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

$\sum F=ma$

$-k_{1}x_{1}-k_{2}(x_{1}-x_{2})=m_{1}{\ddot {x_{1}}}$

${\dfrac {-x_{1}(k_{1}+k_{2})-x_{2}k_{2}}{m_{1}}}={\ddot {x_{1}}}$

On the right, we have

$\sum F=ma$

$-k_{2}(x_{2}-x_{1})=m_{2}{\ddot {x_{2}}}$

${\dfrac {-k_{2}(x_{2}-x_{1})}{m_{2}}}={\ddot {x_{2}}}$

So, our state space matrices are

${\begin{bmatrix}{\ddot {x_{1}}}\\{\dot {x_{1}}}\\{\ddot {x_{2}}}\\{\dot {x_{2}}}\\\end{bmatrix}}={\begin{bmatrix}0&-{\dfrac {k_{1}+k_{2}}{m_{1}}}&0&-{\dfrac {k_{2}}{m_{1}}}\\1&0&0&0\\0&{\dfrac {k_{2}}{m_{2}}}&0&-{\dfrac {k_{2}}{m_{2}}}\\0&0&1&0\\\end{bmatrix}}{\begin{bmatrix}{\dot {x_{1}}}\\x_{1}\\{\dot {x_{2}}}\\x_{2}\\\end{bmatrix}}$

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

${\begin{bmatrix}{\ddot {x_{1}}}\\{\dot {x_{1}}}\\{\ddot {x_{2}}}\\{\dot {x_{2}}}\\\end{bmatrix}}={\begin{bmatrix}0&-6&0&-5\\1&0&0&0\\0&2.5&0&-2.5\\0&0&1&0\\\end{bmatrix}}{\begin{bmatrix}{\dot {x_{1}}}\\x_{1}\\{\dot {x_{2}}}\\x_{2}\\\end{bmatrix}}$

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

$\lambda _{1}=0.705+j2.179$
$\lambda _{2}=0.705-j2.179$
$\lambda _{3}=-0.705+j2.179$
$\lambda _{4}=-0.705-j2.179$

$k_{1}\varpropto {\begin{bmatrix}0.748\\0.101-j0.311\\-0.262-j0.460\\-0.226+j0.047\\\end{bmatrix}}$
$k_{2}\varpropto {\begin{bmatrix}0.748\\0.101+j0.311\\-0.262+j0.460\\-0.226-j0.047\\\end{bmatrix}}$
$k_{3}\varpropto {\begin{bmatrix}0.748\\-0.101-j0.311\\-0.262+j0.460\\0.226+j0.047\\\end{bmatrix}}$
$k_{4}\varpropto {\begin{bmatrix}0.748\\-0.101+j0.311\\-0.262-j0.460\\0.226-j0.047\\\end{bmatrix}}$

The full solution is

${\bar {x}}=c_{1}k_{1}e^{\lambda _{1}t}+c_{2}k_{2}e^{\lambda _{2}t}+c_{3}k_{3}e^{\lambda _{3}t}+c_{4}k_{4}e^{\lambda _{4}t}$

Eigenmodes

The four distinct eigenvalues mean there will be four eigenmodes (though two are similar). They are

1. The two masses moving in the same direction at all times (i.e. one up and the other down).
2. The two masses moving in opposite directions at all times (i.e. both up or down).
3. and 4. The masses moving at plus or minus 90 degrees of phase offset.