\documentclass[%
 reprint,
 amsmath,amssymb,
 aps,
]{revtex4-2}

\usepackage{graphicx}% Include figure files
\usepackage{dcolumn}% Align table columns on decimal point
\usepackage{bm}% bold math
\setlength{\parskip}{\baselineskip}%

\usepackage[inline]{asymptote}
\usepackage{multirow}
\usepackage{tikz}
\usepackage{pgfplots, pgfplotstable}

% Data
\pgfplotstableread{
X Y
0.184 2.07
0.384 3.73
0.435 4.27
0.485 4.70
0.702 7.22
}\datatable

\pgfplotstableread{
X Y
2.449 12.2
2.196 13.5
1.645 17.3
}\omegatable


\begin{document}

\preprint{APS/123-QED}

\title{Acceleration on an Incline: Eddy Current Breaking and Magnetic Oscillations}% Force line breaks with \\

\author{QiLin Xue}

\date{\today}% It is always \today, today,
             %  but any date may be explicitly specified

% \begin{abstract}
% An article usually includes an abstract, a concise summary of the work
% covered at length in the main body of the article.
% \end{abstract}

%\keywords{Suggested keywords}%Use showkeys class option if keyword
                              %display desired
\maketitle

%\tableofcontents

\section{Observations}
Using the markings on the aluminum track, the height of the track at the $1 \text{ m}$ mark was measured and recorded
\begin{table}[h!]
\centering
\begin{tabular}{|c|c|c|}
\hline

Height of Ramp ($m$) & Height $\div$ Length & Acceleration ($\mathrm{m/s^2}$) \\ \hline
$0.184$ & $0.184$ & $2.07 \pm 0.0018$ \\ \hline
$0.384$ & $0.384$ & $3.73 \pm 0.0070$ \\ \hline
$0.435$ & $0.435$ & $4.27 \pm 0.0052$ \\\hline
$0.485$ & $0.485$ & $4.70 \pm 0.0088$ \\\hline
$0.702$ & $0.702$ & $7.22 \pm 0.0088$ \\ \hline
\end{tabular}
% \caption{Measured acceleration of the cart at gradually increasing angles of inclination.}
\end{table}

These points are plotted below, and they show a strong linear relationship.
\begin{figure}[h!]
\centering
\begin{tikzpicture}
\begin{axis}[legend style={at={(1,1)},anchor=south east},
xlabel={height $\div$ length},
ylabel={acceleration}
]
\addplot [only marks, mark = *] table {\datatable};
\addplot [thick, red] table[
    y={create col/linear regression={y=Y}}
] % compute a linear regression from the input table
{\datatable};
\addlegendentry{%
$\pgfmathprintnumber{\pgfplotstableregressiona} \cdot x
\pgfmathprintnumber[print sign]{\pgfplotstableregressionb}$}
\end{axis}
\end{tikzpicture}
% \caption{Plot of the acceleration of the cart at varying angles. The line of best fit is shown in red.}
\end{figure}

The value of the slope is:
\begin{equation}
    m =9.953
\end{equation}

\section{Discussion}
Figure \ref{fig:fbd} shows the free body diagram for the cart. Although there is static friction exerting a torque on the wheels, we make the assumption that the mass of the wheel is negligible compared to the mass of the rest of the cart.
\newline
\begin{figure}
    \label{fig:fbd}
\centering
\begin{tikzpicture}[
    force/.style={draw=blue,fill=blue},
    axis/.style={densely dashed, gray,font=\small},
    m/.style={rectangle,draw=black,fill=lightgray,minimum size=1.2cm,thin},
    ]
    \begin{scope}[rotate=35]
    \node[m,transform shape] (M) {};
    \draw[axis,->] (0,0) -- (0,-2) node[right] {$+x$};
    \draw[axis,->] (M.east) -- ++(3,0) node[below] {$+y$};
    \draw[force,->] (M.east) -- ++(2,0) node[above] {$N$};
    \end{scope}
    \draw[force,->] (M.center) -- (0,-2) node[below] {$Mg$};
\end{tikzpicture}
\caption{Free Body Diagram for the Cart on an Incline with inclination $\theta$.}
\end{figure}
\section{Results}
\noindent The downwards acceleration is:
\begin{equation}
    m\ddot{x}=mg\sin\theta \implies \ddot{x}=g\sin\theta
\end{equation}
where:
\begin{equation}
    \sin\theta = \frac{\text{height}}{\text{length}}
\end{equation}

Since we are plotting $\sin\theta$ against the acceleration $\ddot{x}$, then the slope $m$ would represent the local gravitational acceleration. Our experimental value of $g=9.953 \text{ m/s}^2$ is just slightly above the accepted value of $9.81 \text{ m/s}^2$ though this number can fluctuate in different parts of the world.

Note that this fluctuation is not caused by the rotation of the Earth creating an equatorial bulge. Even though the gravitational acceleration would be higher at the poles than the equator, we cannot measure that difference with our experimental setup due to Earth being an equipotential surface after factoring in the rotation. It is this reason why a clock running near the poles and in the equator would tick at the same speed (gravitational and special relativistic time dilation cancel out!) It is also not because of slight changes in elevation as since the gravitational acceleration is proportional to $g\propto \frac{1}{r^2}$, a first order change to $r$ would not greatly affect it. What does affect the gravitational acceleration is that Earth's crust does not have a spherically uniform density and this causes areas of low and high gravity.

However, we believe that even without fluctuations, we can explain the deviation from the standard value. The slope itself has some uncertainty, and if we use the max-min method to determine the minimum possible slope of the regression line we come up with a value of $g=9.55 \text{ m/s}^2$, which shows that we are within experimental error. A full error analysis is not necessary for our purposes.

Theoretically, if the height of the ramp is zero, then the acceleration would be $\ddot{x}=g\sin 0=0$. We believe these to be caused by random errors due to measurement uncertainty. We suspect an uncertainty of $\delta = 0.005 \text{ m}$ in each measurement. This means that the minimum slope have instead been $b=0.0384-9.95(0.005)=-0.01$ instead. This means that the y-intercept of $(0,0)$ is certainly within the range that our uncertainty margins have set, with just the height measurements alone.

We do not believe that friction, air drag, or even inductive effects play a major role. If they were present, they would actually predict a lower gravitational acceleration as they are all resistive forces.

As the angle approaches $90^\circ$, the ramp is essentially in free fall where the only force acting upon it is gravity. Therefore, the acceleration would be $9.8 \text{ m/s}^2$, which is predicted by the formula: $a=g\sin\theta$, which reaches its maximum value of $g$ at $\theta=90^\circ$.
\section{Oscillations}
The PASCO Smart Car has built in magnets attached. When one cart is placed at the bottom of the ramp and the other car is let go from the top, it can act as a quick breaking system that forces one cart to undergo oscillatory motion. In this section, we will discuss the behaviour of this motion and analyze how it changes via the angle of incline.

The two magnets can be approximated as magnetic dipoles, and the force between two dipoles roughly drop as: $1/r^4$. We can roughly create a model by making connections with electrostatics by assuming magnetic monopoles exist. If they do, then the modified version of Gauss's Law tells us:
\begin{equation}
    \oint \vec{B} \cdot d\vec{A} = \mu_0q_m \implies B = \frac{\mu_0q_m}{4\pi r^2}
\end{equation}
where $q_m$ is the enclosed ``magnetic charge''. Similarly, the magnetic field due to a dipole moment would be:
\begin{equation}
    B = \frac{\mu m}{2\pi d^3}
\end{equation}
where $m$ is the dipole moment and $r$ is measured from the center of the dipole. The force on a second dipole along its long axis and whose center is $r$ away is:
\begin{equation}
    F = \frac{\mu_0 mq_m}{2\pi}\left(\frac{1}{(d-\ell/2)^3}-\frac{1}{(d+\ell/2)^3}\right)
\end{equation}
where $\ell$ is the effective distance between our imaginary monopoles at the ends. Using the first order expansion
$$(d-\ell/2)^{-3}=d^{-3}-(-3)d^{-4}(\ell/2)$$
gives us:
\begin{equation}
    F = \frac{\mu_0 mq_m}{2\pi}\left(\frac{3\ell}{2r^4}\right) = \frac{3\mu_0 Q}{4\pi r^4}
\end{equation}
where $Q\equiv 2q_m^2\ell^2$ is the quadrupole moment. Newton's second law gives:
\begin{equation}
    m\ddot{r}=-\frac{3\mu_0 Q}{4\pi r^4}+mg\sin\theta
\end{equation}
This is a second-order nonlinear ordinary differential equation with no elementary solution. However, given that the amplitude of oscillations is small, we can approximate it to first order. For convenience, let us set $r=d$ as the equilibrium location. Thus:
\begin{equation}
    (d-x)^{-4}=d^{-4}+d^{-5}(-4)(-x) = \frac{1}{d^4}+\frac{4x}{d^5}
\end{equation}
and:
\begin{equation}
    F = -\frac{12\mu_0 Qx}{4\pi d^5}-\frac{3\mu_0 Q}{4\pi d^4}
\end{equation}
Now, Newton's second law gives us:
\begin{equation}
    m\ddot{x} = -\frac{3\mu_0 Qx}{\pi d^5}-\frac{3\mu_0 Q}{4\pi d^4}+mg\sin\theta
\end{equation}
The angular frequency is:
\begin{equation}
    \label{eqn:angularw}
    \omega^2 =\frac{3\mu_0 Q}{\pi m d^5}
\end{equation}
The equilibrium distance $d$ will depend on the angle $\theta$. Balancing forces, we get:
\begin{equation}
    \label{eqn:force-balance}
    mg\sin\theta = \frac{3\mu_0 Q}{4\pi d^4} \implies d = \left(\frac{3\mu_0 Q}{4\pi mg\sin\theta}\right)^{1/4}
\end{equation}
To test our expression, we can do an order-of-magnitude check. The quadrupole moment is given by:
\begin{equation}
    Q = 2(q_m\ell)^2=2\left(\frac{B_rV}{\mu_0}\right)^2
\end{equation}
The volume $V$ is on the order of $5.0\times 10^{-6} \text{ m}^3$, residual flux density $B_r$ is on the order of $B_r=1$, and $\mu_0=1.257\times 10^{-6} \text{N/A}^2$. This gives $Q \approx 31.6 \text{ J/T}$. Solving for $d$ for an average angle of $\sin\theta\approx 0.4$ gives $d = 5.58 \text{ cm}$ which agrees with experiment. Using these numbers, we get a typical angular velocity of: $\omega = 16.7 \text{Hz}$ which also agrees with experiment.
\begin{center}
\begin{tikzpicture}
    \begin{axis}[
        axis lines = left,
        xlabel = $t (s)$,
        ylabel = {$v (m/s)$},
    ]
    \addplot [
        domain=-0:3,
        samples=100,
        color=green,
    ]
    {exp(-0.145*x) * sin(57.3*12.2*x)};
    \addlegendentry{$v=Ae^{-0.145t}\sin(12.2t)$ ($\theta = 13.8^\circ$)}
    \end{axis}
\end{tikzpicture}

\begin{tikzpicture}
    \begin{axis}[
        axis lines = left,
        xlabel = $t (s)$,
        ylabel = {$v (m/s)$},
    ]
    %Below the red parabola is defined
    \addplot [
        domain=-0:3,
        samples=100,
        color=red,
    ]
    {exp(-0.239*x) * sin(57.3*13.5*x)};
    \addlegendentry{$v=Ae^{-0.239t}\sin(13.5t)$ ($\theta = 16.5^\circ$)}
    \end{axis}
\end{tikzpicture}

\begin{tikzpicture}
    \begin{axis}[
        axis lines = left,
        xlabel = $t (s)$,
        ylabel = {$v (m/s)$},
    ]
    \addplot [
        domain=-0:3,
        samples=100,
        color=blue,
    ]
    {exp(-0.553*x) * sin(57.3*17.3*x)};
    \addlegendentry{$v=Ae^{-0.553t}\sin(17.3t)$ ($\theta = 26.8^\circ$)}
    \end{axis}
\end{tikzpicture}
\end{center}
Combining equations \ref{eqn:force-balance} and \ref{eqn:angularw}, we get the proportionality:
\begin{equation}
    \omega \propto \sin(\theta)^{5/8}
\end{equation}
We can test this experimentally by plotting the angular frequency $\omega$ against $\sin(\theta)^{5/8}$.
\begin{figure}[h!]
    \centering
    \begin{tikzpicture}
    \begin{axis}[legend style={at={(1,1)},anchor=south east},
    xlabel={$\sin(\theta)^{5/8}$},
    ylabel={$\omega$}
    ]
    \addplot [only marks, mark = *] table {\omegatable};
    \addplot [thick, red] table[
        y={create col/linear regression={y=Y}}
    ] % compute a linear regression from the input table
    {\omegatable};
    \end{axis}
    \end{tikzpicture}
    % \caption{Plot of the acceleration of the cart at varying angles. The line of best fit is shown in red.}
\end{figure}
A strong positive correlation with $r^2=0.996$ can be found, supporting our hypothesis.

However, when the experiment is performed, a damping factor is observed. As previously discussed, the cart does not experience significant friction, and we can expect the magnetic force at least in this context, to be conservative. We propose that this damping is caused by induced eddy currents creating an opposing magnetic field. Faraday's Law tells us that:
\begin{equation}
    \oint \vec{E} \cdot d\vec{\ell} =  \oint \frac{\partial \vec{B}}{\partial t} \cdot d\vec{A}
\end{equation}
There is some nonzero magnetic flux passing through perpendicular to the aluminum track as a result of the interactions of the two dipoles. This flux density will increase rapidly as the two carts come closer together. The induced current, and thus the magnetic field and the force from that current can vary greatly. However, since we are approximating the motion of the cart to first order, it would make sense to model this resistive force as being proportional to $F_d \propto -v$. Newton's second law then gives:
\begin{equation}
    m\ddot{x}=-b\dot{x} -\frac{3\mu_0 Qx}{\pi d^5}-\frac{3\mu_0 Q}{4\pi d^4}+mg\sin\theta
\end{equation}
To investigate the damping behaviour, we can ignore the constant terms as they only shift the location of the equilibrium and don't affect the velocity. This differential equation is thus in the form of:
\begin{equation}
    \ddot{x}+2\gamma\dot{x}+\omega^2x=0
\end{equation}
It can be shown that the general solution to this differential equation is:
\begin{equation}
    v(t) = e^{-\gamma t}\left(Ae^{\Omega t}+Be^{-\omega t}\right)
\end{equation}
where $2\gamma \equiv b/m$, $\omega$ is as determined before, and
\begin{equation}
    \Omega = \sqrt{\gamma^2-\omega^2}
\end{equation}
We assume $\gamma^2<\omega^2$, which means $\Omega$ has a nonzero complex component. This is a valid assumption as evident per the graphs that the system exhibits underdampening. Thus, we can rewrite our velocity as:
\begin{equation}
    v(t) = Ce^{-\gamma t}\cos(\tilde{\omega}t+\phi)
\end{equation}
where $\tilde{\omega}\equiv-i\Omega$. If we make the assumption that $\gamma^2 \ll \omega^2$ we can say that $\tilde{\omega}\approx\omega$. As a result, it is erroneous to compare the computed theoretical value of $\omega$ against the line of best curve, which actually gives $\tilde{\omega}$.

Experiment shows a typical value to be $\gamma=0.3$, which is small enough that it does not affect the general shape of the graph, when accounted for.

\end{document}
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