2024-04-05 General Coupled Oscillations

System described by

$$k=1,2,3,\dots s$$

With generalized coordinates described by $q_k(t)$

Finding the Lagrangian

$$L=T-u$$

and

$$\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_k}-\frac{\partial L}{\partial q_k}=0$$

Where we get $s$# of equations

For a system where a stable equilibrium exist:

$$q_k=q_{k_0},{\dot{q}}_k=0,{\ddot{q}}_k=0$$

Find $u(q_1,q_2,q_3\dots q_s)$ for coupled oscillations by expansion of the appropriate Taylor series about the equilibrium

Taylor Series, Potential and Equilibrium

$$u(q_1,q_2,\dots q_s)=u_0+\sum _k^s\frac{\partial u}{\partial q_k}{|}_{q_k}+\frac{1}{2}\sum _{j,k}^s\frac{{\partial }^{2}u}{\partial q_j\partial q_k}{|}_0{q}_j{q}_k$$

This is cumbersome. Defining the following makes it less so

$$\frac{{\partial }^{2}u}{\partial q_j\partial q_k}{|}_0=A_{jk}$$ $$u=\frac{1}{2}\sum _{jk}^s{A}_{jk}{q}_j{q}_K$$

Kinetic Energy

$$T=\frac{1}{2}\sum _{j,k}^s{m}_{jk}{\dot{q}}_j{\dot{q}}_k$$

$m_{jk}$ will depend on actual mass of the system and a coordinate transformation.

Quadratic function of generalized velocity.

Coordinate transform into Cartesian coordinates:

$$T=\frac{1}{2}\sum _{\alpha }^n\sum _{i=1}^3{m}_{\alpha }{\dot{x}}_{\alpha i}^2$$ $$x_{\alpha i}=\sum _{k=1}^s\frac{\partial x_{\alpha i}}{\partial q_k}{\dot{q}}_k$$ $$T=\frac{1}{2}\sum _{\alpha }\sum _{i,j,k}^s{m}_{\alpha }\frac{\partial x_{\alpha i}}{\partial q_j}\frac{\partial x_{\alpha i}}{\partial q_k}{\dot{q}}_j{\dot{q}}_k$$ $$m_{jk}=\sum _{\alpha }\sum _i{m}_{\alpha }\frac{\partial x_{\alpha i}}{\partial q_j}\frac{\partial x_{\alpha i}}{\partial q_k}$$

Expanding $m_{jk}$ about the equilibrium

$$m_{jk}=m_{jk}(q_{l0})+\sum _l\frac{\partial m_{jk}}{\partial q_i}{|}_0{q}_l$$

We can ignore the last term because its cubic $\therefore$ small

$$T=\frac{1}{2}\sum _{j,k}^s{m}_{jk}{\dot{q}}_j{\dot{q}}_k$$ $$L=T-U$$ $$=\frac{1}{2}\sum _{jk}^s{m}_{jk}{\dot{q}}_j{\dot{q}}_k-\frac{1}{2}\sum _{j,k}^s{A}_{jk}{q}_j{q}_k$$ $$=\frac{1}{2}\sum _{j,k}^s(m_{j,k}{\dot{q}}_j{\dot{q}}_k-A_{jk}{q}_j{q}_k)$$ $$-\frac{\partial L}{\partial q_k}+\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_j}=0$$

Remove $\frac{1}{2}$

$$\frac{\partial L}{\partial q_k}=-\sum _j{A}_{jk}{q}_j$$ $$\frac{\partial L}{\partial {\dot{q}}_k}=\sum _j{m}_{jk}{\dot{q}}_j$$ $$\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_k}=\sum _j{m}_{jk}{\ddot{q}}_j$$

Equation of motion:

$$\sum _j(A_{jk}{q}_j+m_{jk}{\ddot{q}}_j)=0$$

Solutions of the form

$$q_j(t)=q_j{e}^{i(\omega t-\delta )}$$

Plug in and get

$$\sum _J^s(A_{jk}-{\omega }^{2}m{}_{jk})a_j=0$$