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\begin{document}

\title{A2 Physics, Circular Motion and Oscillation, 1-2}



\author[1]{Noah Phipps}%
\affil[1]{Affiliation not available}%


\vspace{-1em}



  
  \date{May 05, 2024}


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\tableofcontents
\newpage
\section{Circular Motion}


\subsection{Motion in a Circular Path}
Angular measurements are normally made using the SI unit of radians, where a radian is defined as the angle subtended at the centre of a circle by an arc that is equal in length to the radius. This gives the equation
\begin{equation*}
\theta = \frac{s}{r},
\end{equation*}
where $s$ is the arc length. $\theta$ is equal to the angular displacement. Also, $T=\frac{1}{f}$ and vice versa. \\\\The angular velocity, $\omega$, is the angular displacement per unit time, in rad s\textsuperscript{-1}. This is given by
\begin{equation*}
\omega = \frac{2\pi}{T} = 2\pi f.
\end{equation*}
Linear/orbital speed, $v$, is the distance moved per unit time, and is in a direction tangential to the circular path.
\begin{equation*}
v=\omega r
\end{equation*}

\subsection{Centripetal Acceleration and Force}
Due to Newton's law, there must be a constant force for an object to be travelling in a circular path. Thus, the acceleration of an object moving at a constant speed in a circular path is the \textit{centripetal acceleration}.\\\\The direction of the acceleration is dependent on the vector representing the velocity change. This is always towards the centre of the circle, and this is the \textit{centripetal force}. This is given by:
\begin{gather*}
a=\frac{v^2}{r}=\omega^2 r\\
F=\frac{mv^2}{r}=m\omega^2r.
\end{gather*}
The centripetal force is the sum of forces acting in the direction of the centre of the circle.

\subsubsection{Motion of a Conical Pendulum}
Centripetal force is given by the horizontal component of tension in the string. By equation the weight of the mass, and both components of the tension, the following can be attained:
\begin{equation*}
T\text{sin}\theta = \frac{mv^2}{r} \tag{1}\\
\end{equation*}
\begin{equation*}
T\text{cos}\theta = mg \tag{2}\\
\end{equation*}
\begin{equation*}
\frac{\text{(1)}}{\text{(2)}} = \text{tan}\theta = \frac{v^2}{rg}
\end{equation*}
\newpage
\subsubsection{Banked Tracks}
A banked track makes curving easier and allows for a greater speed. The centripetal force will be due to the horizontal components of the normal reaction and lateral friction.\\\\
The principle of banked tracks can also be applied to a banked aircraft. Assuming no air resistance, the centripetal force is given by the horizontal component of the lift force. If this is denoted as $L$, then:
\begin{equation*}
L\text{sin}\theta = \frac{mv^2}{r} \tag{1}\\
\end{equation*}
\begin{equation*}
L\text{cos}\theta = mg \tag{2}\\
\end{equation*}
\begin{equation*}
\frac{\text{(1)}}{\text{(2)}} = \text{tan}\theta = \frac{v^2}{rg}
\end{equation*}

\subsection{Motion in a Vertical Circle}
For an object travelling in a vertical circle, the weight must be considered. If it is a bucket on a string, then the centripetal force will be greatest at the top of the circle, as the force will be due to tension and weight. At the bottom, it will be lowest, as it will be the tension minus the weight. For a whirling bucket, the water will fall out if the tension becomes zero.\\\\
For a car or similar, then the circle could be considered to be a bumb, bridge, or similar. In this case, the centripetal force is due to the weight, minus the contact force. If the contact force becomes zero, the car will leave the road - this would be due to an increase in velocity.

\newpage
\section{Oscillation}
\subsection{Simple Harmonic Motion}
Oscillations are a type of periodic motion, repeating regularly in a given time interval. Mechanical oscillation requires a resultant force in the direction of the equilibrium position.\\ 
In simple harmonic motion (SHM), the restoring force is directly proportional to the displacement, and in the opposite direction - towards the equilibrium position.\\
When $x=$ displacement, and $A=$ amplitude,
\begin{gather*}
x=A\cos(\omega t)\\
\omega = \frac{2\pi}{T}=2\pi f
\end{gather*}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.49\columnwidth]{figures/shm-grpahs/shm-grpahs}
\caption{{SHM graphs%
}}
\end{center}
\end{figure}

If the condition ($t=0,x=0$) is used, then the $\sin$ function must be used when graphing velocity. In SHM, the condition $a \propto -x$ is always true. To obtain the above graphs, differentiation should be used:
\begin{gather*}
x=A\cos(\omega t)\\
\frac{\mathrm{d} x}{\mathrm{d} t}=v=-A\omega\sin(\omega t)\\
\frac{\mathrm{d}^2 x}{\mathrm{d} t^2}=a=-A\omega^2\cos(\omega t)\\
\text{Substituting in to eliminate $A$ and }\cos,\\
a=-\omega ^2x\\
\therefore a\propto-x\text{, as given before.}
\end{gather*}

\newpage
\subsubsection{Velocity and Acceleration in SHM}
At minimum displacement,
\begin{itemize}
\item $a=0$
\item $v=v_{\text{max}}$
\end{itemize}
\\
At maximum displacement,
\begin{itemize}
\item $a=a_{\text{max}}$
\item $v=0$
\end{itemize}
\\Since $a=-\omega^2x$,
\begin{equation*}
a_{\text{max}}=\omega^2A
\end{equation*}
Also,
\begin{gather*}
\sin^2\theta=1-\cos^2\theta\\
\therefore \sin(\omega t) =\sqrt{1-\cos(\omega t)}\\
\therefore v=\pm A\omega\sqrt{1-\cos(\omega t)}\\
\therefore v=\pm2\pi f\sqrt{A^2-x^2}=\pm\omega\sqrt{A^2-x^2}\\
\therefore v_{\text{max}}=\omega A
\end{gather*}


\subsection{Simple Harmonic Systems}
\subsubsection{Mass-Spring System}
The force on a spring is given by $F=k\Delta L$. For a spring with mass $m$ attached, the equilibrium position would be at the position the spring extends to. \\
If the mass is then displaced further, it will oscillate when released. If the spring position with no mass is $0$, the equilibrium position $e$, and the extension $x$, the the force is given by $F=k(e+x)-mg$.Since $ke=mg$,
\begin{equation*}
F=-kx
\end{equation*}
Using $F=ma$, it is shown that $a=-\frac{k}{m}x$. By comparing this with the equation of $a=-\omega^2x$, for a mass spring system;
\begin{equation*}
T=2\pi\sqrt{\frac{m}{k}}
\end{equation*}

\subsubsection{Simple Pendulums}
When resolving forces, the restoring force on a simple pendulum is given by $F=mg\sin\theta$. Since the displacement of the bob is along the arc of a circle, $\theta=\frac{x}{l}$, where $l$ is the length of the pendulum and $x$ is the displacement.\\
Using the small angle approximation, $F=-mg\frac{x}{l}$. Thus,
\begin{equation*}
T=2\pi\sqrt{\frac{l}{g}}
\end{equation*}


\subsection{Energy of Oscillating Systems}
Pendulums are constantly transferring energy between $E_k$ and $E_p$. A free oscillation is one in which the amplitude is constant, and the system undergoes no net change in energy;\\
\textit{The total energy of a free oscillation is the sum of the potential and kinetic energy of the system, and is constant throughout the oscillation.}\\
The kinetic energy of anything oscillating with SHM is proportional to $\sin^2(\omega t)$, and the potential proportional to $\cos^2(\omega t)$.
\begin{itemize}
\item For a mass spring system, there is elastic strain energy and gravitational potential
\end{itemize}

\subsubsection{Damping}
Damping is the deliberate removal of energy from an oscillating system, by increasing the resistive force.
\begin{itemize}
\item Light damping will result in gradual removal of energy
\item Heavy damping will make the object move slowly back to the equilibrium position with no oscillations
\item Critical damping will move the object back to the equilibrium position in the shortest time possible, with no oscillations
\end{itemize}

\subsection{Forced Vibrations and Resonance}
After an initial displacement, a system will oscillate at its natural frequency, $f_0$.\\
If there is a constant force being applied to a system, then it will have forced oscillations, in which energy is constantly inputted. the frequency of the forced oscillation is equal to the driving frequency.\\
If the driving frequency is equal to the natural frequency, the energy is transferred with maximum efficiency.\begin{itemize}
\item Amplitude will increase till energy supplied is equal to that lost by damping
\item A system will \textbf{resonate} when the driving frequency is equal to the natural frequency
\end{itemize}

Resonance can cause problems in mechanical systems, such as bridges. However it is crucial for musical instruments, microwaves, and MRI scanners. The frequency at which resonance occurs is the \textbf{resonant frequency}.\\
In a guitar, the vibration of the strings will cause the guitar body to resonate. This will cause the air to vibrate, producing a sound.

\subsubsection{Damping and Resonance}
\begin{itemize}
\item Resonant frequency decreases as damping increases
\item The resonance curve becomes less sharp as damping increases
\end{itemize}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.28\columnwidth]{figures/Damping-and-Resonance/Damping-and-Resonance}
\caption{{The effect of damping on resonance%
}}
\end{center}
\end{figure}

\selectlanguage{english}
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