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\begin{document}

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\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\SPRING MASS FIRST TRY}




\markboth{Spring Mass First Try}{Spring Mass First Try}






\author{
\begin{minipage}[t]{6 true cm}
Keith Landry\\
Georgia Southern University\\
Statesboro GA  30458 USA
\end{minipage}
\  \quad \
\begin{minipage}[t]{6 true cm}
Brian Winkel, Director
SIMIODE\\ 
Chardon OH 44024 USA\\
BrianWinkel@simiode.org
\end{minipage}
}




\makeStitlePDFLaTex


\begin{abstract}
Students build a model based on their perceptions of what the solution should look like for  a simple spring mass system with no damping. 
\end{abstract}

\section*{SCENARIO DESCRIPTION}

Consider a mass hanging at the end of a spring as depicted in Figure~\ref{fig:3-101-SpringMass}.  

\begin{figure}[h]
\begin{center}
\includegraphics[width = 4.5 true cm]{3-101-SpringMass.eps}

\vspace*{-1.3 true cm}
{\caption{\label{fig:3-101-SpringMass} Simple spring mass configuration.}}
\end{center}
 
\end{figure}

If we let the mass hang still, thus extending the spring from its natural length, we will see that the mass comes to rest at what is called {\it static equilibrium\/}.  In attempting to model the vertical motion of this mass we would impose a coordinate system for $y(t)$, the vertical displacement from the static equilibrium. Engineers refer to such a system as a \underline{S}ingle \underline{D}egree \underline{o}f \underline{F}reedom \underline{S}ystem (SDOFS), as we are tracking only one variable, namely, $y(t)$, vertical displacement from the static equilibrium.  For consistency, let us say $y = 0$ is the spring's vertical displacement at the spring's static equilibrium and is the  distance of mass displacement from that static equilibrium and denote positive in the downward direction and negative in the upward direction. That is, if we extend the spring 3 cm downward then  $y = 3$, while if we compress the spring upward 2 cm then $y = -2$.


We are going to use a Free Body Diagram (Figure~\ref{fig:3-101-FBD}) to depict all the vertical forces acting on the mass in this case. Figure~\ref{fig:3-101-FBD} depicts the mass pulled down just a bit beyond the  static equilibrium position with the only force acting on it being the restoring force of the spring which acts in an upward vertical direction.

\begin{figure}[h]
\begin{center}
\includegraphics[width = 2.5 true cm]{3-101-FBD.eps}


{\caption{\label{fig:3-101-FBD}Template for Free Body Diagram.}}
\end{center}
 
\end{figure}


 Newton's Second Law of Motion says that for a given body the mass of that body times the acceleration of that mass is equal to the sum of the external forces acting on the mass. Again, we will presume the spring mass system is at rest and this means the downward force due to gravity, $m\cdot g$ (mass $m$ times acceleration due to gravity $g$) cancels out the upward restoring force of the spring when the mass is extended. Thus our Free Body Diagram (Figure~\ref{fig:3-101-FBD}) has only one force when extended, namely the restoring force when the spring is extended beyond its static equilibrium by distance $y(t)$ cm.

In high school you may have conducted an experiment on springs to verify  that when a spring is extended a distance $y$ beyond its equilibrium then there is a force of restoration, $F = k \cdot y$, acting to restore the spring to its equilibrium and that force is proportional to the displacement only. The constant of proportionality, $k$, is called the {\it spring constant\/} and it can be determined experimentally by suspending different masses on the spring, each mass producing a corresponding force. Recall $F = m \cdot g$, where $g$ is the acceleration due to gravity. If we denote the corresponding displacement $y$, then when plotting $F$ vs. $y$ the data is linear and goes through the origin, as 0 units of force displaces the spring 0 units of distance. $k$ can thus be determined by  fitting the line $F = k\cdot y$ to the data $F$ vs. $y$ for a given spring. This equation is called Hooke's Law and is named after 17$^{\rm th}$ century British physicist Robert Hooke. 

A spring is thus said to have {\it stiffness\/} which we can characterize as  a force proportional to its extension from static equilibrium.  

\subsection*{Modeling Activities}

\begin{enumerate}
\item Identify the one force acting on the mass shown in Figure~\ref{fig:3-101-FBD} and if $y$ is positive and the spring is extended downward then in what direction is this force? If $y$ is negative and the spring is compressed upward then   in what direction is this force?
\item Complete the differential equation (\ref{eq:first}) using this force in Newton's Second Law of Motion (again, here $m$ is the mass):
 
 \begin{equation}\label{eq:first}
m\cdot y''(t)  = \, \rule{3 true cm}{.2 pt}
\end{equation}

\item Write out an initial value problem using (\ref{eq:first})  if the spring's initial displacement from static equilibrium is $y_0$ and its initial velocity is $v_0$.
\end{enumerate}


You now have a mathematical model of a mass bobbing up and down on a spring in which no resistance to the motion is assumed, i.e. there is no damping. Often, we jokingly say the spring was purchased from the ``Ideal Spring Company''   because without damping, the mass will never return to a position of static equilibrium once released.

We shall study this simple spring model in some depth before moving on to a spring mass model which experiences damping and  eventually comes to rest.

\begin{enumerate}
\setcounter{enumi}{3}
\item If we believe our model without damping will just oscillate up and down forever without any decay then what kind of function(s) does that suggest for our vertical displacement of the mass, $y(t)$? Why, some kind of sine or cosine function, perhaps something like $y(t) = A \sin(\omega t)$ or $y(t) = B \cos(\omega t)$.  But if we were to pick the sine function, then since it always has $y(0) = 0$ we could never model a mass with a non-zero initial displacement, so we would also need the cosine function in our solution. Similarly, if we were only to use the cosine function then since the derivative of cosine is sine we would always have $y'(0) = 0$ which would limit our model as well. So why not have a little of each, i.e. $a$ of  $\sin(\omega t)$ and $b$ of $\cos(\omega t)$. Thus our candidate or conjectured solution for the vertical displacement from static equilibrium for our mass on a spring without resistance might look like
 
\begin{equation}\label{eq:tot} 
y(t) = a \sin(\omega t) + b \cos(\omega t)\, .
\end{equation}

One way to tell how good this conjecture is would be to just check it by substituting it into your differential equation model you built from (\ref{eq:first}). Go ahead and do that to see where it leads. Remember you are trying to figure out if (\ref{eq:tot}) is a solution and then if it is a solution what might $a$, $b$, and $\omega$ have to be to fulfill the initial conditions $y(0) = y_0$ and $y'(0) = v_0$?
\item What information does $k$ and $m$ give you in your confirmed conjectured solution?  Explain.
\item Let us take an actual spring with a mass and use your model with some initial conditions to find a complete solution. Use $m = 3$ gram, stiffness coefficient or spring constant $k = 27$ dyne/cm, $y(0) = 1$ cm and $y'(0) = 2$ cm/sec. Units are in grams, centimeters,  seconds, and dynes appropriately. Solve the differential equation in the above manner and graph its motion over  a reasonable interval of time. Explain what you see.
\item If you have access to a computer algebra system then solve the differential equation with the appropriate initial conditions and compare that solution with your solution from the method above.
\end{enumerate}

\subsection*{Supporting Theory}

There is supporting theory (called existence and uniqueness theorems) in mathematics which says that there will be a solution and if we find a candidate solution for our differential equations which also satisfies the initial conditions then this solution is the unique solution and we need look no further. Thus our intuition has helped us to find the unique solution. We shall develop other means of finding ``candidate'' solutions in our further studies.

\teacher{
\section*{NOTES FOR TEACHERS}


We attempt to permit the intuitive notion students possess of what happens when we ``release" an ideal spring, namely it oscillates. And how do we model this?  We do this with trigonometric functions, sine and cosine. We attempt to lead the students to this and then let them loose on what the consequences are, namely, what the coefficients have to be to meet the initial conditions.  We only briefly mention the existence and uniqueness guarantees of theoretical mathematics to support student confidence that once we have a function which satisfies both the differential equation itself and the initial conditions then we have THE solution in which we can be confident.

This approach is more supportive than the traditional ``guess" a solution to a simple case of a pure oscillator  in which there is only sine or cosine in the solution. This approach gets students set for combinations of sine and cosine functions used in building a general solution. Here we do not entertain combining these two solution strategies, nor do we employ the ``trick'' of guessing $y(t) = e^{r t}$ which will eventually lead to the characteristic (quadratic) equation for the second order constant coefficient differential equation - not yet.


We offer an analysis using Mathematica in a notebook 3-101-Mma-T-SpringMassFirstTry-Teacher Version.nb and accompanying pdf file for those who do not use Mathematica. But any computer algebra system will do these operations and analyses.

We use this FirstTry approach in the case of no damping and then we move on to the case where there is damping. However, another approach might be to move on to resonance and beats, which play out in the pure oscillator case with no damping, whereas in the case with damping one needs to do a fuller analysis of frequency response, which we believe important as well.

}



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