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\usepackage{quiz}
\newif\ifsolutions
 \solutionstrue
\begin{document}
\section*{Sample Quiz 11}
\rule{\textwidth}{0.5ex}
{\bf Sample Quiz 11, Problem 1}. {\bf Vertical Motion Seismoscope}

The 1875 \BF{horizontal motion seismoscope} of F.\ Cecchi (1822-1887) reacted to an earthquake. It started a clock, and then it started motion of a recording surface, which ran at a speed of $1$ cm per second for $20$ seconds. The clock provided the observer with the earthquake hit time.

\begin{tabular}{@{}lcr@{}}
\includegraphics[scale=1.0]{figures/seismoscope.png}
&
\mbox{~}
&
\begin{minipage}[b]{0.6\textwidth}
\begin{center}\bf A Simplistic Vertical Motion Seismoscope\end{center}

\bigskip

The apparently stationary heavy mass on a spring writes with the attached stylus onto a rotating drum, as the ground moves up. The generated trace is $x(t)$.

\vspace*{2pc}
\end{minipage}
\end{tabular}
% Source book {fig4.9.6}

The motion of the heavy mass $m$ in the figure can be modeled initially by a forced spring-mass system with damping. The initial model has the form
$$mx''+cx'+kx=f(t)$$
where $f(t)$ is the vertical ground force due to the earthquake. In terms of the vertical ground motion $u(t)$, we write via Newton's second law the force equation $f(t)=-mu''(t)$ (compare to falling body $-mg$). The final model for the motion of the mass is then
\begin{equation}\label{4-8.eq.seismograph-vertical}
\left\{\begin{array}{l}
x''(t) + 2\beta \Omega_0 x'(t) + \Omega_0^2 x(t) = -u''(t), \\[1ex]
\dd\frac{c}{m}=2\beta\Omega_0,\quad \frac{k}{m}=\Omega_0^2,\\[2ex]
x(t) = \mbox{center of mass position measured from equilibrium}, \\[1ex]
u(t) = \mbox{vertical ground motion due to the earthquake}.
\end{array}\right.
\end{equation}

Terms {\bf seismoscope, seismograph, seismometer} refer to the device in the figure. Some observations:% about equation \rf{4-8.eq.seismograph-vertical}:
\begin{itemize}\sf
\item[] Slow ground movement means $x'\approx 0$ and $x''\approx 0$, then \rf{4-8.eq.seismograph-vertical} implies   $\Omega_0^2 x(t) = -u''(t)$. The
seismometer records ground acceleration.
\item[] Fast ground movement means $x\approx 0$ and $x'\approx 0$, then \rf{4-8.eq.seismograph-vertical} implies
      $x''(t)  = -u''(t)$. The
seismometer records ground displacement.

\item[] A \BF{release test} begins by starting a vibration with $u$ identically zero. Two successive maxima $(t_1,x_1), (t_2,x_2)$ are recorded. This experiment determines constants $\beta,\Omega_0$.

\item[] The objective of \rf{4-8.eq.seismograph-vertical} is to determine $u(t)$, by knowing $x(t)$ from the seismograph.
\end{itemize}
\begin{large}\rm
{\bf The Problem}.

\BF{(a)}~ Explain how  a \BF{release test} can find values for $\beta,\Omega_0$ in the model $x''+2\beta\Omega_0 x' + \Omega_0^2 x=0$.

\BF{(b)}~  Assume the seismograph trace can be modeled at time $t=0$ (a time after the earthquake struck) by $x(t)=Ce^{-at}\sin(bt)$ for some positive constants $C,a,b$. Assume a release test determined $2\beta\Omega_0=12$ and $\Omega_0^2=100$. Explain how to find a formula for the ground motion $u(t)$, then provide a formula for $u(t)$, using technology.

\end{large}
\rule{\textwidth}{0.5ex}

\ifsolutions
\vfill\newpage
{\bf Solution}.

\BF{(a)}~ A \BF{release test} is
an experiment which provides initial data $x(0)>0$, $x'(0)=0$  to the seismoscope mass.
The model is $x''+2\beta\Omega_0 x' + \Omega_0^2 x=0$ (ground motion zero). The recorder graphs
$x(t)$ during the experiment,  until two successive maxima $(t_1,x_1), (t_2,x_2)$ appear in the graph. This is enough information to find values for $\beta,\Omega_0$.

In an under-damped oscillation, the characteristic equation is $(r+p)^2+\omega^2=0$
corresponding to complex conjugate roots $-p\pm\omega i$. The phase-amplitude form is $x(t)=Ce^{-pt}\cos(\omega t - \alpha)$, with period $2\pi/\omega$.

The equation $x''+2\beta\Omega_0 x' + \Omega_0^2 x=0$ has characteristic equation $(r+\beta)^2+\Omega_0^2 = 0$. Therefore $x(t)=Ce^{-\beta t}\cos(\Omega_0 t - \alpha)$.

The period is $t_2-t_1 = 2\pi/\Omega_0$. Therefore, $\Omega_0$ is known. The maxima occur when the cosine factor is $1$, therefore
$$\frac{x_2}{x_1} = \frac{Ce^{-\beta t_2}\cdot 1}{Ce^{-\beta t_1}\cdot 1}=e^{-\beta(t_2-t_1)}.$$
This equation determines $\beta$.

\BF{(b)}~ The equation $-u''(t) = x''(t) + 2\beta \Omega_0 x'(t) + \Omega_0^2 x(t)$ (the model written backwards) determines $u(t)$ in terms of $x(t)$. If $x(t)$ is known, then this is a quadrature equation for the ground motion $u(t)$.

For the example $x(t)=Ce^{-at}\sin(bt)$, $2\beta\Omega_0=12,\Omega_0^2=100$, then the quadrature equation is
$$-u''(t) = x''(t) + 12 x'(t) + 100 x(t).$$
After substitution of $x(t)$, the equation becomes
$$-u''(t) =C{{\rm e}^{-at}} \left(
\sin \left( bt \right) {a}^{2}-\sin \left( bt \right) {b}^{2}-2\,\cos
 \left( bt \right) ab-12\,\sin \left( bt \right) a+12\,\cos \left( bt
 \right) b+100\,\sin \left( bt \right)  \right)
$$
which can be integrated twice using Maple, for simplicity. All integration constants will be assumed zero. The answer:
$$\begin{array}{lcl}
\dd u \left( t \right) &=& \dd {\frac {C{{\rm e}^{-at}} \left( 12\,{a}^{2}b+12
\,{b}^{3}-200\,ab \right) \cos \left( bt \right) }{ \left( {a}^{2}+{b}
^{2} \right) ^{2}}}
\\[2pc]
& & \dd - {\frac {C{{\rm e}^{-at}} \left( {a}^{4}+2\,{a}^{2}
{b}^{2}+{b}^{4}-12\,{a}^{3}-12\,a{b}^{2}+100\,{a}^{2}-100\,{b}^{2}
 \right) \sin \left( bt \right) }{ \left( {a}^{2}+{b}^{2} \right) ^{2}
}}
\end{array}
$$
The Maple session has this brief input:
\begin{verbatim}
de:=-diff(u(t),t,t) = diff(x(t),t,t) + 12*diff(x(t),t) + 100* x(t);
x:=t->C*exp(-a*t)*sin(b*t);
dsolve(de,u(t));subs(_C1=0,_C2=0,%);
\end{verbatim}





%Cecchi's seismograph was built in 1875. It was designed to record the relative motion of a pendulum and the Earth as a function of time. For horizontal vibrations, two common pendulums were used, one vibrating in a north-south plane and the other vibrating in an east-west plane. The pendulums "beat seconds"; their motion was magnified three times by a thread-and-pulley apparatus. For vertical motions, a mass on a spiral spring was used. A machine that recorded rotary motions, was incorporated into the seismograph. This consisted of a cross bar with weights at both ends, much like a dumbbell, which was pivoted at its center of mass so as to rotate in a horizontal plane. Restoring force was applied to the dumbbell by springs, so that it oscillated with a period of one second.
%
%Cecchi arranged a seismoscope to start a clock and to start into motion the recording surface at the time of an earthquake. The recording surface would translate under the indicating needles at a speed of one centimeter-per-second for twenty seconds. From the time on the clock, an observer arriving at the seismograph would determine how long before his arrival the earthquake had occurred.
%
%Cecchi's instrument was apparently installed in several observatories. Tbe earliest date found for a seismogram obtained with a Cecchi instrument is February 23, 1887, when one of these seismographs recorded a large earthquake which occurred in the French-Italian border region. In that earthquake, only the east-west component was recorded, although the seismograph was located in a zone of strong shaking. However, the seismogram record appears to us to be as accurate a representation of earth movement as possible by an early seismograph.
\vfill\newpage
\rule{\textwidth}{0.5ex}
{\bf Sample Quiz 11, Problem 2}. {\bf Laplace Theory}

\begin{quote}\Large\sf
Laplace theory implements the {\em method of quadrature} for higher order differential equations, linear systems of differential equations, and certain partial differential equations.

Laplace's method {\bf solves differential equations}.
\end{quote}

{\bf The Problem}. Solve by table methods or Laplace's method.

\begin{itemize}
\item[{\bf (a)}]
Forward table. Find $\L(f(t))$ for $f(t)= te^{2t} + 2t\sin(3t)+ 3e^{-t}\cos(4t)$.

\item[{\bf (b)}]
Backward table. Find $f(t)$ for $$\L(f(t))= \dd\frac{16}{s^2+4} + \frac{s+1}{s^2 - 2s + 10} + \frac{2}{s^2+16}.$$

\item[{\bf (c)}]
Solve the initial value problem $x''(t) + 256 x(t) = 1$, $x(0)=1$, $x'(0)=0$.

\end{itemize}

{\bf Solution (a)}.

$$\begin{array}{lcll}
\L(f(t))&=& \L(te^{2t} + 2t\sin(3t)+ 3e^{-t}\cos(4t)) & \\
        &=& \L(te^{2t}) + 2\L(t\sin(3t))+ 3\L(e^{-t}\cos(4t)) & \mbox{Linearity} \\
        &=& -\frac{d}{ds}\L(e^{2t}) - 2\frac{d}{ds}\L(\sin(3t))+ 3\L(e^{-t}\cos(4t)) & \mbox{Differentiation rule} \\
        &=& -\frac{d}{ds}\L(e^{2t}) - 2\frac{d}{ds}\L(\sin(3t))+ 3\left.\L(\cos(4t))\right|_{s=s+1} & \mbox{Shift rule} \\
        &=& -\frac{d}{ds}\frac{1}{s-2} - 2\frac{d}{ds}\frac{3}{s^2+9}+ 3\left.\frac{s}{s^2+16}\right|_{s=s+1} & \mbox{Forward table} \\
        &=& \frac{1}{(s-2)^2} + \frac{12s}{(s^2+9)^2}+ 3\frac{s+1}{(s+1)^2+16} & \mbox{Calculus}
        \end{array}
        $$

{\bf Solution (b)}.

$$
\begin{array}{lcll}
\L(f(t))&=& \frac{16}{s^2+4} + \frac{s+1}{s^2 - 2s + 10} + \frac{2}{s^2+16} &  \\
        &=& 8\frac{2}{s^2+4} + \frac{s+1}{(s - 1)^2 + 9} + \frac12\frac{4}{s^2+16} & \mbox{Prep for backward table} \\
        &=& 8\L(\sin 2t) + \frac{s+1}{(s - 1)^2 + 9} + \frac12\L(\sin 4t) & \mbox{backward table} \\
        &=& 8\L(\sin 2t) + \left.\frac{s+2}{s^2 + 9}\right|{s=s-1} + \frac12\L(\sin 4t) & \mbox{shift rule} \\
        &=& 8\L(\sin 2t) + \left.\L(\cos 3t +\frac23\sin 3t)\right|{s=s-1} + \frac12\L(\sin 4t) & \mbox{backward table} \\
        &=& 8\L(\sin 2t) + \L(e^t\cos 3t +e^t\frac23\sin 3t) + \frac12\L(\sin 4t) & \mbox{shift rule} \\
        &=& \L(8\sin 2t) + e^t\cos 3t +e^t\frac23\sin 3t + \frac12\sin 4t) & \mbox{Linearity} \\
  f(t) &=&  8\sin 2t + e^t\cos 3t +e^t\frac23\sin 3t + \frac12\sin 4t & \mbox{Lerch's cancel rule}      \end{array}
  $$

{\bf Solution (c)}.

$$
\begin{array}{lcll}
\L(x''(t) + 256 x(t)) &=& \L(1) & \mbox{$\L$ acts like matrix mult} \\
s\L(x') -x'(0)+ 256\L(x) &=& \L(1) & \mbox{Parts rule} \\
s(s\L(x)-x(0)) -x'(0)+ 256\L(x) &=& \L(1) & \mbox{Parts rule} \\
s^2\L(x)-s+ 256\L(x) &=& \L(1) & \mbox{Use $x(0)=1,x'(0)=0$} \\
(s^2+ 256)\L(x) &=& s+\L(1) & \mbox{Collect $\L(x)$ left}
\end{array}
$$
$$
\begin{array}{lcll}
\L(x) &=& \frac{s+\L(1)}{(s^2+ 256)} & \mbox{Isolate $\L(x)$ left} \\
\L(x) &=& \frac{s+1/s}{(s^2+ 256)} & \mbox{Forward table} \\
\L(x) &=& \frac{s^2+1}{s(s^2+ 256)} & \mbox{Algebra} \\
\L(x) &=& \frac{A}{s} + \frac{Bs+C}{s^2+ 256} & \mbox{Partial fractions} \\
\L(x) &=& A\L(1) + B\L(\cos 16t)+\frac{C}{16}\L(\sin 16t) & \mbox{Backward table} \\
\L(x) &=& \L(A + B\cos 16t+\frac{C}{16}\sin 16t) & \mbox{Linearity} \\
x(t) &=& A + B\cos 16t+\frac{C}{16}\sin 16t & \mbox{Lerch's rule} \\
\end{array}
$$
The partial fraction problem remains:
$$
\frac{s^2+1}{s(s^2+ 256)}=
\frac{A}{s} + \frac{Bs+C}{s^2+ 256}
$$
This problem is solved by clearing the fractions, then swapping sides of the equation, to obtain
$$
A(s^2+ 256) + (Bs+C)(s) = s^2+1.
$$
Substitute three values for $s$ to find 3 equations in 3 unknowns $A,B,C$:
$$
\begin{array}{llcl}
s=0 &  256A &=& 1 \\
s=1 & 257 A + B + C &=& 2 \\
s=-1 & 257 A + B - C &=& 2
\end{array}
$$
Then $A=1/256, B=255/256, C=0$ and finally
$$x(t)= A + B\cos 16t+\frac{C}{16}\sin 16t = \frac{1+255\cos 16t}{256}$$

\subsection*{Answer Checks}
\begin{verbatim}
# Sample quiz 11
# answer check problem 2(a)
f:=t*exp(2*t)+2*t*sin(3*t)+3*exp(-t)*cos(4*t);
with(inttrans): # load laplace package
laplace(f,t,s);
# The last two fractions simplify to 3(s+1)/((s+1)^2+16).
# answer check problem 2(b)
F:=16/(s^2+4)+(s+1)/(s^2-2*s+10)+2/(s^2+16);
invlaplace(F,s,t);
# answer check problem 2(c)
de:=diff(x(t),t,t)+256*x(t)=1;ic:=x(0)=1,D(x)(0)=0;
dsolve([de,ic],x(t));
# answer check problem 2(c), partial fractions
convert((s^2+1)/(s*(s^2+256)),parfrac,s);
\end{verbatim}

The output appears on the next page

\noindent
\includegraphics[scale=0.95]{quiz-11-answer-check-crop.pdf}
\fi
\end{document}
\endinput 