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\usepackage{tikz}
%\usepackage{pgfplots}
\usepackage{circuitikz}
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\newcommand{\LRI}{\quad\Leftrightarrow\quad}
\newcommand{\LI}{\quad\Leftarrow\quad}
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\newcommand{\tipl}{\mbox{{\tt {\tiny +}}}}
\newcommand{\timi}{\mbox{{\tt {\tiny -}}}}
\newcommand{\tizero}{\mbox{{\tiny\tt{0}}}}
\newcommand{\tione}{\mbox{{\tiny \tt{1}}}}
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\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\LC-Circuits, Beats, and Wireless Telegraphy}


\markboth{LC-Circuits}{Beating and Wireless Telegraphy}

\author{Gabriel Nagy\\
Michigan State University\\
East Lansing MI  48824 USA\\
gnagy@msu.edu
}

\makeStitlePDFLaTex


\begin{abstract}
This project has three parts, the first part to be done at home, the last two parts in class. In the home part we recall how to solve second order, linear, differential equations with constant coefficients and simple source functions. In the first in-class part we analyze the solutions found at home and try to understand what is a resonant solution, the beats phenomenon, and how both arise. The last in-class part is to apply the beats phenomenon to send an audible tone across the Atlantic Ocean using radio waves.
\end{abstract}

\teacher{
\section*{NOTES FOR TEACHER}

This project is for students who know how to solve second order linear differential equations with simple source functions. The project starts with a pre-project activity to be done at home by the students. They need to compute the current circulating in an $LC$-series circuit---there is no resistance---with a sinusoidal voltage source. The equation for the current $y(t)$ of such circuit is
\[
y''(t) + \omega_{\tizero}^2 \, y(t) = c_{\tizero} \cos(\nu t),
\]
where $\omega_{\tizero}$ is called the {\em natural frequency}, and $\nu$ is called the {\em driver frequency}. In class we briefly go over the solution formulas and we introduce these names. The home part of the project should be a refresher of concepts already covered in class.

The project really begins with the in-class parts. We start analyzing the graph of the solutions found at home as the driver frequency approaches the natural frequency. This is when students {\em should understand resonance and discover beats}. They should realize that beats actually happens when the solution is the difference of two cosine functions with similar frequencies. Finally, in the last part of the project students should use the insight learned about beats to transmit an audible pulses through large distances using radio waves. 

The second and third parts are supposed to be done in class, with the students working in groups of three, maybe four. The students need to open an {\em interactive graph} which graphs the different solutions the students should have found at home. The interactive graph has a free parameter---the driver frequency---that can be varied in a certain range, from zero to close to the natural frequency. Students can see what the resonant solution looks like, what the non-resonant solution looks like, and how the non-resonant solution approaches the resonant one as the driver frequency approaches the natural frequency.

The interactive graph does not require any paid software, such as Mathematica or Maple. The interactive graph uses a free software that works in any browser, MathStudio, \href{http://mathstud.io}{{\bf http://mathstud.io}}. MathStudio is a computational software for iPhone, iPad, Apple Watch, Mac and any web browser. It feels like a light version of Mathematica or Maple. Since it works in any web browser, students do not need to install additional software in their computers. The link itself, provided later on in this project, contains all the instructions in Javascript needed to create the code we run. The browser itself performs the calculations. And the calculations are not intensive; they can be carried out in any browser, in any laptop, and in any cellphone. 

The third part of the project is a simple application of the beats phenomenon containing only trivial algebraic calculations. The idea is to rediscover how people sent the first Morse code audible tones across long distances using radio waves. Simple algebraic calculations show that to receive radio waves at the human hearing frequencies requires a 100 km long antenna. That is why people used carrier waves of high frequency---high enough to be received by a 1 meter antenna---so that they have beats on the human audible range. Summarizing:
\begin{itemize}
\item
This is a project for groups of three or four students.

\item
It requires one day of class, and may be a week of calculations done {\em before} the class meets.

\item
In the home part, the students compute solutions to second order linear differential equations. In the class part, students answer a series of qualitative short answer questions.

\item
All the data in this project is generated in the interactive graphic software, no data input is needed.

\item
Upon completion of the project the students hand in a finished worksheet, one per group, or one per student, as the teacher prefers.

\item
To assess the project the teacher should grade the qualitative content of the written work.
\end{itemize}

{\cb\em In Problems, material written in this blue italics font are for Teachers only.}
}

\section*{SCENARIO DESCRIPTION}

This project has three parts, the first part is done at home and the other two parts are done in class. The home part is purely computational---we solve second order linear differential equations having constant coefficients with and without sources. In the second part, done in class, we analyze the solutions found at home. We try to visualize in an interactive graph the resonant solution. We also try to understand what is the beats phenomenon and how it arises. In the last part, also done in class, we apply what we have learned about the beats phenomenon to solve a new problem---to use radio waves to send an audible tone across the Atlantic Ocean.

\section*{Part 1 (at home): Finding the Electric Current in an LC-Circuit}

\vspace{-0.3cm}
\begin{figure}[!h]
\begin{minipage}{0.49\textwidth}
In this project we study the current oscillations in an $LC$-series circuit, which is an electric circuit containing a inductor $L$ and a capacitor $C$. We assume that there is no resistance in the circuit, i.e. $R=0$. We also introduce in the circuit a time-dependent input voltage $V(t)$. See Figure~\ref{F:Circuit}. 

\end{minipage}
\begin{minipage}{0.5\textwidth}
\centering


\begin{circuitikz} \draw
(0,1.5) to[L, l_=$L$] (2,1.5) to[C, l_=$C$] (3,1.5) to[sV, l_=$V(t)$] (5,1.5) -- (5,0) -- (0,0) -- (0,1.5)
;
\node [below] at (2.5,0) {$I(t) = \mbox{electric current}$};
\draw [<->,very thick] (2,0) -- (3,0);

\end{circuitikz}

%\includegraphics[height=2cm,width=5cm]{figs/so-cc-circuit}
\caption{An LC circuit.}
\label{F:Circuit}
\end{minipage}
\end{figure}

This system is described by an integro-differential equation found by Kirchhoff, now called Kirchhoff's voltage law,
\begin{equation}\label{sole-a-c1}
L \,I'(t) + \frac{1}{C}\,\int_{t_{\tizero}}^{t} I(s)\,ds =V(t).
\end{equation}
If we take one time derivative in the equation above we obtain a second order differential equation for the electric current,
\[
L \,I''(t) + \frac{1}{C}\, I(t) = V'(t).
\]
This equation is usually rewritten as
\[
I''(t) + \frac{1}{LC}\, I(t) = \frac{V'(t)}{L}.
\]
If we introduce the {\em natural frequency} $\omega_{\tizero} = \dfrac{1}{\sqrt{LC}}$, then Kirchhoff's law can be expressed as 
\begin{equation}\label{sole-a-c2}
I''+ \omega_{\tizero}^2\,I =\frac{V'(t)}{L}.
\end{equation}
We are now ready to solve the following example.

\vspace{0.4cm}
\noindent
{\bf Problem 1: Homogeneous Case (No Source).}

\noindent
Consider an $LC$-series circuit with capacitor $C$, inductor $L$, but no voltage source, i.e. $V(t) = 0$ for all $t$. 
\begin{enumerate}[{\bf ({1}a)}]
\item
Write the Kirchhoff's equation for the electric current $I$ of this circuit.
\item
Find all possible solutions for the electric current.
\end{enumerate}

\vspace{0.4cm}
\teacher{{\cb\em
\noindent
The Kirchhoff equation for this problem is
\[
I'' + \omega_{\tizero}^2 I = 0.
\]
The characteristic equation is $r^2+\omega_{\tizero}^2 =0$, the roots are $r_{\pm} = \pm \omega_{\tizero} i$, so fundamental solutions are
\[
I_{\tione} = \cos(\omega_{\tizero}t),
\qquad
I_{\titwo} = \sin(\omega_{\tizero}t).
\]
Therefore, all solutions of the homogeneous differential equation above are
\[
\cb
I(t) = c_{\tione} \cos(\omega_{\tizero}t) + c_{\titwo}  \sin(\omega_{\tizero}t),
\]
where $c_{\tione}$, $c_{\titwo}$ are arbitrary constants.
}}



\vspace{0.4cm}
\noindent
{\bf Problem 2: Non-Homogeneous Resonant Case.}

\noindent
Consider an $LC$-series circuit with capacitor $C$, inductor $L$ and voltage source 
\[
V(t) = V_0\,\sin(\omega_{\tizero} t), 
\qquad
V_0 \neq 0,
\qquad 
\omega_{\tizero} = \frac{1}{\sqrt{LC}}.
\]
If the electric current $I$ flowing in this circuit satisfies the initial conditions 
\[
I(0) =0,
\qquad 
I'(0)=0,
\]
then show that such current is given by a solution of equation~(\ref{sole-a-c2}),
\[
I(t) = \frac{v_{\tizero} \,t}{2} \sin(\omega_{\tizero}t),
\qquad
v_{\tizero} = \dfrac{V_{\tizero}}{L}.
\]

\noindent
{\bf Note:} The frequency in the Voltage source $V(t)$ is the natural frequency of the circuit, $\omega_{\tizero}$.



\vspace{0.4cm}
\teacher{{\cb\em
\noindent
The Kirchhoff equation for this problem is
\[
I'' + \omega_{\tizero}^2 I = f(t)
\]
where the source function has the form
\[
f(t) = v_{\tizero} \, \omega_{\tizero} \, \cos(\omega_{\tizero} t)
\]
and we denoted $v_{\tizero} = \dfrac{V_{\tizero}}{L}$. We know from Problem 1 that the fundamental solutions to the homogeneous problem are
\[
I_{\tione} = \cos(\omega_{\tizero}t),
\qquad
I_{\titwo} = \sin(\omega_{\tizero}t).
\]
We can now use the variation of parameters method to find a particular solution of the non-homogeneous problem using the formula
\[
I_p(t) = u_{\tione}(t) \, I_{\tione}(t) + u_{\titwo}(t)\, I_{\titwo}(t).
\]
We know that the $u_{\tione}$ and $u_{\titwo}$ are given by
\[
u_{\tione}'(t) = - \frac{I_{\titwo}(t)\, f(t)}{W(t)},
\qquad
u_{\titwo}'(t) =  \frac{I_{\tione}(t)\, f(t)}{W(t)},
\]
where we have introduced the Wronskian 
\[
W(t) = I_{\tione}(t) \, I_{\titwo}'(t) - I_{\tione}'(t) \, I_{\titwo}(t).
\]
It is not hard to see that in our case $W(t) = \omega_{\tizero}$, and 
\[
u_{\tione}(t) = \frac{v_{\tizero}}{4\omega_{\tizero}}\cos(2\omega_{\tizero}t)
\qquad
u_{\titwo}(t) = \frac{v_{\tizero}}{4\omega_{\tizero}}\sin(2\omega_{\tizero}t) + \frac{v_{\tizero}}{2}\, t.
\]
Therefore, {\bf\em after some lengthy algebra} we get
\[
I_p(t) = \frac{v_{\tizero}}{4\omega_{\tizero}} \cos(\omega_{\tizero} t) + \frac{v_{\tizero} \,t}{2} \sin(\omega_{\tizero}t).
\]
Since the first term above is a solution of the homogeneous equation, we can always choose
\[
I_p(t) = \frac{v_{\tizero} \,t}{2} \sin(\omega_{\tizero}t).
\]
Therefore, the general solution for the non-homogenous problem is
\[
I(t) = c_{\tione} \cos(\omega_{\tizero}t) + c_{\titwo}  \sin(\omega_{\tizero}t) + \frac{v_{\tizero} \,t}{2} \sin(\omega_{\tizero}t).
\]
It is simple to see that the initial condition $I(0) =0$ implies $c_{\tione} =0$, and then the other condition $I'(0)=0$ imples $c_{\titwo}=0$. So the solution of the initial value problem is $I_p$, that is,
\[
I(t) = \frac{v_{\tizero} \,t}{2} \sin(\omega_{\tizero}t).
\]
}} 



\vspace{0.5cm}
\noindent
{\bf Problem 3: Non-Homogeneous Non-Resonant Case.}

\noindent
Consider an $LC$-series circuit with capacitor $C$, inductor $L$ and voltage source 
\[
V(t) = V_0\,\sin(\nu t),
\qquad
\nu \neq \omega_{\tizero} = \dfrac{1}{\sqrt{LC}}.
\]
If the electric current $I$ flowing in the electric circuit satisfies the initial conditions
\[
I(0) =0,
\qquad 
I'(0)=0,
\]
then show that such current is given by
\[
I(t) = \frac{v_{\tizero}\nu}{(\omega_{\tizero}^2-\nu^2)} \bigl(\cos(\nu t) - \cos(\omega_{\tizero} t)\bigr),
\qquad
v_{\tizero} = \dfrac{V_{\tizero}}{L}.
\]


\noindent
{\bf Note:} The frequency $\nu$ in the Voltage source $V(t)$ is different from the natural frequency $\omega_{\tizero}$ of the circuit.

\vspace{0.5cm}
\teacher{{\cb\em
\noindent
The Kirchhoff equation for this problem is
\[
I'' + \omega_{\tizero}^2 I = v_{\tizero} \nu \, \cos(\nu t)
\]
where we denoted $v_{\tizero} = \dfrac{V_{\tizero}}{L}$. 
We know from Problem 1 that the general solution of the homogeneous problem is
\[
\cb
I_h(t) = c_{\tione} \cos(\omega_{\tizero}t) + c_{\titwo}  \sin(\omega_{\tizero}t),
\]
where $c_{\tione}$, $c_{\titwo}$ are arbitrary constants. For $\nu \neq \omega_{\tizero}$ the source function is not solution of the homogeneous equation, so the correct guess for a particular solution of the non-homogeneous equation is
\[
I_p= c_{\tione}\cos(\nu t) + c_{\titwo}\sin(\nu t).
\]
If we put this function $I_p$ into the non-homogeneous equation we get
\[
-\nu^2 (c_{\tione}\cos(\nu t) + c_{\titwo}\sin(\nu t)) + \omega_{\tizero}^2 (c_{\tione}\cos(\nu t) + c_{\titwo}\sin(\nu t)) = v_{\tizero} \nu \cos(\nu t).
\]
If we reorder terms we get
\[
\bigl(c_{\tione}(\omega_{\tizero}^2 - \nu^2) -v_{\tizero}\nu \bigr)\cos(\nu t) + c_{\titwo} (\omega_{\tizero}^2-\nu^2)\sin(\nu t) =0.
\]
From here we get that
\[
c_{\tione}(\omega_{\tizero}^2 - \nu^2) -v_{\tizero}\nu =0
\quad\mbox{and}\quad
c_{\titwo} (\omega_{\tizero}^2-\nu^2) =0.
\]
Since we are studying the case $\nu \neq \omega_{\tizero}$, we conclude that
\[
c_{\tione} = \frac{v_{\tizero}\nu}{(\omega_{\tizero}^2-\nu^2)},
\qquad
c_{\titwo} =0.
\]
So, the particular solution is
\[
I_p(t) = \frac{v_{\tizero}\nu}{(\omega_{\tizero}^2-\nu^2)} \cos(\nu t).
\]
The general solution of the non-homogeneous equation is
\[
I(t) = c_{\tipl}I_{\tipl} + c_{\timi} I_{\timi} + I_p
\RI
I(t) = c_{\tipl}\cos(\omega_{\tizero} t) + c_{\timi}\sin(\omega_{\tizero} t) + \frac{v_{\tizero}\nu}{(\omega_{\tizero}^2-\nu^2)} \cos(\nu t).
\]
We now look for the solution that satisfies the initial conditions $I(0) =0$, and $I'(0)=0$. For the first condition we get
\[
0 = I(0) = c_{\tipl} + \frac{v_{\tizero}\nu}{(\omega_{\tizero}^2-\nu^2)}
\RI
c_{\tipl} = -\frac{v_{\tizero}\nu}{(\omega_{\tizero}^2-\nu^2)}.
\]
The other boundary condition implies
\[
0 = I'(0) = c_{\timi}\, \omega_{\tizero}
\RI
c_{\timi} =0.
\]
So, the solution of the initial value problem for the electric current is
\[
I(t) = \frac{v_{\tizero}\nu}{(\omega_{\tizero}^2-\nu^2)} \bigl(\cos(\nu t) - \cos(\omega_{\tizero} t)\bigr).
\]
}}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Part 2 (in class): Graphical Analysis of the Solutions, Resonance, and Beats}

Consider an $LC$-circuit where 
\[
v_{\tizero} = 1, 
\qquad
\omega_{\tizero} =5,
\qquad
\nu \neq 5.
\]
In this case we have that the solution of Problem 2, called the {\bf Resonant} solution, is
\begin{equation}\label{res-sol}
I_R(t) = \frac{t}{2} \sin(5t),
\end{equation}
while the solution of Problem 3, called the {\bf Non-Resonant} solution, is
\begin{equation}\label{nonres-sol}
I_{NR}(t) = \frac{\nu}{(25-\nu^2)} \bigl(\cos(\nu t) - \cos(5t)\bigr).
\end{equation}
Notice that the Non-Resonant solution is the {\bf sum} of two solutions, $I_{NR} = I_p + I_h$, where 
\begin{equation}\label{nonres-sol-ab}
I_p(t) = \frac{\nu}{(25-\nu^2)} \cos(\nu t),
\qquad
I_h(t) = - \frac{\nu}{(25-\nu^2)} \cos(5t).
\end{equation}
We call $I_p$ the {\bf Particular} solution and $I_h$ the {\bf Homogeneous} solution.



\vspace{0.4cm}
\noindent
{\bf Problem 4.}

\noindent
An oscillatory function has {\bf beats} when the function has a periodic modulation in amplitude with frequency smaller than the function frequency, as shown in the Figure~\ref{F:Beating}.

\begin{figure}[h!]
\centering
\begin{tikzpicture}[xscale=1,yscale=1]

\draw [<->,thick] (0,2.5) -- (0,0) -- (6.75,0);
\draw [-,thick] (-0.5,0) -- (0,0);
\draw [-,thick] (0,0) -- (0,-2.25);
\node [below right] at (6.75,0) {$t$};

\node [blue, left] at (0,2.5) {$I(t)$};

%\node [teal, right,] at (5.3,1) {$e^{-\omega_{d}t}$};

\draw [blue,domain=0:6,very thick, samples=200] plot(\x, {(2/3)* (cos(15*\x r) - cos(17*\x r) )});

\end{tikzpicture}
\caption{The $I$ for $\nu$ close to $\omega_{\tizero}$, showing the beats phenomena.}
\label{F:Beating}
\end{figure}

\begin{figure}[h!]
\begin{minipage}{0.5\textwidth}
In order to answer the questions below, recall the solutions names given above: 
\begin{itemize}
\item
$I_R$ in (\ref{res-sol}) is the {\bf Resonant} solution;
\item
$I_{NR}$ in (\ref{nonres-sol}) is the {\bf Non-Resonant} solution; 
\item
$I_p$ and $I_h$ in (\ref{nonres-sol-ab}) are the {\bf Particular} and {\bf Homogeneous} solutions respectively. 
\end{itemize}
\end{minipage}
\begin{minipage}{0.49\textwidth}
\centering
\includegraphics[height=2.5cm, width=6cm]{InteractiveGraph}
\caption{Picture of the interactive graph.}
\label{F:igs}
\end{minipage}
\end{figure}

These solution names appear at the bottom of the interactive graph, which at the start will look as in Figure~\ref{F:igs}. When you click on these buttons in the interactive graph, the corresponding solution will show up, {\bf\color{DarkBlue} Resonant} in blue, {\bf\color{Purple}Non-Resonant} in purple, {\bf\color{DarkRed}Particular} in red, and {\bf\color{DarkGreen}Homogeneous} in green. The last button on the bottom right in Figure~\ref{F:igs} is a slider that changes the value of the driver frequency $\nu$ in the interval $[0,5]$, where $5$ is the natural frequency of the system.

Now click on the following Interactive Graph Link: 

\begin{center}
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}{{\bf Beats Phenomenon.}} 
}
\end{center}

\begin{enumerate}[{\bf(4a)}]
\item
In the {\bf interactive graph link} above turn {\bf on} the {\bf\color{DarkBlue} Resonant} solution. 
\begin{enumerate}[\bf ({4a}i)]
\item
Look at the graph of the resonant solution $I_R$, and describe what happens with the amplitude of the solution as time grows. 
\item
Identify the precise part in the formula for $I_R$ that produces the behavior you observed in the graph.
\end{enumerate}

\vspace{0.3cm}
\teacher{{\cb\em
\begin{enumerate}[\bf ({4a}i)]
\item
As $t$ grows, we see in the graph that the amplitude grows without bounds, that is, diverges as time grows. This divergence is linear in $t$. 
\item
This behavior in the graph of $I_R$ is due to the factor $\dfrac{t}{10}$, the amplitude of the oscillation, which grows linearly in time and diverges at time grows.
\end{enumerate}
}}

\item
Let's go back at the behavior of the Resonant solution as $t$ grows. 
\begin{enumerate}[\bf ({4b}i)]
\item
What does that behavior in $I_R$ mean for the current in the circuit? Are the electrons in the circuit moving at the same speed, or slower, or faster as $t$ grows? 
\item
Do you think that this mathematical model predicts accurately the behavior of an actual physical circuit as $t\to\infty$? Why? 
\item
How would you improve the mathematical model to get a better description of an electrical circuit for large $t$?
\end{enumerate}

\vspace{0.3cm}
\teacher{{\cb\em
\begin{enumerate}[\bf ({4b}i)]
\item
The current in the circuit is made of electrons moving through the wires. Since the amplitude of the current increases with time, that means the electrons must be moving faster and faster as time grows. In the limit $t\to\infty$ the model predicts the electrons moving infinitely fast. 
\item
This behavior as $t\to\infty$ in the mathematical model does not happen in the physical system. A physical variable diverging to infinity only occurs in mathematical models, not in nature. The appearance of an infinity in a mathematical model means that at least one of the hypotheses in the model does not hold anymore. In the case of the circuit, the electrons cannot move infinitely fast. They would if there were no resistance at all in the circuit, as we assumed in the mathematical model. But there always is a bit of resistance in the cables. And that resistance becomes important when the electrons move fast enough. Then, nontrivial amounts of heat are produced by the electrons moving at high speeds and they slow down. This heat is not taken into account in our mathematical model.
\item
We could improve the accuracy of the mathematical model at large $t$ if we include a small resistor in the circuit to account for the resistance in the cables themselves. So we would need to study an RLC-Series circuit.
\end{enumerate}
}}


\item
Go back to the {\bf interactive graph link} above. Keep the Resonant solution on and now turn {\bf on} the {\bf Non-Resonant} solution. Vary the driving frequency {\bf nu} from zero to five. Describe what you see. Why is {\bf nu} equal five so special?

\vspace{0.3cm}
\teacher{{\cb\em
When the driving frequency $\nu$ approaches five the Non-Resonant solution approaches the Resonant solution. The number five is special, because we chose the natural frequency of the $LC$-circuit to be five, i.e. $\omega_{\tizero} = 5$.
}}


\item
Go back to the {\bf interactive graph link} above. Turn off the Resonant solution and turn {\bf on} the {\bf Non-Resonant} solution. Vary the driving frequency {\bf nu} from 0 to 5. Find the minimum value of the driving frequency {\bf nu} such that the solution $I_{NR}$ displays beating.

\vspace{0.3cm}
\teacher{{\cb\em
The minimum driving frequency such that beating appear in $y_{\rm NonResonant}$ is $\nu =4$. Correct answers are accepted in the interval $[4, 4.4]$.
}}

\item
Go back to the {\bf interactive graph link} above. Keep the Non-Resonant solution on, and now turn {\bf on} both the {\bf Particular} and the {\bf Homogeneous} solutions. Vary the driving frequency {\bf nu}. Based on what you see, describe in your own words why the beating phenomenon occurs.

\vspace{0.2cm}
\noindent
{\bf Note:} Feel free to experiment with the graph. Turn on and off different functions, move the driving frequency. Play around with the graph.

\vspace{0.3cm}
\teacher{{\cb\em
The Non-Resonant solution, in purple, is the {\bf\em sum} of the particular and the Homogeneous solutions. When the driving frequency gets close to the natural frequency periodic regions form where $I_p \sim I_h$, and other regions where $I_p \sim - I_h$. In the former regions the $I_{NR}$ is large, in the latter regions is small. This behavior is periodic. This is beating.
}}


\item
The beating phenomena occurs when there is a sinusoidal behavior that satisfies the superposition property. This may happen with waves propagating in a medium; for example sound---air pressure waves propagating through the air. If there is beating in a sound wave, what would it sound like?

\vspace{0.3cm}

\teacher{{\cb\em
If my voice has a given frequency and I increase and decrease the volume in a periodic fashion, this would be similar to beating.
}}


\item
Watch the short video ``Beat Frequency'', from SMUPhysics,

 \begin{center}
 \url{https://www.youtube.com/watch?v=V8W4Djz6jnY}.
\end{center}

What are the frequencies of the two tuning forks? Explain why beating occurs.

\vspace{0.3cm}

\teacher{{\cb\em
One fork has a frequency of 250Hz and the other has a frequency of 238Hz. These frequencies are sufficiently close to cause beating.
}}


\end{enumerate}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Part 3 (in class): Application of Beating: Wireless Transmission of an Audible Tone}

Imagine this is 1887. Heinrich Hertz has just shown that radio waves are real, not something that James Maxwell had imagined in 1861 when he modified Michael Faraday's equations of electromagnetism. Maxwell modified Faraday's laws so that the electric charge would be conserved. Faraday's laws predicted electric charge creation from nothing. Maxwell equations fixed that. But many people doubted Maxwell, because he changed the equations without any experimental data. Even more, this modification also predicted the existence of radio waves and hypothesized that light is a radio wave. Maxwell's modification was a matter of heated debate, until Hertz silenced everybody when he showed that radio waves actually exist. Maxwell became a hero overnight; unfortunately he didn't live to see it.

Hertz transmitted information without the need of any cables. Up to then, communications used wires. People even laid submarine cables across the Atlantic ocean as early as 1858. However, no communications with ships at sea were possible. As soon as Hertz transmitted radio noise without cables, people wanted to use radio waves to transmit Morse code as audible sounds without cables across long distances. In this way you wouldn't need expensive submarine cables and you could reach ships at sea. We now summarize a few facts needed for the last questions.

\begin{itemize}
\item
Humans are able to hear frequencies in the range 20 Hz to 20,000 Hz. 
\item
Radio waves are electromagnetic waves that travel at the speed of light, $c$, given by
\[
c = 300,000 \; {\rm km/sec}.
\]
\item
The frequency $\gamma$ and the wavelength $\lambda$ of a radio wave satisfy the relation
\[
\gamma \lambda = c.
\]
So, given the frequency one can compute the wavelength, and vice versa. 
\item
To capture a radio signal of wavelength $\lambda$ one needs an antenna of length no smaller than $\lambda/3$.
\end{itemize}

%\vspace{0.2cm}
\noindent
{\bf Problem 5:}

\begin{enumerate}[{\bf (5a)}]
\item
It turns out that antennae that capture radio waves in the audible wavelength range are too long to be built. Find out how long an antenna must be so that we can capture radio waves of 1 kHz.

\teacher{{\cb\em
The wavelength of a $\gamma_{\tizero} =1000$ Hz radio wave is
\[
\lambda_{\tizero} = \frac{c}{\gamma_{\tizero}} = 300 \; {\rm km}.
\]
The smallest antenna to capture such a radio wave must be $\ell_{\tizero} = \lambda_{\tizero}/3 = 100$ km.
}}

\item
Radio waves that can be transmitted and captured by small antennae do not have audible frequencies. Find the minimum frequency $\gamma_{\tione}$ of a radio wave that can be captured by an antenna of $\ell_{\tione} = 1$ meter.

\teacher{{\cb\em
A one meter long antenna can capture radio waves $3$ meters length or shorter, that is $\lambda_{\tione} \leqslant 3$ meters. That means that the radio wave frequency is
\[
\gamma_{\tione} = \frac{c}{\lambda_{\tione}} \geqslant  100,000,000 \; {\rm Hz} = 100 \;{\rm MHz}.
\]
So the frequency of the radio wave must be $\gamma_{\tione} \geqslant 100$ MHz.
}}


\item
Use what you have learned about beating to send an audible tone using radio waves that can be captured by a 1 meter antenna. How many radio waves do you need to transmit? What should their frequencies be?

\teacher{{\cb\em
Use two radio waves with frequencies close to $100$ MHz. One frequency is $\gamma_{\tione} = 100$ MHz. The second radio wave has a frequency 1 kHz apart from $\gamma_{\tione}$, for example $\gamma_{\titwo} = 100.001$ MHz. The addition of the two frequencies will have a beating phenomena of about $1$ kHz. The beating will be in the audible range.
}}


\item
Can you think of a way to transmit an audible tone by transmitting only one radio wave?

\teacher{{\cb\em
Beating requires a minimum of two signals. If transmitter and receiver agree on what is the base signal, say $\gamma_{\tione} = 100$ MHz, then the transmitter needs to send only one radio wave, the one with $\gamma_{\titwo} = 100.001$ MHz. The receiver will combine the receiver wave with the $\gamma_{\tione}$ generated locally at the receiver to get the beating and so the audible information.
}}



\end{enumerate}

\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Appendix: MathStudio Code}

We reproduce here the MathStudio code used to create the interactive graph {\bf Beats Phenomenon} in Part 2. If the link shown in Part 2 does not work, use any browser to go to MathStudio, \href[pdfnewwindow=true]{http://mathstud.io}{{\bf http://mathstud.io}} and then type the code below.

\vspace{0.2cm}
{\tt
\noindent
/*\\
No Damping\\
Beats for nu close to omega\\
And Resonance for nu = omega\\
*/ \\

\noindent
Button(Resonant,[0,2,4],0)\\
Button(NonResonant, [0,2,4],0)\\
Button(Particular,[0,2,4],0)\\
Button(Homogeneous,[0,2,4],0)\\

\noindent
slider(nu,0..4.999->0.001,0)\\
omega = 5\\

\noindent
yp(x) =  nu/(omega$^{\wedge}$2-nu$^{\wedge}$2)*cos(nu*x)\\
yh(x) = -nu/(omega$^{\wedge}$2-nu$^{\wedge}$2)*cos(omega*x)\\
yt(x) = yh(x) + yp(x)\\

\noindent
yr(x)=(x/2)*sin(omega*x)\\

\noindent
Plot(yp(x),color=red,linewidth=Particular)\\
Plot(yh(x),color=green,linewidth=Homogeneous)\\

\noindent
Plot(yt(x),color=purple,linewidth=NonResonant)\\
Plot(yr(x),color=blue, x=[-30,30],y=[-4,4], height=600, width=1400, linewidth=Resonant)
}

%\begin{thebibliography}{98}

%\bibitem{Hutchinson1978} Hutchinson, G. Evelyn. 1978. {\it Introduction to Population Ecology\/}. New Haven CT:  Yale University Press.


%\end{thebibliography}


\end{document}
