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\begin{document}

\newpage
\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\Two Spring Mass}





\markboth{Two Spring Mass}{Two Spring Mass}







\author{Brian Winkel, Director SIMIODE\\
Chardon OH 44024 USA\\
BrianWinkel@simiode.org}

\makeStitlePDFLaTex


\begin{abstract}
We ask students to build a Free Body Diagram for a vertical two mass situation in which the two masses are held fixed at the tip and at the bottom.  The mass holds the springs together at the join of the two springs in between. The motion of the mass is tracked under differing situations, e.g., different spring parameters.  
\end{abstract}



\section*{SCENARIO DESCRIPTION}

We consider two springs (coils - top one thin as it has smaller spring constant and bottom one thicker as it has larger spring constant) and a mass (blue plate) in a vertical mode configuration as depicted below. The figures below in Figure~\ref{fig:SecondSpringPicture} show on the left when the spring system is in static equilibrium, i.e. at rest, while the figure on the right has the mass displaced vertically down from static equilibrium. 


\begin{figure}[t]
\begin{center}

\begin{minipage}[t]{ 4 true cm}
\includegraphics[width=6cm]{5-14-FirstSpringPicture.eps}

\end{minipage} \ \hspace*{2.8 true cm} \
\begin{minipage}[t]{ 4 true cm}


\includegraphics[width=6cm]{5-14-SecondSpringPicture.eps}



\end{minipage}

\vspace*{-.8 true cm}
\parbox{12 true cm}{\caption{\label{fig:SecondSpringPicture}  Two  spring mass system with thin spring on top when  the mass is at the static equilibrium, i.e. at rest (left) and the mass is displaced vertically down from the static equilibrium (right).}}
\end{center}
\vspace*{-.8 true cm}

\end{figure}

Here is your set of Activities:

\begin{enumerate}
\item Build a mathematical model for the displacement $y(t)$ of the mass in this configuration,  using Newton's Second Law with a Free Body Diagram  and the named parameters above. Show that you come up with the  second order differential equation (\ref{eq:model}) for $y(t)$ the displacement from static equilibrium of the mass:
\begin{equation}\label{eq:model}
m y''(t) + (c_1 + c_2) y'(t) + (k_1 + k_2) y(t) = 0
\end{equation}
where $m$ is the mass at the point where the two masses join, $k_1$ is the spring constant of Spring 1, $k_2$ is the spring constant of Spring 2, $c_1$ is the coefficient of resistance constant of Spring 1, and $c_2$ is the coefficient of resistance constant of Spring 2,
\item Convert the second order differential equation into a system of linear first order differential equations using the `key'' $x(t) = y'(t)$.
\item Use these values for the parameters in the problem, $m = 10$, $k_1 = 20$, $k_2 = 30$, $c_1 = 5$, $c_2 = 5$, $y(0) = y_0 = -18$, and $y'(0) = v_0 = 0$. Solve this system using an eigenvalue and eigenvector approach completely!
\item Solve this model for $y(t)$ from the second order differential equation you obtained in (1).
\item Using  the numbers in (3) compare the eigenvalue eigenvector solution for $y(t)$ found in (3) with that found in (4).
\item Using some tool to get a plot of the displacement, $y(t)$, over the time interval $[0, 20]$ sec.

\item Here are some further questions to consider. Take on as many as assigned.

\begin{itemize}
  \item [a)] What would happen if we changed both $c_1$ and $c_2$ (equally, differently)? 
  \item [b)] What if both $c_1$ and $c_2$ were 0?  Could they be in reality?
  \item [c)] What would happen if we changed both $k_1$ and $k_2$ (equally, differently)? 
  \item [d)] What would happen if we changed $y_0$?  Say making it positive?
  \item [e)] What would happen if we changed the mass, $m$?
  \item [f)] What would happen if we used a $y'(0)$ NOT 0?
\end{itemize}   
\end{enumerate}

\teacher{
\section*{NOTES FOR TEACHERS}


We offer comments and solution strategies in a Mathematica notebook 5-014-TwoSpringMass.nb and a pdf version for those who do not use Mathematica as the nature of the narrative is the same.

We begin here to highlight the strategies. We need to develop a Free Body Digram from which we can extract a differential equation using Newton's Second Law of Motion. Whenever we can, we make sure the students understand the applications of first principles from science and this is a case in point.

\begin{figure}[h]
\begin{center}

\includegraphics[width=7cm]{5-14-TwoSpringMassFBD.eps}




\parbox{12 true cm}{\caption{\label{fig:TwoSpringMassFBD}  Free Body Diagram for our Two Spring Mass system.  $y = y(t)$ is the displacement in cm below the static equilibrium. Suppose the mass is moving downward and is below the static equilibrium position while  the positive direction is downward for position and velocity. }}
\end{center}
\end{figure}


We consider a Free Body Diagram of our Two Spring Mass system Figure~\ref{fig:TwoSpringMassFBD} in which  the mass is moving downward and is below the static equilibrium position while  the positive direction is downward for position and velocity. Then spring 1 exerts an upward force of $k_1 y$ dynes because spring 1 has been stretched, while spring 2 exerts an upward force of $k_2 y$ dynes because spring 2 has been compressed. Both the $c_1 y$ and $c_2 y$ resistance terms are upward as the motion of the mass is downward.



From the Free Body Diagram of our Two Spring Mass system Figure~\ref{fig:TwoSpringMassFBD} we use Newton's Second Law of Motion which says that the sum of the forces acting on a body are equal to the product of the body's mass and acceleration to produce an equation in which one side is $m y''(t)$ where $y(t)$ is the displacement in cm below the static equilibrium at time $t$ (in sec) and $m$ is the mass of our object in grams. We can construct our differential equation model (\ref{eq:FBDmodel})We note, again that the positive direction is downward for position and velocity; hence all these forces are negative.




\begin{equation}\label{eq:FBDmodel}
m y''(t) = - c_1 y'(t) - c_2 y'(t) - k_1 y(t) - k_2 y(t) \, .
\end{equation}

We put this in the familiar form (\ref{eq:FBDmodel1})



\begin{equation}\label{eq:FBDmodel1}
m y''(t) +  (c_1 + c_2) y'(t) + (k_1 y+ k_2) y(t)  = 0  \, .
\end{equation}

Further, we can convert this second order differential equation into a system of two first order differential equations by identify in $x(t) = y'(t)$ and thereby produce (\ref{eq:MatrixForm})


\begin{equation}\label{eq:MatrixForm}
\left(
\begin{array}{c}
 x'(t) 
   \\
 y'(t) \\
\end{array}
\right)
 =
\left(
\begin{array}{cc}
0 & 1 \\
   \frac{-c_1-c_2}{m} & \frac{-k_1-k_2}{m}  
\\
\end{array}
\right)
\cdot
\left(
\begin{array}{c}
 x(t) 
   \\
 y(t) \\
\end{array}
\right) \, .
\end{equation}

Here

$$A 
 =
\left(
\begin{array}{cc}
0 & 1 \\
   \frac{-c_1-c_2}{m} & \frac{-k_1-k_2}{m}  
\\
\end{array}
\right) \, $$
is the matrix of our system. We proceed to determine the eigenvalues and eigenvectors and build the eigensolution, using initial conditions to determine  the unknown constants.  We obtain the same solution as when we do the solution of the second order differential equations.

From time to time we use two approaches for a solution to review a solution strategy and to compare solutions. We do this here in using the system and second order approach.

We encourage students to play with the parameters and see their meanings and relationships. To this end technology is invaluable. We call the reader's attention to a Mathematica notebook 5-014-T-Mmma-TwoSpringMass.nb and a pdf version for those who do not use Mathematica as the nature of the narrative is the same for more details.
}




\end{document}
