Coupled Oscillator: Pulley - Revision history

Nathan.reeves: /* Eigenmodes */

2009-12-15T04:49:35Z

Eigenmodes

← Older revision		Revision as of 21:49, 14 December 2009
Line 88:		Line 88:
	The four distinct eigenvalues mean there will be four eigenmodes (though two are similar). They are		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).		1. The two masses moving in the same direction at all times (i.e. one up and the other down).<br>
	2. The two masses moving in opposite directions at all times (i.e. both up or down).		2. The two masses moving in opposite directions at all times (i.e. both up or down).<br>
	3. and 4. The masses moving at plus or minus 90 degrees of phase offset.		3. and 4. The masses moving at plus or minus 90 degrees of phase offset.

Nathan.reeves: /* Coupled Oscillator: Pulley System */

2009-12-15T04:49:18Z

Coupled Oscillator: Pulley System

← Older revision		Revision as of 21:49, 14 December 2009
Line 17:		Line 17:
	<math>\sum F=ma </math>		<math>\sum F=ma </math>

	<math>-k_1x_1-~~k_2x_1~~=m_1\ddot{x_1}</math>		<math>-k_1x_1-k_2(x_1-x_2)=m_1\ddot{x_1}</math>

	<math>\dfrac{-x_1(k_1+k_2)}{m_1}=\ddot{x_1}</math>		<math>\dfrac{-x_1(k_1+k_2)-x_2k_2}{m_1}=\ddot{x_1}</math>

	On the right, we have		On the right, we have
Line 25:		Line 25:
	<math>\sum F=ma </math>		<math>\sum F=ma </math>

	<math>~~k_2x_2~~=m_2\ddot{x_2}</math>		<math>-k_2(x_2-x_1)=m_2\ddot{x_2}</math>

	<math>\dfrac{~~k_2x_2~~}{m_2}=\ddot{x_2}</math>		<math>\dfrac{-k_2(x_2-x_1)}{m_2}=\ddot{x_2}</math>

	So, our state space matrices are		So, our state space matrices are
Line 37:		Line 37:
	=		=
	\begin{bmatrix}		\begin{bmatrix}
	0 & -\dfrac{k_1+k_2}{m_1} & 0 & 0 \\		0 & -\dfrac{k_1+k_2}{m_1} & 0 & -\dfrac{k_2}{m_1} \\
	1 & 0 & 0 & 0 \\		1 & 0 & 0 & 0 \\
	0 & 0 & 0 & \dfrac{k_2}{m_2} \\		0 & \dfrac{k_2}{m_2} & 0 & -\dfrac{k_2}{m_2} \\
	0 & 0 & 1 & 0 \\		0 & 0 & 1 & 0 \\
	\end{bmatrix}		\end{bmatrix}
Line 60:		Line 60:
	=		=
	\begin{bmatrix}		\begin{bmatrix}
	0 & -6 & 0 & 0 \\		0 & -6 & 0 & -5 \\
	1 & 0 & 0 & 0 \\		1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 2.5 \\		0 & 2.5 & 0 & -2.5 \\
	0 & 0 & 1 & 0 \\		0 & 0 & 1 & 0 \\
	\end{bmatrix}		\end{bmatrix}
Line 71:		Line 71:
	Using any number of different computer utilities, we can calculate the eigenvalues and eigenvectors of the coefficient matrix quickly and easily.		Using any number of different computer utilities, we can calculate the eigenvalues and eigenvectors of the coefficient matrix quickly and easily.

	<math>\lambda_1=j2.45</math><br>		<math>\lambda_1=0.705+j2.179</math><br>
	<math>\lambda_2=-j2.45</math><br>		<math>\lambda_2=0.705-j2.179</math><br>
	<math>\lambda_3=1.58</math><br>		<math>\lambda_3=-0.705+j2.179</math><br>
	<math>\lambda_4=-1.58</math><br><br>		<math>\lambda_4=-0.705-j2.179</math><br><br>

	<math>k_1 \varpropto \begin{bmatrix} 0.~~926~~ \\ -j0.~~378~~ \\ 0 \\ 0 \\ \end{bmatrix}</math><br>		<math>k_1 \varpropto \begin{bmatrix} 0.748 \\ 0.101-j0.311 \\ -0.262-j0.460 \\ -0.226+j0.047 \\ \end{bmatrix}</math><br>
	<math>k_2 \varpropto \begin{bmatrix} 0.~~926~~ \\ j0.~~378~~ \\ 0 \\ 0 \\ \end{bmatrix}</math><br>		<math>k_2 \varpropto \begin{bmatrix} 0.748 \\ 0.101+j0.311 \\ -0.262+j0.460 \\ -0.226-j0.047 \\ \end{bmatrix}</math><br>
	<math>k_3 \varpropto \begin{bmatrix} 0 \\ 0 \\ 0.~~845~~ \\ 0.~~535~~ \\ \end{bmatrix}</math><br>		<math>k_3 \varpropto \begin{bmatrix} 0.748 \\ -0.101-j0.311 \\ -0.262+j0.460 \\ 0.226+j0.047 \\ \end{bmatrix}</math><br>
	<math>k_4 \varpropto \begin{bmatrix} 0 \\ 0 \\ -0.~~845~~ \\ 0.~~535~~ \\ \end{bmatrix}</math>		<math>k_4 \varpropto \begin{bmatrix} 0.748 \\ -0.101+j0.311 \\ -0.262-j0.460 \\ 0.226-j0.047 \\ \end{bmatrix}</math>

			The full solution is

			<math>\bar{x}=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

			===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.

Nathan.reeves: New page: ==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. ...

2009-12-15T04:30:33Z

New page: ==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. ...

New page

==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 <math>k_2</math>.

[[Image:PulleySystem.jpg|200px|thumb|left]]
 
----

===Solution===

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

On the left, we have

<math>\sum F=ma </math>

<math>-k_1x_1-k_2x_1=m_1\ddot{x_1}</math>

<math>\dfrac{-x_1(k_1+k_2)}{m_1}=\ddot{x_1}</math>

On the right, we have

<math>\sum F=ma </math>

<math>k_2x_2=m_2\ddot{x_2}</math>

<math>\dfrac{k_2x_2}{m_2}=\ddot{x_2}</math>

So, our state space matrices are

<math>
\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 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 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} </math>

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

We now have

<math>
\begin{bmatrix}
\ddot{x_1} \\ \dot{x_1} \\ \ddot{x_2} \\ \dot{x_2} \\
\end{bmatrix}
=
\begin{bmatrix}
0 & -6 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2.5 \\
0 & 0 & 1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
\dot{x_1} \\ x_1 \\ \dot{x_2} \\ x_2 \\
\end{bmatrix} </math>

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

<math>\lambda_1=j2.45</math> 
<math>\lambda_2=-j2.45</math> 
<math>\lambda_3=1.58</math> 
<math>\lambda_4=-1.58</math> 

<math>k_1 \varpropto \begin{bmatrix} 0.926 \\ -j0.378 \\ 0 \\ 0 \\ \end{bmatrix}</math> 
<math>k_2 \varpropto \begin{bmatrix} 0.926 \\ j0.378 \\ 0 \\ 0 \\ \end{bmatrix}</math> 
<math>k_3 \varpropto \begin{bmatrix} 0 \\ 0 \\ 0.845 \\ 0.535 \\ \end{bmatrix}</math> 
<math>k_4 \varpropto \begin{bmatrix} 0 \\ 0 \\ -0.845 \\ 0.535 \\ \end{bmatrix}</math>