Coupled Oscillator: Pulley

Coupled Oscillator: Pulley System

Problem

Given the system shown, with all initial positions set so that gravity is accounted for and no motion is occurring, find the response. The spring around the pulley is modeled as a single spring with constant $k_2$.

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Solution

We begin by writing the sum of forces equations for both blocks.

On the left, we have

$\sum F=ma$

$-k_1{x}_1-k_2{x}_1=m_1\ddot{x_1}$

$\frac{-x_1(k_1+k_2)}{m_1}=\ddot{x_1}$

On the right, we have

$\sum F=ma$

$k_2{x}_2=m_2\ddot{x_2}$

$\frac{k_2{x}_2}{m_2}=\ddot{x_2}$

So, our state space matrices are

$\left[\begin{array}{c}\ddot{x_1}\\ \dot{x_1}\\ \ddot{x_2}\\ \dot{x_2}\\ \end{array}\right]=\left[\begin{array}{cccc}0& -\frac{k_1+k_2}{m_1}& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& \frac{k_2}{m_2}\\ 0& 0& 1& 0\\ \end{array}\right]\left[\begin{array}{c}\dot{x_1}\\ x_1\\ \dot{x_2}\\ x_2\\ \end{array}\right]$

Now, suppose we impose the initial conditions as follows:
$m_1$=10 kg
$m_2$=20 kg
$k_1$=10 N/m
$k_2$=50 N/m

We now have

$\left[\begin{array}{c}\ddot{x_1}\\ \dot{x_1}\\ \ddot{x_2}\\ \dot{x_2}\\ \end{array}\right]=\left[\begin{array}{cccc}0& -6& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 2.5\\ 0& 0& 1& 0\\ \end{array}\right]\left[\begin{array}{c}\dot{x_1}\\ x_1\\ \dot{x_2}\\ x_2\\ \end{array}\right]$

Using any number of different computer utilities, we can calculate the eigenvalues and eigenvectors of the coefficient matrix quickly and easily.

${\lambda }_1=j2.45$
${\lambda }_2=-j2.45$
${\lambda }_3=1.58$
${\lambda }_4=-1.58$

$k_1\propto \left[\begin{array}{c}0.926\\ -j0.378\\ 0\\ 0\\ \end{array}\right]$
$k_2\propto \left[\begin{array}{c}0.926\\ j0.378\\ 0\\ 0\\ \end{array}\right]$
$k_3\propto \left[\begin{array}{c}0\\ 0\\ 0.845\\ 0.535\\ \end{array}\right]$
$k_4\propto \left[\begin{array}{c}0\\ 0\\ -0.845\\ 0.535\\ \end{array}\right]$