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\begin{document}
%\section*{Quiz 6}
%\rule{\textwidth}{0.5ex}
{\bf Problem 5}. {\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}.

Assume the seismograph trace can be modeled at time $t=0$ (a time
after the earthquake struck) by $x(t)=10\cos(3t)$. Assume a release
test determined $2\beta\Omega_0=16$ and $\Omega_0^2=80$. Explain how
to find a formula for the ground motion $u(t)$, then provide details
for the answer $u(t) = \frac{710}{9}\cos(3t)-\frac{160}{3}\sin(3t)$
(assume integration constants are zero).

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

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

Background:
\begin{quote}\sf
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 $\beta$.
\end{quote}

\BF{Details}.
 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 $x(t)=10\cos(3t)$, $2\beta\Omega_0=16,\Omega_0^2=80$, the quadrature equation is
$$-u''(t) = x''(t) + 16 x'(t) + 80 x(t).$$
After substitution of $x(t)$, the equation becomes
$$-u''(t) =710\cos(3t)-480\sin(3t),$$
which can be integrated twice by hand. All integration constants will be assumed zero. The answer:
$$u(t) = \frac{710}{9}\cos(3t)-\frac{160}{3}\sin(3t).$$
A Maple answer check has this brief input:
\begin{verbatim}
de:=-diff(u(t),t,t) = diff(x(t),t,t) + 16*diff(x(t),t) + 80* x(t);
x:=t->10*cos(3*t);
dsolve(de,u(t));subs(_C1=0,_C2=0,%);
\end{verbatim}

\fi



%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 Problem 6}. {\bf Resistive Network} with 2 Loops and DC Sources.

\includegraphics[scale=1.0]{figures/resistor-circuit-DC-2loops-edit2.png}


The {\bf Branch Current Method} can be used to find a $3\times 3$ linear system for the {\bf branch currents} $I_1,I_2,I_3$.
$$\begin{array}{rcrcrcrl}
  I_1 &-&   I_2 &-& I_3 &=& 0    &\mbox{~~~KCL, upper node} \\
3 I_1 &+& 2 I_2 & &     &=& 18   &\mbox{~~~KVL, left loop} \\
      && 2 I_2 &-& 2I_3  &=& 5    &\mbox{~~~KVL, right loop}
\end{array}$$
Symbol {\bf KCL} means {\em Kirchhoff's Current Law}, which says the algebraic sum of the currents at a node is zero. Symbol {\bf KVL} means {\em Kirchhoff's Voltage Law}, which says the algebraic sum of the voltage drops around a closed loop is zero.

(a) Solve the equations to find the currents $I_1, I_2, I_3$ in the figure.
 % $I_1=\frac{62}{16}, I_2=\frac{51}{16}, I_3=\frac{11}{16}$ Amperes.

(b) Compute the voltage drops across resistors $R_1,R_2,R_3$. Answer: $\frac{93}{8}, \frac{51}{8},\frac{11}{8}$ volts.

(c) Replace the $5$ volt battery by a $4$ volt battery. Solve the system again, and report
the new currents and voltage drops.
% Currents are $I_1=4, I_2=3, I_3=1$ amperes. The voltage drops are $12,6,2$.

% A:=<1,-1,-1|3,2,0|0,2,-2>^+;b:=<0,18,5>;X:=A^(-1).b;<3,0,0|0,2,0|0,0,2>.X;

{\bf References}. Edwards-Penney 3.7, electric circuits. All About Circuits Volume I -- DC, by T. Kuphaldt:
\newline
\href{http://www.allaboutcircuits.com/}{http://www.allaboutcircuits.com/}. 
\newline
Course slides on Electric Circuits:
\newline
\href{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/electricalCircuits.pdf}{\scriptsize\url{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/electricalCircuits.pdf}}
\newline
Solved examples of electrical networks can be found in the lecture notes of Ruye Wang:\newline
\href{http://fourier.eng.hmc.edu/e84/lectures/ch2/node2.html}{http://fourier.eng.hmc.edu/e84/lectures/ch2/node2.html}.

\vfill\newpage
%\rule{\textwidth}{0.5ex}
{\bf Problem 7}. {\bf $RLC$-Circuit}\newline
\begin{center}
\includegraphics[scale=1.0]{figures/rlc-circuit.png}
\end{center}
{\large\bf The Problem}.
\begin{large}
Suppose $E=\sin(40t)$, $L=1$ H, $R = 50\,\Omega$ and $C=0.01$ F. The model for the charge $Q(t)$ is $LQ''+RQ'+\frac{1}{C}Q=E(t)$.
\begin{itemize}
\item[{\bf (a)}] Differentiate the charge model and substitute $I=\frac{dQ}{dt}$ to obtain the current model $I''+50I'+100I=40\cos(40t)$.

\item[{\bf (b)}] Find the {\bf reactance} $S=\omega L - \frac{1}{\omega C}$, where $\omega=40$ is the input frequency, the natural frequency of $E=\sin(40t)$ and $E'=40\cos(40t)$. Then find the {\bf impedance} $Z=\sqrt{S^2+R^2}$.

\item[{\bf (c)}]  The steady-state current is
 $I(t)=A\cos(40t)+B\sin(40t)$ for some constants $A,B$.
 Substitute $I=A\cos(40t)+B\sin(40t)$ into the current model (a) and solve for $A,B$.

Answers: $A=-\frac{6}{625}$, $B=\frac{8}{625}$.

\item[{\bf (d)}] Write the answer in (c) in phase-amplitude form $I=I_0\sin(40t-\delta)$ with $I_0>0$ and $\delta \ge 0$. Then compute the {\bf time lag} $\delta/\omega$.

    Answers: $I_0=0.016$, $\delta=\arctan(0.75)$, $\delta/\omega=0.0160875$.

% arctan(x)+arctan(1/x)=Pi/2
\end{itemize}
\end{large}

\rule{\textwidth}{0.5ex}
\subsection*{References}
Course slides on Electric Circuits:
\newline
\href{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/electricalCircuits.pdf}{\scriptsize\url{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/electricalCircuits.pdf}}
\newline
Edwards-Penney {\em Differential Equations and Boundary Value Problems}, sections 3.4, 3.5, 3.6, 3.7.


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