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\begin{document}
\section*{Sample Quiz 10}
\rule{\textwidth}{0.5ex}
{\bf Sample Quiz 10, 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 $\beta=6$ and $\Omega_0=10$. 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}
\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 $x(t)=Ce^{-at}\sin(bt)$, $\beta=6,\Omega_0=10$, 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.



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