2} k [(y_{2}-y_{02})-(y_{1}-y_{01})]^{2} - mgy_{1} - mgy_{2}$$

But this can't be right! Since they're both conservative forces (gravity and the springs), doesn't the derivative need to vanish when evaluated at the initial values? It doesn't for the gravity potential parts. Help!

2. May 8, 2008

### Shooting Star

The negative of the derivative of V is the force. Why should the force vanish at any point?

Where does it reflect in your equation that the masses are joined in the manner described? This looks like the total PE of any two masses hung by different springs.

Consider the fact that the force on the upper spring is the weight of both the masses, while that on the lower one is just the weight of one mass. Equate the elastic forces to the weights on each spring. Take the point to which the top spring is connected to be baseline for gravitational PE.

3. May 9, 2008

### logic smogic

Because we're evalutating the system in equilibrium, right?

Well, I'm saying that the total potential energy contribution from the springs is,

$$V_{springs} = \frac{1}{2} k (y_{1} - y_{01})^{2} + \frac{1}{2} k [(y_{2}-y_{02})-(y_{1}-y_{01})]^{2}$$

...where you can see I've accomodated for the coupling in the second term. I've checked this with a horizontal 2-spring/mass system, and it seems consistent.

Wouldn't I be double counting the mass on the bottom? If I get what you're suggesting, the potential energy from gravity would be,

$$V_{gravity} = -2mgy_{1} - mg(y_{2}-y_{1})$$

Here's the trick, though - this contribution is still linear in $y_{1}$ and $y_{2}$ (so I can't form a tensor with it), and it doesn't vanish after differentiation.

4. May 9, 2008

### Shooting Star

You are correct. If you evaluate $\partial V/\partial y_{1}$ and equate it to zero, you will get an equation in y1. But you don't get to put in the value to make it zero, which I think you were trying to do. Same for y2. Then you can solve for y1 and y2.

If we keep things a bit simple, the solution becomes easy. If the lower spring stretches by a length x due to mg, how much would the upper one stretch due to 2mg? This will give you the positions of the two masses, and hence the total PE.