# Three masses two strings system: lagrange and eigenvalues

1. May 16, 2013

### kejal

1. The problem statement, all variables and given/known data
We have a three mass two strings system with:
$m_1$ string M string $m_2$

The end masses are not attached to anything but the springs, the system is at rest, and k is equal for both strings and $m_1$ and $m_2$ are equal. The distance between to $m_1$ and $m_2$, on both sides of M, is b.

It appears to be the same problem as in this thread:

But we want to find the Lagrange equation, use the Euler-Lagrange equation to find the equations for movement and then finally the eigenvalues.

2. Relevant equations
L=T-V

$\frac{\partial L}{\partial x} = \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}$

3. The attempt at a solution
T=$\frac{m\dot{x}^{2}_{1}}{2}+\frac{M\dot{x}^{2}_{2}}{2}+\frac{m\dot{x}^{2}_{3}}{2}$

V=$\frac{k}{2}((x_2 -x_1)-b)^2 + \frac{k}{2}((x_3-x_2)-b)^2$

L=T-V=$\frac{m\dot{x}^{2}_{1}}{2}+\frac{M\dot{x}^{2}_{2}}{2}+\frac{m\dot{x}^{2}_{3}}{2}+\frac{k}{2}(b- (x_2 -x_1))^2 + \frac{k}{2}(b - (x_3-x_2))^2$

Using the Euler-Lagrange we then got:

$m_1 \ddot x_1 +k ((x_2 - x_1)-b)=0$

$m_2 \ddot x_2 + k (x_3 - x_1)=0$

$m_3 \ddot x_3 +k (b - (x_3 - x_2))=0$

And then we're pretty much stuck with what to do to find the eigenvalues, especially with b, but an idea was:

Using $\omega^{2}=\frac{k}{m}$:

$\ddot x_1 = -\omega^{2}x_2+\omega^{2}x_1+\omega^{2}b$

$\ddot x_2 = -\omega^{2}x_3 +\omega^{2}x_1$

$\ddot x_3 = -\omega^{2}b + \omega^{2}x_3 -\omega^{2}x_2$

Then we assumed $x=Ae^{iwt}$, so $\ddot x=-\omega^{2}Ae^{iwt}=-\omega^{2}x$

Then $λ=-\omega^{2}$

$A\bar{x}=λ\bar{x}$

Calculating $\left|A+ \omega^{2}I\right|=0$ gave us $10 \omega^{6}b+3 \omega^{4}=0$ which is... how about no.

Any hint on what to do with b? Can we just exclude it or how should we think? All and any help/hints would be highly appreciated!

2. May 16, 2013

### TSny

Hello, kejal. Welcome to PF!

Note that in a normal mode, each mass oscillates about its equilibrium position. Things will be easier if you let $x_1$ denote the displacement of $m_1$ from its equilibrium position. Similarly for the other masses. You will then find that $b$ will not appear in the Lagrangian.