%This is a template for LaTeXing homework for
%the Winter 2008 classical mechanics course by John Baez
%Please fill in the number and date of the lecture the homework
%came from in the slot below, and say who you are!
%Please fill in the number and date of the lecture in the appropriate
%slot below, and say who is taking these notes.
%Also: use the macros below for common physics and math symbols!
%
%In particular: use \R for the real numbers!
%
%Use \maps for the colon in the notation for functions:
%instead of f: X \rightarrow Y, please write f \maps X \to Y.
%
%Use \define for terms being defined,
%as in: ``We define the \define{position} to be...''
\documentclass{article}
\usepackage{amsfonts,amssymb}
%\usepackage{latexsym}
\hfuzz=6pt
% common physics symbols - use these macros!
\newcommand{\q}{q} %position
\newcommand{\p}{p} %momentum
\newcommand{\E}{E} %energy
\newcommand{\T}{T} %kinetic energy
\newcommand{\V}{V} %potential energy
\newcommand{\J}{J} %angular momentum
% common math symbols - use these macros
\newcommand{\maps}{\colon} %correct symbol for colon in f: X -> Y
%write this as: f \maps X \to Y
\newcommand{\R}{{\mathbb R}} %real numbers
\newcommand{\C}{{\mathbb C}} %complex numbers
\newcommand{\Z}{{\mathbb Z}} %integers
\renewcommand{\O}{{\rm O}} %orthogonal group
\newcommand{\SO}{{\rm SO}} %special orthogonal group
\newcommand{\so}{{\frak so}} %special orthogonal Lie algebra
% use \define for defined terms:
\newcommand{\define}[1]{{\bf #1}}
\newtheorem{thm}{Theorem}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{rem}[thm]{Remark}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{defn}[thm]{Definition}
\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}
\newcommand{\ba}{\begin{eqnarray}}
\newcommand{\ea}{\end{eqnarray}}
\newcommand{\ban}{\begin{eqnarray*}}
\newcommand{\ean}{\end{eqnarray*}}
\newcommand{\barr}{\begin{array}}
\newcommand{\earr}{\end{array}}
\textwidth 6in
\textheight 8.5in \evensidemargin .25in
\oddsidemargin .25in
\topmargin .25in
\headsep 0in
\headheight 0in
\footskip .5in
\pagestyle{plain}
\pagenumbering{arabic}
\begin{document}
\begin{center}
{\large Classical Mechanics Homework\\}
{\small January 17, 2008 \\}
{\small John Baez}
{\small homework by John Huerta}
\end{center}
\section*{Homework 1}
Solve Newton's second law $F = ma$ for $q(t)\in\R^3$ for
\[ F(t) = (0, 0, -mg), \]
the force felt by a particle near the Earth's surface only under the influence
of gravity. Find $q(t)$ in terms of the initial position $q(0)$ and
$\dot{q}(0)$.
\subsection*{Solution}
In this case, Newton's second law $F = ma = m\ddot{q}$ just says that
\[ \ddot{q} = \frac{F}{m} = (0, 0, -g). \]
In other words, acceleration $\ddot{q}$ is a constant. Since there's no increase in difficulty, we'll just solve all constant acceleration problems in any dimension, i.e. differential equations of the form
\[ \ddot{q} = a \]
where $a \in \R^n$ is a constant. Then we'll set $n = 3$ and $a = (0, 0, -g)$ to solve this special case.
It's straightforward to integrate
\[ \ddot{q} = a \]
once and get
\[ \dot{q} = v + ta \]
where our constant of integration $v \in \R^n$ is seen, upon setting $t = 0$, to be the initial velocity
\[ \dot{q}(0) = v \]
so we have
\[ \dot{q} = \dot{q}(0) + ta \]
thus far, and we integrate again to get
\[ q = x + t\dot{q}(0) + \frac{1}{2}t^2 a \]
where, just as before, our constant of integration $x \in \R^n$ is seen, upon setting $t = 0$, to be the initial positon
\[ q(0) = x \]
So our full solution is thus
\[ q = q(0) + t\dot{q}(0) + \frac{1}{2}t^2 a \]
Now back to our original problem, where $n = 3$ and $a = (0, 0, -g)$. Taking
the initial position and velocity to be
\[ q(0) = (x_0, y_0, z_0) \]
and
\[ \dot{q}(0) = (v_x, v_y, v_z) \]
we get that our position at time $t$ is
\[ q(t) = ( x_o + v_x t, y_o + v_y t, z_o + v_z t - \frac{1}{2}gt^2 ) \]
in terms of its components. In particular, if we take the $z$-axis to point in
the vertical direction, we have that the particle's height at time $t$ is
\[ q_z(t) = z_o + v_z t - \frac{1}{2}gt^2, \]
the familiar formula from any freshman physics text.
\section*{Homework 2}
Solve Newton's second law for $q(t)\in\R$ when the force is given by Hooke's
law
\[ F(t) = -kq(t) \]
in terms of $m$, $k$, $q(0)$, and $\dot{q}(0)$, and then find the period $P$ of
the oscillation and the frequency $\omega$, where $\displaystyle \omega =
\frac{2\pi}{P}$
\subsection*{Solution}
Plugging this force into Newton's second law, we get
\[ \ddot{q} = -\frac{k}{m} q \]
or
\[ \ddot{q} + \frac{k}{m} q = 0 \]
a slightly more difficult differential equation than the one in the first
problem.
To solve it, we turn to the arcane techniques of the theory of ordinary
differential equations: we find the \define{characteristic polynomial} of this
equation,
\[ \lambda^2 + \frac{k}{m}, \]
which is the polynomial that looks like the differential equation, if we
replace each $n$th derivative of $q$ with the $n$th power of $\lambda$. Now we
find the roots of this polynomial,
\[ \lambda = \pm i\sqrt{\frac{k}{m}} = \pm i \omega \]
where we've defined $\omega = \sqrt{\frac{k}{m}}$. Since the characteristic
polynomial has no repeated roots, the most general complex-valued solution to
our differential equation is of the form
\[ \tilde{q}(t) = A e^{i\omega t} + Be^{-i\omega t} \]
where $A$ and $B$ are complex constants to be determined by initial conditions.
But we don't want a complex-valued solution; we want a real solution. This is
easy to obtain, because the real part of $\tilde{q}$,
\[ q(t) = \mbox{Re}( \tilde{q}(t) = A'\cos(\omega t) + B'\sin(\omega t) \]
where $A'$ and $B'$ are real constants, satisfies the same differential
equation as $\tilde{q}$, the reason being that Re is a real linear operator
from complex-valued funtions to real-valued functions that commutes with
differentiation!
So now that we've found the most general solution,
\[ q(t) = A \cos \omega t + B \sin \omega t, \]
it only remains to determine $A$ and $B$. We have
\[ q(0) = A \cos 0 + B \sin 0 = A \]
and
\[ \dot{q}(0) = -\omega A \sin 0 + \omega B \cos 0 = \omega B. \]
Our solution is thus
\[ q(t) = q(0) \cos \omega t + \frac{\dot{q}(0)}{\omega} \sin \omega t. \]
Both of the trig functions in our solution have period
\[ P = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}} \]
so $q$ also has this period. $q$ therefore oscillates with frequency
\[ \omega = \sqrt{\frac{k}{m}}. \]
\end{document}