# Using Matrices to Solve Systems of Masses and Springs

1. May 26, 2014

### kq6up

I am going through Mary Boas' "Mathematical Methods in the Physical Sciences 3rd Ed". I finished the chapter 3 section 12 problem set, but I do not understand how she gets eq. 12.39. These don't seem obviously equal to each other. Here is the equation:

$$\lambda Tr=Vr$$ Where T is a matrix representing the kinetic energy, and V is a matrix representing the stiffness of the springs (potential energy).

Any help would be appreciated.

Thanks,
Chris Maness

2. May 26, 2014

### UltrafastPED

Ideal springs transform potential to kinetic energy, and back again.

When averaged over a full cycle the kinetic & potential energy each equal 1/2 of the total energy.

The problem description will provide further detail for the details of the expression.

3. May 26, 2014

### kq6up

Ironically, I only see the equation pop into the text ex nihilo. Would this be an equivalent expression?:

$$k\left< { r }|{ V }|{ r } \right> =m\left< { \dot { r } }|{ T }|{ \dot { r } } \right> \quad if\quad r={ e }^{ i\omega t }$$

This seems to be equivalent to the original equation in my mind. I used the BRA/KET just to reflect the transpose since it seems to be an inner product with a the matrices sandwiched in between. The equation above does not seem to work. That is why I am scratching my head.

Thanks,
Chris Maness

Last edited: May 26, 2014
4. May 27, 2014

### AlephZero

Your equation isn't quite right. Using matrix notation (I'm an engineer not a physicist!) the potential energy is $\frac 1 2 r^T V r$ and the kinetic energy is $\frac 1 2 \dot r^T T \dot r$. The T matrix includes the mass so you don't need another $m$ term.

If you assume harmonic motion $r = e^{i\omega t}$, then $\dot r = i\omega r$ so $T = -\omega^2 \frac 1 2 r^T T r$, or $T = -\lambda \frac 1 2 r^T T r$.

By conservation of energy, $-\lambda \frac 1 2 r^T T r + \frac 1 2 r^T V r$ is constant.

If you consider this as a function of $r$, the physical solutions are when the energy does not change for small perturbations of $r$, i.e. when $\frac {\partial}{\partial r}(-\lambda \frac 1 2 r^T T r + \frac 1 2 r^T V r) = 0$ which gives your equation $\lambda Tr = Vr$.

You probably want a longer explanation of the last step of taking $\frac {\partial}{\partial r}$, but you should be able to find that in a textbook. I don't have Boas so I can't give you a page reference.

5. May 27, 2014

### kq6up

Ah, I now see where I get my omega squared. There are two r dots there.

Chris