# Rate at Which 2 Masses & Springs Exchange Energy

1. Apr 8, 2007

### dimensionless

I ran some simulations with a two mass-three springs oscillator. I found that one mass would oscillate for a while and then almost stop. When that happened the other would start moving back and forth. Is there a way I can predict at what rate the masses will exchange energy?

2. Apr 9, 2007

### lpfr

Yes. If you write down the two differential equations, you can verify that the solution is a mix of two modes of oscillation.
a) One with the two masses en phase. In this case the central spring does nothing and the frequency is given just by the spring constant and the mass.
b) Another with the two masses in opposition of phase. This time the frequency is higher.
When you add the two modes, you should obtain something like: $$[y_1]=A_1\cos\omega_1t + A_2\cos\omega_2t$$
$$[y_2]=A_1\cos\omega_1t - A_2\cos\omega_2t$$
You can work this result in the form:
$$\cos\left({\omega_1+\omega_2\over 2}t\right) \cos\left({\omega_1-\omega_2\over 2}t\right)$$
This gives and oscillation at the mean frequency, modulated by a sinusoid at half the frequency difference.
As you kown the dependency of energy with amplitude, you can work the power transfer between the two masses.

Last edited: Apr 9, 2007
3. Apr 9, 2007

### dimensionless

It looks like your using the identity
$$cos{C} + cos{B} = 2 cos{\frac{C+B}{2}}cos{\frac{C-B}{2}}$$
Would this identity still work if my coefficient for $$y_{n}$$ have values other than $$A_{n} = 1$$?

Last edited: Apr 9, 2007
4. Apr 9, 2007

### lpfr

Not directly, but you can always split things as:
$$[y_1]=A_1\cos\omega_1t + A_2\cos\omega_2t= A_1\left(\cos\omega_1t + \cos\omega_2t \right) + (A_2-A_1) \cos\omega_2t$$
There is a constant amplitude term and a variable one.

5. Apr 9, 2007

### dimensionless

If I implement the identity from here, won't it more or less bring me back to where I started? My goal is to figure out the rate at which the masses exchange energy.

6. Apr 9, 2007

### lpfr

Do as you like. But I know what I said.