Three masses two strings system: lagrange and eigenvalues

[kejal]

Homework Statement

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:
https://www.physicsforums.com/showthread.php?t=397347

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

Homework Equations

L=T-V

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

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!

 

[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.