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\begin{document}
\title{Singapore Physics Olympiad Training (SPOT) - Oscillations~~~}
\author{Garett Tok\and Zhiming Darren TAN}%
\vspace{-1em}
\date{\today}
\begingroup
\let\center\flushleft
\let\endcenter\endflushleft
\maketitle
\endgroup
\sloppy
\section*{Oscillations}
{\label{254337}}
\subsection*{Introduction}
{\label{342160}}
Oscillations are a common phenomenon in the mechanics of classical
objects, and the concepts involved have proven to be fundamental in our
modern understanding of physics. Mathematically, the quadratic form is
the lowest order approximation of the potential function at its local
minimum point (where the first derivative vanishes).
\par\null
In these notes, we will attempt to showcase some of the diversity and
beauty of the concepts involved, and introduce some interesting problems
and solutions. We first look at the simple harmonic oscillator, then
introduce concepts of damping (energy dissipation), resonance
(characteristic response to external forces), and normal modes
(behaviour of coupled systems).
\subsection*{1. Derivation for an idealised spring-mass
system}
{\label{448316}}
Consider a system composed of a horizontal spring attached to a block of
mass \(\)\(m\) (supported on a frictionless
flat surface). Let~\(x\) be the extension of the spring,
and let~\(k\) be the spring constant.
\subsubsection*{1.1 Using Newton's second law of motion, write down the
equation of motion for the mass in terms of~\(x\) and its
time derivatives. Solve the differential equation (there will be two
arbitrary constants involved, since this is a second order differential
equation). Plot~\(x\left(t\right)\), and label the amplitude and period
on the
graph.}
{\label{297107}}
\vskip 2in
We notice something very interesting about the expression obtained. When
an object's motion is governed by this form, we say that it is
undergoing simple harmonic motion (SHM). For an object to undergo SHM,
it must have the equation of motion in the form:
\begin{center}$\ddot{x} + \omega^2 x = 0$\end{center}
where \(\omega=\sqrt{\frac{k}{m}}\) is a real positive constant.
\subsubsection*{1.2 What if~\(\omega\) was purely imaginary? Can
you think of a physical system that would give such an equation of
motion?}\label{what-ifmathplaceholder1id-was-purely-imaginary-can-you-think-of-a-physical-system-that-would-give-such-an-equation-of-motion}
\subsubsection*{}\label{section}
\subsubsection*{\texorpdfstring{1.3 What is the relationship
between~\(\omega\) and the period of motion? Due to this
relationship, we call~\(\omega\) the angular frequency. The
``angle'' involved is the phase angle of the
oscillation.}{1.3 What is the relationship between~\(\omega\) and the period of motion? Due to this relationship, we call~\(\omega\) the angular frequency. The angle involved is the phase angle of the oscillation.}}\label{what-is-the-relationship-betweenmathplaceholder2id-and-the-period-of-motion-due-to-this-relationship-we-callmathplaceholder3id-the-angular-frequency.-the-angle-involved-is-the-phase-angle-of-the-oscillation.}
\par\null
\subsubsection*{1.4 Determine the velocity and
acceleration~\(v\left(t\right)\) and~\(a\left(t\right)\) accordingly. Plot
and label these
graphs.}
{\label{355080}}\par\null\par\null
\vskip 2in
\subsubsection*{1.5 Hence or otherwise, determine~\(v\left(x\right)\)
and~\(a\left(v\right)\). Plot these functions as
graphs.}
{\label{344993}}
\vskip 2in
\par\null
Note that the representation of the relationship
between~\(v\) and~\(x\) is a timeless map of
all the possible states of the system. The two-dimensional ``abstract
space'' of~\(\left(x,v\right)\) is known as ``phase space'' and the
system's ``trajectory'' in this space is a collection of points. Each
such point represents a possible state of the system. Thinking about
phase space can provide important insight into system behaviour.
\subsubsection*{}
{\label{344993}}
\subsubsection*{1.6 Determine the potential energy~\(U\),
and the kinetic energy~\(K\), as a function of
\(x\).}
{\label{625087}}
\vskip 2in
\subsection*{2. Derivation for a spring-mass system with
damping}
{\label{730495}}
Now consider the same scenario, where the effect of air resistance is
not neglected in the motion. The viscous medium provides a drag force
proportional to the velocity,~\(F_{drag\ }=\ -bv\).
Initial conditions:~Assume that at~\(t=0\), we
have~\(v=0\)~and~\(x=A>0\).
\subsubsection*{2.1 Write down the equation of motion of the mass in
terms of \(x\) and its time derivatives. Arrange it in a
form~ \(\ddot{x} + 2 \gamma \dot{x} + \omega^2 x = 0\). Find expressions for \(\gamma\) and
\(\omega\). Note that both~\(\gamma\)
and~\(\omega\) have dimensions of inverse
time.}
{\label{274590}}\par\null\par\null
\vskip 4in
\par\null
\subsubsection*{2.2 For the equation~ ~\(\ddot{x} + 2 \gamma \dot{x} + \omega^2 x = 0\) , there are
three qualitatively different solutions
for~\(\)\(x\), corresponding to the cases
(i)~\(\gamma > \omega\), (ii)~\(\gamma = \omega\),
(iii)~\(\gamma<\omega\). These three cases correspond to the
over-damped, critically damped, and under-damped scenarios. By recalling
how we defined~\(\gamma\), which case corresponds to which
condition?}
{\label{534836}}\par\null
\subsubsection*{2.3 Solve the equations for over-damped and critically
damped cases. There will be two arbitrary constants in both cases as we
are dealing with a second-order differential equation, and these can be
determined from the initial conditions. Sketch their corresponding plots
of \(x\left(t\right)\) on the same
graph.~}
{\label{831662}}\par\null\par\null
\vskip 6in
\par\null
\subsubsection*{2.4 Show that the under-damped case has a
solution~\(x = Ce^{-\gamma t} \sin(\omega t + \phi)\). The arbitrary constants \(C\)
and \(\phi\) are related to the initial conditions. Sketch the
corresponding plot of
\(x\left(t\right)\).~}
{\label{122651}}\par\null\par\null
\vskip 4in
\par\null
\subsection*{3. Derivation of resonance
response}
{\label{282446}}
Now, add on an external force~\(F_{ext}=F\cos(\Omega t)\). We have assumed that
this external force is a periodic driving force, with angular
frequency~\(\Omega\)~that is generically different
from~\(\omega\).~ Note that~\(F\) is a constant.
\par\null
Initial conditions:~Assume that at~\(t=0\), we
have~\(v=0\)~and~\(x=A>0\).
\par\null
\subsubsection*{3.1 Write down the equation of motion of the mass in
terms of \(x\) and its time derivatives. Arrange it in a
form~ \(\ddot{x} + 2 \gamma \dot{x} + \omega^2 x = P(t)\), where \(P\left(t\right)\) is some function of
time only (and not~a function of
\(x\)).}
{\label{542922}}\par\null
\vskip 1in
\par\null
We need some mathematical tricks to solve such a differential equation
where~\(P\left(t\right)\ne0\). This is an inhomogeneous equation, because
if~\(x_1\left(t\ \right)\) is a solution and if~\(x_2\left(t\right)\) is another
solution, then~\(x_3=\left(x_1+x_2\right)\) is~\emph{not} a solution.
What we can do is to consider an \emph{associated homogeneous equation}
of the form:
\begin{center}$\ddot{x} + 2 \gamma \dot{x} + \omega^2 x = 0$\end{center}
which is basically the same differential equation but with the
right-hand-side set to zero. For such a homogeneous equation, if you
have a solution, you can scale it by any constant and still get a
solution. You can add it to any other solution and still get a solution.
\par\null
You might see how this helps us get the general solution for our
original inhomogeneous equation.
\begin{enumerate}
\tightlist
\item
Find the general solution~\(x_h\left(t\right)\) for the associated
homogeneous equation, which is what you did in the previous section
(without the external driving force).
\item
Find a particular solution~\(x_p\left(t\right)\) for the actual
inhomogeneous equation.
\item
The general solution for the inhomogeneous equation
is~\(x\left(t\right)=x_h\left(t\right)+x_p\left(t\right)\). {[}It makes sense, and more rigorous proof is
left to the interested reader.{]}
\end{enumerate}
\par\null
\subsubsection*{3.2 We have the solution for the associated homogeneous
equation from the previous section (consider only the under-damped
case). We just need to guess a particular solution.
Try~\(x_p=\alpha\cos\left(\Omega t\right)+\beta\sin\left(\Omega t\right)\), since we expect the system to move according to
the external driving force (physically, this is certainly the
expectation in the steady state, after any transient effects have been
damped away). By making the appropriate substitutions into the
inhomogeneous differential equation, obtain an equation that involves
terms in~\(\cos(\Omega t)\)
and~\(\sin(\Omega t)\).}
{\label{901093}}\par\null\par\null
\vskip 4in
\par\null
\subsubsection*{3.3 Note that the equation must hold no matter for all
time~\(t\). What does this tell you? (Hint:
if~\(\Omega t=\frac{\pi}{2}\), what will the equation become? And
for~\(\Omega t=\pi\)?) This allows you to derive a set of simultaneous
equations to solve for the other unknowns.
~}
{\label{792034}}\par\null\par\null
\vskip 2in
\par\null
\subsubsection*{3.4 Hence, solve for the unknowns.
Suppose~\(\omega\) for the spring-mass system is spring-mass
system is fixed and~\(\Omega\)~for the driving force can be
varied. Notice what happens to the solution to the equation of motion
as~\(\Omega\)
changes.}
{\label{613886}}
\subsubsection*{}
{\label{613886}}
\subsubsection*{}
{\label{613886}}
\vskip 3in
The phenomenon of large values for the amplitude of oscillation is known
as resonance.
\par\null\par\null
\subsection*{4.~ Coupled Oscillations}
{\label{937490}}
Consider two masses~\(m_1\) and~\(m_2\) on a
frictionless horizontal plane, attached to each other by a spring with
spring constant~\(k\). Let~\(x_{1}\)
and~\(x_2\) be the displacement of the masses from their
respective reference positions when the spring is unstretched. Note that
we are assuming only one-dimensional motion of the masses in the
longitudinal direction (along the line of the spring).
\par\null
\subsubsection*{4.1 Write down the equations of motion for both masses.
Be extremely careful with
direction.}
{\label{404623}}\par\null
\vskip 2in
\par\null
The use of complex numbers when working with oscillations is often
convenient. However, before we confidently exploit any such convenience,
we have to be convinced that the process is mathematically legitimate.
Let's use the current example as a concrete illustration.
\par\null
For the physical system, we have two equations
involving~\(x_1,\ x_2\) that we have to solve simultaneously. For
each of the equations, we can write down a copy of the equations that is
identical, except writing~\(y_1,\ y_2\) for the displacements
instead, which is just a notational change. Now imagine writing the same
equations yet again but with~\(z_1=x_1+iy_1\) and correspondingly
for~\(z_2=x_2+iy_2\). Now we have equations involving~\emph{complex
numbers}, but there is a direct mapping (both ways) between solutions
for the real-valued~\(x_1,\ x_2\) and the
complex-valued~\(z_1,\ z_2\). In treating the equation in
real-variables as an equation in complex-variables, we will basically be
solving the same equations in both the real-part and the imaginary-part.
After obtaining the solution in complex-variables, the final step would
be to take either the real-part or the imaginary-part of the solution,
both of which should be solutions to the original problem in
real-variables (but corresponding to different initial conditions).
\par\null
The ``convenience'' of moving to complex numbers comes from the way
sines and cosines become exponentials. We exploit the mathematical
connection (de Moivre's theorem)
\begin{center} $e^{i\theta} = \cos \theta + i\sin \theta$ \end{center}
that allows us to interpret complex numbers naturally as two-dimensional
vectors in Argand diagrams, and the geometrical nature becomes more
transparent. Both the real part and imaginary part of a complex variable
can represent the real-valued physical coordinate of a one-dimensional
system.
\par\null
When performing differentiation and integration, it is easier to deal
with exponential functions for complex numbers rather than trigonometric
functions for real numbers. Also, instead of using the addition and
subtraction formulas for trigonometric functions, we just use familiar
rules related to exponential functions.~Multiplication by a unit complex
number ``phase factor'' is just a rotation in the complex plane. For a
generic complex number, we can write in polar form~\(z=r\ e^{i\theta}\),
where~\(r\) and~\(\theta\) are both real-valued.
Multiplication by a unit complex number gives~\(\left(r\ e^{i\theta}\right)\ e^{i\phi}=r\ e^{i\left(\theta+\phi\right)}\).
\par\null
\subsubsection*{4.2 Let~\(z_1,\ z_2\) be complex numbers such
that~\(x_1,\ x_2\) are the real parts. Consider the original
equations of motion in real-variables to be equations for
complex-variables, and assume these equations are solved
by~\(z_1\ =\ Ae^{i\omega t}\)and~~\(z_2\ =\ Be^{i\omega t}\). Note that we
assume~\(A,\ B\)~are constants, but we allow them to be
complex-valued (and are not simply assumed to be real-valued). So there
could be different constant phase factors for \(z_1\) and
\(z_2\). With the form of the solutions as assumed, plug
into the differential equations obtained in 4.1, which then become
algebraic equations since taking a time derivative is equivalent to
multiplication
by~\(i\omega\).}
{\label{769171}}\par\null\par\null
\vskip 4in
\par\null
\subsubsection*{4.3 Re-write the equations obtained in 4.2 to obtain a
matrix equation, in the form~\(M\vec{z}=\vec{0}\). Note
that~\(M\) is a 2-by-2 complex-valued matrix,
and~\(\vec{z}=\begin{pmatrix}z_1 \\ z_2\end{pmatrix}\) is a 2-component vector with complex-values.
If~\(M\) is (left-)invertible, meaning that we can find the
(left-)inverse~\(M^{-1}\), then we will
obtain\(M^{-1}\left(M\vec{z}\right)=M^{-1}\left(\vec{0}\right)=\vec{0}\). But matrix multiplication is associative,
meaning that~\(M^{-1}\left(M\vec{z}\right)=\left(M^{-1}M\right)\vec{z}=\vec{z}\), and thus only the trivial
solution~\(z_1=0,\ z_2=0\) exists.~Thus, we want to solve for the case
where the determinant of~\(M\) equals zero, so that we have
non-trivial
solutions.}
{\label{522904}}\par\null\par\null
\vskip 3in
\par\null
Congratulations, you have just derived the modes of oscillation present
in this coupled oscillator system! Allow us to explore what you have
done in this question. The primary leap of faith was done in 4.2, where
we guessed the form of the solution in order to proceed. From physical
intuition, we could argue that if we choose the initial conditions
carefully, we might be able to find the system in a single oscillation
frequency as we only have springs and masses connected up without any
other moving parts. Therefore we set both~\(\omega\) in the
assumed solution to be the same (but of course, if we run into
contradictions later on then we should realise we were being too
simplistic!). ~
\par\null
As a general rule of thumb, the number of modes present in a
one-dimensional system is determined by the number of coupled masses
present. Note that in 4.3 you have derived a mode
where~\(\omega=0\). This corresponds to the case where both masses
move together with constant velocity, resulting in a ``trivial''
oscillation.
\par\null
\subsubsection*{4.4 Create your own three mass system and solve
accordingly.}
{\label{197390}}\par\null\par\null
\vskip 6in
\par\null
\subsection*{5. Oscillations and Potential
Energy}
{\label{354365}}
Consider a particle with potential energy \(U(x)\). Let an
equilibrium position be denoted by~\(x_0\).
Using a Taylor series approximation, we have:
\begin{center}$U\left(x\right)\ =\ U\left(x_0\right)\ +\ U'\left(x_0\right)\left(x-x_0\right)\ +\frac{1}{2}\ U''\left(x_0\right)\left(x-x_0\right)^{2\ }+\ ...$\end{center}
{[}When~\(x_0\ =\ 0\), this is known as a Maclaurin series.{]}~Note
that since the system is in equilibrium at~\(x=x_0\), we
have~\(U'\left(x_0\right)\ =\ 0\). If we consider ``small'' oscillations, we can
ignore terms in~\(\left(x-x_0\right)^3\) or higher powers, as long as the
coefficient of the~\(\left(x-x_0\right)^2\) term is non-zero. For a stable
equilibrium, we need \(U''\left(x_0\right)>0\). If so, our approximation
simplifies to a parabolic potential:
\begin{center}$U\left(x\right)\ =U\left(x_0\right)+\frac{1}{2}\ U''\left(x_0\right)\left(x-x_0\right)^{2\ }$\end{center}
In this approximation, the force acting on the body is a restoring force
that is positive for \(xx_0\):
\begin{center}$F = - \frac{dU}{dx} = - U''(x_0) (x-x_0)$\end{center}
Comparing this expression with the elastic restoring force from a
spring, we see that~\(U''\left(x_0\right)\) plays the role of the spring
constant \(k\), and we thus have:
\begin{center}$\omega = \sqrt{\frac{U''(x_0)}{m}}$\end{center}
This is a generic result that applies when a mass is close to a local
minimum in a potential energy function, such that the function is
approximated by a parabola. Of course, including higher order terms like
the cubic term allows for asymmetry in this potential function (i.e. one
side is ``steeper'' than another side).
\par\null
Extending these ideas, if we think about a potential function in two
spatial dimensions, we could also have ``saddle-shaped'' potentials,
where the local equilibrium is stable in one direction but unstable in
the perpendicular direction.
\par\null
\subsection*{6. Concluding Remarks}
{\label{723356}}
The mathematics of classical oscillations tends to involve similar
techniques, even if the equations may arise out of ``different
physics''. For example, in RLC circuits, we have charges (rather than
masses) oscillating due to Maxwell's laws in capacitors and inductors. A
balloon may experience oscillations measured in terms of its volume as
opposed to a linear dimension.~
\par\null
Tips: Work through enough problems to know how to arrive at the
mathematical equations based on the physics.~For most questions, you
will likely make reasonable approximations (``small oscillations'') so
that the resulting equations simplify, to leading order, into the form
of the SHM equation.
\par\null
\subsection*{7. Supplementary Questions}
{\label{479086}}
\subsubsection*{7.1 The diagram below depicts a part of an instrument
made to measure earthquakes. A rigid rod of mass~\(M\),
length~\(l\) ~is constrained to only rotate about a
pivot~\(O\)~at its bottom. On the top of the rod sits a
ball of mass , connected by two light springs of spring constant to
fixed walls. In its equilibrium position, the rod is perfectly vertical,
and both springs are unstretched. Now we give it a slight impulse to
kickstart small oscillations about its equilibrium position. What
conditions must be satisfied so that the resulting oscillations are
simple harmonic? Determine the period of small
oscillations.}
{\label{221964}}
\subsubsection*{}
{\label{221964}}
\subsubsection*{~ ~ ~ ~~}
{\label{221964}}\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.28\columnwidth]{figures/Untitled/Untitled}
\caption{{Figure for Question 7.1
{\label{657297}}%
}}
\end{center}
\end{figure}
\subsubsection*{7.2 Two identical masses~\(m\) are
constrained to move on a horizontal hoop. Two identical springs with
spring constant \(k\) connect the masses and wrap around
the hoop. One mass is subjected to a driving force~\(F_{d\ }\cos\left(\omega_dt\right)\).
Find the particular solution for the motion of the
masses.}
{\label{481263}}\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.28\columnwidth]{figures/Untitled1/Untitled1}
\caption{{Figure for Question 7.2
{\label{953740}}%
}}
\end{center}
\end{figure}
\subsubsection*{7.3 Two identical masses are constrained to move in a
horizontal hoop. Two identical springs with spring constant
\(k\) connect the masses and wrap around the hoop. Find the
normal modes. Do it again for \(N=3\) identical masses and
springs. Do it again for general~\(N\) identical springs
and
masses.~}
{\label{943574}}
\subsubsection*{7.4 Consider a double pendulum made of two
masses,~\(m_1\) and~\(m_2\), and two rods of
lengths~\(l_1\)and~\(l_2\). Find the equations of
motion. For small oscillations, find the normal modes and their
frequencies for the special case~\(l_{1\ }=\ l_{2\ }\). Do the same for the
case~\(m_1\ =\ m_2\).}
{\label{997084}}\par\null
\subsubsection*{7.5 A pendulum consists of a mass \(m\) at
the end of a massless stick of length \(l\). The other end
of the stick is made to oscillate vertically with a position given by
\(y\left(t\right)\ =\ A\ \cos\ \left(\omega t\right)\), where \(A<