{
"cells": [
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"import numpy as np\n",
"import sympy as smp\n",
"from scipy.integrate import odeint\n",
"import matplotlib.pyplot as plt\n",
"from matplotlib import animation\n",
"from mpl_toolkits.mplot3d import Axes3D\n",
"from matplotlib.animation import PillowWriter"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Preliminary\n",
"\n",
"Suppose we have the Lagrangian\n",
"\n",
"$$L = \\frac{1}{2} m_1 (\\dot{x}_1^2+\\dot{y}_1^2) + \\frac{1}{2} m_2 (\\dot{x}_2^2+\\dot{y}_2^2) -m_1gy_1 - m_2gy_2 $$\n",
"\n",
"We can remove some degrees of freedom by dividing by $m_1$ and some characteristic length $A$ of the problem\n",
"\n",
"$$\\frac{L}{A^2 m_1} = \\frac{1}{2} \\left(\\left(\\dot{\\frac{x_1}{A}}\\right)^2+\\left(\\dot{\\frac{y_1}{A}}\\right)^2 \\right) + \\frac{1}{2} \\frac{m_2}{m_1}\\left(\\left(\\dot{\\frac{x_2}{A}}\\right)^2+\\left(\\dot{\\frac{y_2}{A}}\\right)^2 \\right) -\\frac{g}{A}\\frac{y_1}{A} - \\frac{m_2}{m_1}\\frac{g}{A}\\frac{y_2}{A} $$\n",
"\n",
"This might look more complicated, by defining $m = m_2/m_1$ and putting primes such that $x' = x/A$ we get\n",
"\n",
"$$\\frac{L}{A^2 m_1} = \\frac{1}{2} (\\dot{x'}_1^2+\\dot{y'}_1^2) + \\frac{1}{2} m (\\dot{x'}_2^2+\\dot{y'}_2^2) -g'y_1' - mg'y_2' $$\n",
"\n",
"Note that $\\frac{L}{A^2 m_1}$ produces the same equations of motion as $L$ (for any $C$, $CL$ produces the same equations of motion as $L$). Thus we can solve the Lagrangian problem above, and then put in any values of $A$ and $m_1$ we want afterwards, and scale appropriately."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## The Problem\n",
"\n",
"The driven double pendulum with masses $m_1=1$ and $m_2=m$ and arm-lengths $L_1$ and $L_2$ has it upper arm connector to a motor that displaces the base with at $\\cos(\\omega t)$ where $\\omega$ is the driving frequency.\n",
"\n",
"
![\"drawing\"](\"../images/vid15_p1.PNG\")
\n",
"