Question

Three masses are connected by a spring to each other and to fixed positions as shown in the figure below. Find the eigenfrequencies and describe the normal modes 2

Answer 1

The kinetic energy is

$T =\frac{[m(dx_{1}/dt)^2+ m(dx_{2}/dt)^2 + m(dx_{3}/dt)^2]}{2}$
and the potential energy is

$U =\frac{k[(x_{2} - x_{1})^2 + (x_{3} - x_{2})^2 + x_{1}^2 + x_{3}^2]}{2}$
$U = \frac{[2x_{1}^2 + 2x_{2}^2 + 2x_{3}^2 - x_{1}x_{2} - x_{2}x_{1} - x_{2}x_{3} - x_{3}x_{2}]}{2}$
The x_i are the displacements from the equilibrium positions.

let use the x_i as our generalized coordinates q_i.
The Lagrangian is L = T - U.

This can be put into the form

$L =\frac{\sum _{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}]}{2}$

where

$T_{ij} = T_{ji}, k_{ij} = k_{ji}.$
$Here T_{11} = T _{22} = T_{33} = m, T_{ij}(i\neq j) = 0,$

$k_{11} =k_{22} =k_{33} = 2k, k_{12} =k_{21} =k_{23} =k_{32} = -k, k_{13}= k_{31} = 0.$

Solutions of the form x_j = Re(A_je^i?t) can be found. For solutions of this form the equations of motion reduce to

$\begin{pmatrix} 2k-\omega ^2m &-k &0 \\ -k& 2k-\omega ^2m &-k\\ 0 & -k & 2k-\omega ^2m \end{pmatrix}$$\begin{pmatrix} A_{1}\\ A_{2} \\ A_{3} \end{pmatrix}=0$.

We can find the ?² from det(k_ij - ?²T_ij) = 0.

For a system with n degrees of freedom, n characteristic frequencies ?_a can be found. Our system has 3 degrees of freedom.

$\begin{vmatrix} 2k-\omega ^2m & -k &0\\ -k& 2k-\omega ^2m&-k \\ 0 & -k & 2k-\omega ^2m \end{vmatrix}=0$.

$\therefore (2k - \omega ^2m)^3 - 2k^{2}(2k -\omega ^2m) = 0.$

1. $(2k - \omega ^2m) = 0, \omega _{1}^2 = 2k/m, A_{2} = 0, A_{1} = -A_{3}$ .

2. $(2k - \omega ^2m)^2 - 2k^2 = 0, \omega _{2}^2 = (2 + \sqrt{2})k/m, A_{3} = A_{1}, A_{2} = - \sqrt{2A_{1}}.$

3. $\omega_{3}^2 = (2 - \sqrt{2})k/m, A_{3} = A_{1}, A_{1} = \sqrt{2A_{1}}$