Nothing works!!! I just updated all my old packages (I had an old version of TexLive, and the updater was configured wrong, so it's good I took care of that, anyway). I also changed all my \eqnarray to \align, but the problem is still there. Here's my new file list, in case someone can find something wrong with it.
Also, here's the entire file... As I said before, sometimes the problem shows up only after so many pages, so maybe cutting it short changed it last time. I realize this is not exactly a MWE, but I don't understand what's going on here... This time, it's only page 6 that has an issue.
I'm not entirely sure about this, but going over my files changing eqnarray to align, i think I have finally found something in common among all the problem files. The issue seems to happen after the first figure in the document. I'm not sure it's always like this, cause I haven't checked all files yet, but, could this be the cause of the problem? And if so, what am I doing wrong? I just changed my image file to a jpg, to see if that made a difference, but it didn't.
Code: Select all
\documentclass[a4paper,11pt,answers]{exam}
\usepackage[pdftex]{graphicx}
\printanswers
\qformat{\textbf{Problem \thequestion}\hfill}
\noaddpoints
\renewcommand{\solutiontitle}{\textbf{Solution:}\par\noindent}
\renewenvironment{TheSolution}
{
\vspace{\parskip}
\leftskip=0pt
\rightskip=0pt
\solutiontitle
\ignorespaces
}
\usepackage{amsmath}
\usepackage{amssymb}
%\usepackage{setspace}
\usepackage{enumerate}
\usepackage{paralist}
\usepackage{verbatim}
\newcommand{\dis}{\displaystyle}
\usepackage[left=2cm,top=2cm,right=3cm,nohead,nofoot]{geometry}
\footskip=1cm
\author{me}
\title{PHYSICS}
\date{}
\begin{document}
\maketitle
\centering{\LARGE{Homework 6 Solutions}}
\begin{questions}
\question \begin{inparaenum} [\bfseries(a)] \item Find the normal frequencies for small oscillations of the double pendulum for arbitrary values of the masses and lengths.
\item Check that your answers are correct for the special case that $m_1=m_2$ and $L_1=L_2$.
\item Discuss the limit that $m_2\rightarrow 0$
\end{inparaenum}
\begin{solution}
\begin{inparaenum}[\bfseries(a)]
\item The entire first part of this problem is actually worked out in Taylor as an example (pages 431-434), so I won't re-derive it here. Instead, I'll take over from where Taylor left off.
The equation of motion is $\mathbf{M}\boldsymbol{\ddot{\phi}}=-\mathbf{K}\boldsymbol{\phi}$, where $\mathbf{M}$ and $\mathbf{K}$ are given by equation 11.44 in Taylor:
\[ \mathbf{M}=
\begin{pmatrix}
(m_1+m_2)L_1^2&m_2L_1L_2\\
m_2L_1L_2&m_2L_2^2
\end{pmatrix}, \hspace{1cm}
\mathbf{K}=
\begin{pmatrix}
(m_1+m_2)gL_1&0\\
0& m_2gL_2
\end{pmatrix}
\]
The normal frequencies are found by proposing oscillatory solutions, writing them as the real part of an exponential, plugging them back into the equation of motion, and then solving $(\mathbf{K}-\omega^2\mathbf{M})\mathbf{a}=0$, which only has solutions when det$(\mathbf{K}-\omega^2\mathbf{M})=0$. So we just have to find the values of $\omega$ that satisfy that determinant equation. I did this entirely in Mathematica, because it's ugly and messy. I invite you to do the same. The answer is:
$$ \omega^2=\frac{g(L_1+L_2)(m_1+m_2)\pm g\sqrt{(m_1+m_2)\left[(L_1-L_2)^2m_1+(L_1+L_2)^2m_2\right]}}{2L_1L_2m_1}$$
which is just scary.
Notice we can re-write the equation for $\mathbf{a}$ to make it more clear that it's just an eigenvalue equation:
\begin{align*}
(\mathbf{K}-\omega^2\mathbf{M})\mathbf{a}&=0\\
\mathbf{K\, a}&= \omega^2\mathbf{M \,a}\\
\mathbf{M}^{-1}\mathbf{K \,a}&=\omega^2\mathbf{a}
\end{align*}
so another way to find our values of $\omega$ is to see that they are just the eigenvalues of that matrix.
\item When $m_1=m_2$ and $L_1=L_2$, the first term in the root cancels, and the rest is:
$$ \omega^2=\frac{g2L\cdot2m\pm g\sqrt{2m^2\,4L^2}}{2L^2m}=\frac{g2L\cdot2m\pm g\,2m\cdot L\sqrt{2}}{2L^2m}=\frac{g}{L}\left(2\pm\sqrt{2}\right)$$
which is exactly equation 11.47 where Taylor describes that example.
\item In the case when $m_2=0$ we get:
$$\omega^2=\frac{g(L_1+L_2)m_1\pm g\sqrt{m_1^2(L_1-L_2)^2}}{2L_1L_2m_1}=\frac{g\left(L_1+L_2\pm(L_1-L_2)\right)}{2L_1L_2}=\frac{g}{L_{1,2}}$$
which is what we would expect: the pendulum has been reduced to a simple pendulum. The fact that we can get a result with $L_1$ or with $L_2$ simply stems from the fact that $\omega$ is symmetric in the two lengths:it doesn't really matter where the heavier mass is. We would obviously choose $L_1$ in our case, because the other one makes no sense.
\end{inparaenum}
\end{solution}
\question Two equal masses are constrained to move without friction, one on the positive $x$ axis and one on the positive $y$ axis. They are attached to two identical springs (force constant $k$) whose other ends are attached to the origin. In addition, the two masses are connected to each other by a third spring of force constant $k'$. The springs are chosen so that the system is in equilibrium with all three springs relaxed. What are the normal frequencies? Find and describe the normal modes.
\begin{solution}
As usual, the complicated part about this problem is figuring out the potential energy. The issue here is the annoying $k'$ spring, which is diagonal. Now, drawing springs is really hard, so just pay attention to my description.
Let $L$ be the length of the $k'$ spring when the other two are stretched a distance $x$ and $y$ respectively, and let $\ell$ be the rest length of the horizontal springs. From basic trigonometry, the rest length of the $k'$ spring is $\sqrt{2}\ell$. Now, the distance we're actually interested in in $\Delta L=L-\sqrt{2}\ell$, since this is what determines the force this spring exerts. In what follows, I'll assume $x$ and $y$ are small, so I'll only keep terms up to second order in them.
We can express $\Delta L$ as:
\begin{align*}
\Delta L&=\sqrt{(\ell+x)^2+(\ell+y)^2}-\sqrt{2}\ell\\
&= \sqrt{2\ell^2+x^2+y^2+2\ell x+2\ell y}-\sqrt{2}\ell\\
&=\sqrt{2}\ell\sqrt{1+\frac{x^2+y^2}{2\ell^2}+\frac{x+y}{\ell}}-\sqrt{2}\ell\\
&\approx \sqrt{2}\ell\left(1+\frac{x^2+y^2}{4\ell^2}+\frac{x+y}{2\ell}-1\right)\\
\Delta L&=\frac{\sqrt{2}}{2}\left(\frac{x^2+y^2}{2\ell}+x+y\right)
\end{align*}
With that, our potential energy is
\begin{align*}
U&==\frac{1}{2}k(x^2+y^2)+\frac{1}{2}k'\Delta L^2\\
&=\frac{1}{2}k(x^2+y^2)+\frac{1}{4}k'\left(\frac{x^2+y^2}{2\ell}+x+y\right)^2\\
&\approx \frac{1}{2}k(x^2+y^2)+\frac{1}{4}k'\left(x^2+y^2+2xy\right)
\end{align*}
where in the last step I got rid of all terms of order higher than 2.
With that potential energy, our Lagrangian becomes $$\mathcal{L}=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-\frac{1}{2}k(x^2+y^2)-\frac{1}{4}k'\left(x^2+y^2+2xy\right)$$ so our equations of motion are
\begin{align*}
m\ddot{x}&=-x\left(k+\frac{1}{2}k'\right)-\frac{1}{2}k'y\\
m\ddot{y}&=-y\left(k+\frac{1}{2}k'\right)-\frac{1}{2}k'x
\end{align*}
or $\mathbf{M\ddot{x}}=-\mathbf{Kx}$, where
\[\mathbf{M}=
\begin{pmatrix} m&0\\0&m \end{pmatrix}
,\hspace{0.5cm}\text{and}\hspace{0.5cm} \mathbf{K}=
\begin{pmatrix}
k+\frac{1}{2}k'&\frac{1}{2}k'\\
\frac{1}{2}k'&k+\frac{1}{2}k'
\end{pmatrix}\]
Now we do our usual thing: assume oscillatory solutions and re-write the matrix equation in terms of that, which will lead us to require that det$(\mathbf{K}-\omega^2\mathbf{M})=0$:
\[\text{det}
\begin{pmatrix}
k+\frac{k'}{2}-\omega^2 m&\frac{k'}{2}\\
\frac{k'}{2}&k+\frac{k'}{2}-\omega^2 m
\end{pmatrix}=
\left(k+\frac{k'}{2}-\omega^2m\right)^2-\frac{k'^2}{4}=0\]
Working on the equation:
$$k^2+kk'+\omega^4m^2-2\omega^2m\left(k+\frac{k'}{2}\right)=0$$
which gives us
$$\omega_1=\pm\sqrt{\frac{k}{m}}, \hspace{1cm} \omega_2=\pm\sqrt{\frac{k+k'}{m}}$$
This is already giving us a hint about the normal modes. In the first one, the diagonal spring plays no role, so we expect it isn't stretched. Let's see...
Using $\omega_1$ first, our eigenvector equation becomes:
\[\mathbf{K}-\omega^2\mathbf{M}=\frac{k'}{2}
\begin{pmatrix}
1&1\\
1&1
\end{pmatrix} \cdot \begin{pmatrix} a_1\\a_2\end{pmatrix}=\frac{k'}{2}
\begin{pmatrix}
a_1+a_2\\
a_1+a_2
\end{pmatrix}=0\]
This means that in this mode $a_1=-a_2$. That is, the horizontal springs are oscillating exactly out of phase and with the same amplitude. As we expected, this means that the diagonal spring is always kept at the same length, so it plays no role in the equations of motion.
Using $\omega_2$ we get:
\[\mathbf{K}-\omega^2\mathbf{M}=\frac{k'}{2}
\begin{pmatrix}
-1&1\\
1&-1
\end{pmatrix}\cdot\begin{pmatrix}a_1\\a_2\end{pmatrix}=\frac{k'}{2}
\begin{pmatrix}
-a_1+a_2\\
a_1-a_2
\end{pmatrix}=0\]
This means that $a_1=a_2$. In this case, the springs are oscillating perfectly in phase and with the same amplitude, thus stretching and compressing the diagonal spring, so its spring constant plays a role in the frequency of oscillations.
\end{solution}
\question A bead of mass $m$ is threaded on a frictionless circular wire hoop of radius $R$ and mass $m$ (same mass). The hoop is suspended at the point $A$ and it's free to swing in its own vertical plane. Using the angles $\phi_1$ and $\phi_2$ as generalized coordinates, solve for the normal frequencies of small oscillations, and find and describe the motion in the corresponding normal modes.
\begin{solution}
I actually find Taylor's suggestions for angles unnecessarily complicated, so I'm using a slightly modified version described in the figure. The only difference is that the second angle is with respect to the vertical, not the diameter.
\begin{figure}[h]
\centering
\includegraphics[width=0.3\textwidth]{problem3}
\caption{The variables used in Problem 3}
\label{prob3fig}
\end{figure}
Now, with those angles, the coordinates of the mass $m$ are given by:
\begin{align*}
x_m&=R(\sin\phi_2+\sin\phi_1)\\
y_m&=-R(\cos\phi_2+\cos\phi_1)
\end{align*}
and its kinetic energy is given by
\begin{align*}
T_m&=\frac{1}{2}m(\dot{x}_m^2+\dot{y}_m^2)\\
&= \frac{1}{2}mR^2\left[(\dot{\phi}_1\cos\phi_1+\dot{\phi}_2\cos\phi_2)^2+(\dot{\phi}_1\sin\phi_1+\dot{\phi}_2\sin\phi_2)^2\right]\\
&=\frac{1}{2}mR^2\left[\dot{\phi}_1^2+\dot{\phi}_2^2+2\dot{\phi}_1\dot{\phi}_2\cos(\phi_1-\phi_2)\right]\\
&\approx\frac{1}{2}mR^2(\dot{\phi}_1^2+\dot{\phi}_2^2+2\dot{\phi}_1\dot{\phi}_2)
\end{align*}
where in the last line I used the small angle approximation up to second order.
The kinetic energy of the hoop is only rotational. The only "tricky" part is finding the right moment of inertia, but that's easy, because this is a 2D problem, so $I$ is basically a scalar. We just have to move it form the center to point $A$ by using the parallel axes theorem: $I=I_{\text{cm}}+mR^2=mR^2+mR^2=2mR^2$, so now our total kinetic energy is:
$$T_{\text{tot}}=mR^2\dot{\phi}_1^2+\frac{1}{2}mR^2(\dot{\phi}_1^2+\dot{\phi}_2^2+2\dot{\phi}_1\dot{\phi}_2)$$
To find the potential energy, all we need is the $y$ coordinate of the center of the hoop. This is given by $y_{\text{hoop}}=-R\cos\phi_1$, so our potential energy is $$U_\text{tot}=-2mgR\cos\phi_1-mgR\cos\phi_2\approx \frac{1}{2}mgR(2\phi_1^2+\phi_2^2)+U_0$$
where, again, I approximated up to second order and put all the constants in $U_0$ since they won't show up in the equations of motion.
After all the algebra with the Lagrangian, our equations of motion turn out to be:
\begin{align*}
3\ddot{\phi}_1+\ddot{\phi}_2&=-\frac{2g}{R}\phi_1\\
\ddot{\phi}_1+\ddot{\phi}_2&=-\frac{g}{R}\phi_2
\end{align*}
which we can express in matrix notation by defining
\[\mathbf{M}=
\begin{pmatrix}
3&1\\
1&1
\end{pmatrix},
\hspace{1cm}
\mathbf{K}= \frac{g}{R} \begin{pmatrix}
2&0\\
0&1
\end{pmatrix}\]
%\vspace{6cm}
so now our equation is simply $\mathbf{M}\ddot{\boldsymbol{\phi}}=-\mathbf{K}\boldsymbol{\phi}$. The rest is the same old spiel: guess an exponential solution with frequency $\omega$, plug it back in, write in a form that necessitates a zero determinant, solve for $\omega$. So our equation for $\omega$ is:
\[\text{det}\begin{pmatrix}
2\frac{g}{R}-3\omega^2&-\omega^2\\
-\omega^2&\frac{g}{R}-\omega^2
\end{pmatrix}=(2\frac{g}{R}-3\omega^2)(\frac{g}{R}-\omega^2)-\omega^4=0\]
This gives us values of $\omega$:
$$\omega_1^2=\frac{2g}{R}, \hspace{1cm}\omega_2^2=\frac{g}{2R}$$
which turns our $\mathbf{a}$ equation into, for $\omega_1$
\[\begin{pmatrix}
2\frac{g}{R}-6\frac{g}{R}&-2\frac{g}{R}\\
-2\frac{g}{R}&\frac{g}{R}-2\frac{g}{R}
\end{pmatrix}\cdot
\begin{pmatrix} a_1\\a_2 \end{pmatrix}=-\frac{g}{R}
\begin{pmatrix}
4&2\\
2&1
\end{pmatrix}\cdot\begin{pmatrix}a_1\\a_2\end{pmatrix}=0\Rightarrow\mathbf{a}=
\begin{pmatrix}1\\-2\end{pmatrix}\]
and for $\omega_2$:
\[\begin{pmatrix}
2\frac{g}{R}-\frac{3g}{2R}&-\frac{g}{2R}\\
-\frac{g}{2R}&\frac{g}{R}-\frac{g}{2R}
\end{pmatrix}\cdot
\begin{pmatrix} a_1\\a_2 \end{pmatrix}=\frac{g}{2R}
\begin{pmatrix}
1&-1\\
-1&1
\end{pmatrix}\cdot\begin{pmatrix}a_1\\a_2\end{pmatrix}=0\Rightarrow\mathbf{a}=
\begin{pmatrix}1\\1\end{pmatrix}\]
We can interpret the first solution as movements in opposite directions: the bead goes down when the hoop goes up, and viceversa. Also, the movement of the bead is twice as big as the movement of the loop. This makes some intuitive sense, since we imagine it's easier to move a point particle than an extended one.
The other solution gives movements in phase and with the same amplitude. That is to say, the bead doesn't move with respect to the hoop.
\end{solution}
\question \begin{inparaenum}[\bfseries(a)] A simple pendulum of mass $M$ and length $L$ is suspended from a cart of mass $m$ that moves freely along a horizontal track.
\item What are the normal frequencies?
\item Find and describe the corresponding normal modes
\end{inparaenum}
\begin{solution} \begin{inparaenum}[\bfseries(a)]
The coordinates I'll use are shown in the figure.
\begin{figure}[h]
\centering
\includegraphics[width=0.3\textwidth]{problem4}
\caption{The variables used in problem 4}
\label{prob4fig}
\end{figure}
From the figure we have
\begin{align*}
x_M=x_m+L\sin\phi &\Rightarrow \dot{x}_M=\dot{x}_m+L\dot{\phi}\cos\phi\\
y_M=-L\cos\phi&\Rightarrow\dot{y}_M=L\dot{\phi}\sin\phi
\end{align*}
and let me call $x_m=x$ because I'll get tired really soon of writing the subscripts. With those relationships, and approximating the trig functions up to second order, the kinetic and potential energies become:
\begin{align*}
T&=T_m+T_M=\frac{1}{2}m\dot{x}^2+\frac{1}{2}M(\dot{x}_M^2+\dot{y}_M^2)\\
&=\frac{1}{2}m\dot{x}^2+\frac{1}{2}M(\dot{x}^2+L^2\dot{\phi}^2+2L\, \dot{x}\dot{\phi}\cos\phi)\\
&\approx\frac{1}{2}(m+M)\dot{x}^2+\frac{1}{2}M(L^2\dot{\phi}^2+2L\dot{x}\dot{\phi})\\
U&= Mgy_M\\
&=-MgL\,\cos\phi\\
&\approx -MgL+\frac{1}{2}MgL\,\phi^2
\end{align*}
After calculating the Lagrange equations of motion we end up with:
\begin{align*}
(M+m)\,\ddot{x}+ML\ddot{\phi}&=0\\
ML\,\ddot{x}+ML^2\ddot{\phi}&=-MgL\phi
\end{align*}
or, defining
\[\mathbf{M}=\begin{pmatrix}
M+m&ML\\
ML&ML^2
\end{pmatrix}, \hspace{1cm} \mathbf{K}=
\begin{pmatrix}
0&0\\
0&MgL
\end{pmatrix}\]
we have $\mathbf{M\,\ddot{x}}=-\mathbf{K\,x}$. Now we can calculate all that stuff.
\item The frequencies $\omega$ are, as usual, the eigenvalues of $\mathbf{M}^{-1}\mathbf{K}$, or the solutions of det$(\mathbf{K}-\omega^2\mathbf{M})=0$, which is the same. So:
\[\text{det}
\begin{pmatrix}
-\omega^2(M+m)&-\omega^2ML\\
-\omega^2ML&MgL-\omega^2ML^2
\end{pmatrix}=-\omega^2ML(M+m)(g-\omega^2L)-\omega^4M^2L^2=0\]
Working on that equation:
\begin{align*}
-\omega^2ML\left[g(M+m)-\omega^2L(M+m)+\omega^2ML\right]&=0\\
-\omega^2 ML\left[g(M+m)-\omega^2Lm\right]&=0
\end{align*}
so our two normal frequencies are
$$\omega_1=0\hspace{1cm}\omega_2=\sqrt{\frac{g(M+m)}{Lm}}$$
\item To find the normal modes, we go back to $\mathbf{K}-\omega^2\mathbf{M}=0$ and plug in the values of $\omega$. The first one is easiest:
\[\begin{pmatrix} 0&0\\0&MgL \end{pmatrix}\cdot
\begin{pmatrix} a_1\\a_2 \end{pmatrix}=
\begin{pmatrix} 0\\ a_2 \,MgL\end{pmatrix} =0\hspace{0.5cm}\Rightarrow\hspace{0.5cm} a_2=0\]
Notice this doesn't tell us anything about $a_1$. However, $a_2$ represents $\phi$ in our case, so we can go back to our equations of motion and see what we get when $\phi(t)$ is identically zero. We can see this implies that $\ddot{x}=0$. That means that the cart is moving without acceleration, at a constant velocity. The pendulum doesn't move at all in this case, like we would expect, since it experiences no forces. We would need initial conditions to know more about the movement, but this is certainly a lot coming from a simple zero!
Using our other value for $\omega$ we find:
\[\begin{pmatrix}
-g\frac{(M+m)^2}{Lm} & -g(M+m)\frac{M}{m}\\
-g(M+m)\frac{M}{m} & MgL-gL(M+m)\frac{M}{m}
\end{pmatrix}\cdot
\begin{pmatrix}a_1\\a_2\end{pmatrix}=0\]
So we get an equation that relates $a_1$ to $a_2$:
$$(M+m)a_1+MLa_2=0\hspace{0.5cm}\Rightarrow\hspace{0.5cm}a_1=-\frac{ML}{M+m}a_2$$
Now, if you recall, this whole $a$ business came about by assuming oscillatory motion, so the fact we get a solution means there is, in fact, oscillatory motion. The frequency of the motion is given by $\omega_2$. As we would expect, this is the frequency of a simple pendulum modified by the two masses. Note that if $m\rightarrow\infty$ then the $M$ in the numerator of $\omega_2$ doesn't matter and we can cancel the $m$'s, recovering the simple pendulum frequency. This is to be expected, since an infinitely massive cart wouldn't move.
Also, the amplitudes of the oscillations of the pendulum and the cart are related via the relationship between $a_1$ and $a_2$. Notice first of all that they are oscillating out of phase, due to the minus sign. Also, $a_1\rightarrow0$ when $m\rightarrow\infty$, as we would expect.
\end{inparaenum}
\end{solution}
\question \begin{inparaenum}[\bfseries(a)] As a model of a linear triatomic molecule such as CO$_2$, consider a system with two identical atoms each of mass $m$ connected by two identical springs to a single atom of mass $M$. To simplify matters, assume that the system is confined to move in one dimension.
\item Write down the Lagrangian and find the normal frequencies of the system. Show that one of the normal frequencies is zero.
\item Find and describe the motion in the normal modes whose frequencies are nonzero.
\item Do the same for the mode with zero frequency.
\end{inparaenum}
\begin{solution}
\begin{inparaenum}[\bfseries(a)]
\item Since the system can only move in one dimension, we'll have just one coordinate for each atom. Call them $x_1$, $x_2$ and $x_3$. Now, the kinetic energy is easy enough: $T=\frac{1}{2}(m\dot{x}_1^2+m\dot{x}_3^2+M\dot{x}_2^2)$, but the potential energy is slightly more complicated, because the springs stretch and compress due to the different molecules. But it's not so bad. Basically, motion of the middle molecule to the right will compress the right spring and stretch the left one, so we have to add $x_2$ to one of the coordinates and subtract it from the other one. It doesn't really matter which. I've chosen to make it so that $U=\frac{1}{2}k[(x_2-x_1)^2+(x_3-x_2)^2]$
With that, the Lagrangian becomes $$\mathcal{L}=\frac{1}{2}\left[m(\dot{x}_1^2+\dot{x}_3^2)+M\dot{x}_2^2\right]-\frac{1}{2}k\left[(x_2-x_1)^2+(x_3-x_2)^2\right]$$ and our equations of motion are:
\begin{align*}
m\ddot{x}_1&=k(x_2-x_1)\\
M\ddot{x}_2&=k(x_3+x_1)-2kx_2\\
m\ddot{x}_3&=-k(x_3-x_2)
\end{align*}
which leads us to define
\[\mathbf{M}=
\begin{pmatrix}
m&0&0\\
0&M&0\\
0&0&m
\end{pmatrix}, \hspace{1cm}
\mathbf{K}=k \begin{pmatrix}
1&-1&0\\
-1&2&-1\\
0&-1&1
\end{pmatrix}\]
Thus, we now have the equation:
\[\text{det}\begin{pmatrix}
k-\omega^2m&-k&0\\
-k&2k-\omega^2M&-k\\
0&-k&k-\omega^2m
\end{pmatrix}= (k-\omega^2m)[(2k-\omega^2M)(k-\omega^2m)-k^2]-k^2(k-\omega^2m)=0\]
which we can re-write as $$\omega^2\left[-\omega^4m^2M+2km(M+m)\omega^2-k^2(2m+M)\right]=0$$ so we can see that we do get a zero frequency, as advertised. Now we can also get the other two frequencies. Solving the equation with Mathematica we get:
\begin{align*}
\omega_1&=\pm\sqrt{\frac{k}{m}}\\
\omega_2&=\pm\sqrt{k\left(\frac{1}{m}+\frac{2}{M}\right)}
\end{align*}
\item Now we want to find the values of $\mathbf{a}$. Our equations are, for the first $\omega_1$:
\[\begin{pmatrix}
0&-k&0\\
-k&2k-kM/m&-k\\
0&-k&0
\end{pmatrix}\cdot\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=k
\begin{pmatrix}
-a_2\\
-a_1+(2-M/m)a_2-a_3\\
-a_2
\end{pmatrix}=0\]
So we find that $a_2=0$ and $a_1=-a_3$. This means that the carbon molecule is not moving and the two oxygen molecules are moving in opposite directions. The center of mass of the molecule is not moving, like we would expect.
For $\omega_2$ we get:
\[\begin{pmatrix}
-2km/M&-k&0\\
-k&-k/m&-k\\
0&-k&-2km/M
\end{pmatrix}\cdot\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=-k
\begin{pmatrix}
2a_1\,m/M+a_2\\
a_1+a_2/m+a_3\\
a_2+2a_3m/M
\end{pmatrix}=0\]
which means that $a_1=a_3$ and $a_2=-2a_1\,m/M$. That is, in this case, the two oxygens are moving in phase and with the same amplitudes, and the carbon is moving opposite them and with an amplitude determined by the ratio of the masses.
\item Using $\omega=0$ our equation is simply
\[k\begin{pmatrix}
1&-1&0\\
-1&2&-1\\
0&-1&1
\end{pmatrix}\cdot\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=k
\begin{pmatrix}
a_1-a_2\\
-a_1+2a_2-a_3\\
-a_2+a_3
\end{pmatrix}=0\]
which means that $a_1=a_2=a_3$. This, together with the fact that the frequency of the motion is zero, tells us that this describes a displacement of the entire molecule rather than a vibration of it.
\end{inparaenum}
\end{solution}
\end{questions}
\end{document}
I was hoping so much for this to be an outdated packages issue....
