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\begin{document}
\section*{Chapter 3: Linear Differential Equations and Applications}
\rule{\textwidth}{0.5ex}
{\bf Chapter 3. Sample 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}.

\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 to the Seismoscope Problem}.

\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 Chapter 3. Sample Problem 6}. {\bf Resistive Network} with 2 Loops and DC Sources.

\includegraphics[scale=1.0]{figures/resistor-circuit-DC-2loops-edit1.png}


The {\bf Branch Current Method} can be used to find a $3\times 3$ linear system for the {\bf branch currents} $I_1,I_2,I_3$.
$$\begin{array}{rcrcrcrl}
  I_1 &-&   I_2 &-& I_3 &=& 0    &\mbox{~~~KCL, upper node} \\
4 I_1 &+& 2 I_2 & &     &=& 28   &\mbox{~~~KVL, left loop} \\
      && 2 I_2 &-& I_3  &=& 7    &\mbox{~~~KVL, right loop}
\end{array}$$
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 verify the currents reported in the figure: $I_1=5, I_2=4, I_3=1$ Amperes.

(b) Compute the voltage drops across resistors $R_1,R_2,R_3$. Answer: $20, 8, 1$ volts.

% A:=Matrix([[1,-1,-1],[4,2,0],[0,2,-1]]);LinearAlgebra[LinearSolve] (A,<0,28,7>);

{\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
{\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}.
\section*{Solutions to Problem 6}
{\bf Part (a)}. Write the system as a matrix equation $A\vec{x}=\vec{b}$. Solve for $\vec{x}$ by any method, including technology, to get $\vec{x}=\minicolvector{5\\ 4\\ 1}$, whose components are currents $I_1,I_2,I_3$.

{\bf Part (b)}. Use Ohm's Law $V=RI$ to compute $V$, which is the voltage drop across resistor $R$. Use the answers from part (a).

Handwritten solutions to Problem 6 appear below, after the solution to Problem 7.
\vfill\newpage
\rule{\textwidth}{0.5ex}
{\bf Chapter 3. Sample 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 $E=100\sin(20t)$, $L=5$ H, $R = 250\,\Omega$ and $C=0.002$ 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 $5I''+250I'+500I=2000\cos(20t)$.

\item[{\bf (b)}] Find the {\bf reactance} $S=\omega L - \frac{1}{\omega C}$, where $\omega=20$ is the input frequency, the natural frequency of $E=100\sin(20t)$ and $E'=2000\cos(20t)$.

\item[{\bf (c)}] Substitute $I=A\cos(20t)+B\sin(20t)$ into the current model (a) and solve for $A=\frac{-12}{109}, B=\frac{40}{109}$. Then the steady-state current is
    $$I(t)=A\cos(20t)+B\sin(20t)=\frac{-12\cos(20t)+40\sin(20t)}{109}.$$
%(40/109)*sin(20*t)-(12/109)*cos(20*t)

\item[{\bf (d)}] Write the answer in (c) in phase-amplitude form $I=I_0\sin(20t-\delta)$ with $I_0>0$ and $\delta \ge 0$. Then compute the {\bf time lag} $\delta/\omega$.

    Answers: $I_0=\frac{4}{\sqrt{109}}$, $\delta=\arctan(3/10)$, $\delta/\omega=0.01457$.

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

\rule{\textwidth}{0.5ex}
\subsection*{References}
Course slides on Electric Circuits:
\newline
{\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.
\vfill\newpage
\section*{Solutions to Problem 7}

{\large\bf Part (a)} Start with $5Q''+250Q'+500Q=100\sin(20t)$. Differentiate across to get
$5Q'''+250Q''+500Q'=2000\cos(20t)$. Change $Q'$ to $I$.

\bigskip

{\large\bf Part (b)} $S=(20)(5)-1/(20*0.002)=75$

\bigskip

{\large\bf Part (c)} It helps to use the differential equation $u''+400u=0$ satisfied by both $u_1=\cos(20t)$ and $u_2=\sin(20t)$. Functions $u_1,u_2$ are Euler solution atoms, hence independent. Along the solution path, we'll use $u_1'=-20\sin(20t)=-20u_2$ and $u_2'=20\cos(20t)=20u_1$.
The arithmetic is simplified by dividing the equation first by $5$. We then substitute $I=Au_1+Bu_2$.

$$\begin{array}{l}
I'' + 50I' + 100I = 400\sin(20t)
\\
A(u_1'' + 50u_1' + 100u_1)+B(u_2'' + 50u_2' + 100u_2)=400\sin(20t)
\\
A(-400u_1 + 50(-20u_2) + 100u_1)+B(-400u_2 + 50(20u_1) + 100u_2)=400\sin(20t)
\\
(-400A +100A+ 1000B )u_1 + (-1000A -400B + 100B)u_2=400u_2
\end{array}
$$
By independence of $u_1,u_2$, coefficients of $u_1,u_2$ on each side of the equation must match. The linear algebra property is called {\em unique representation of linear combinations}. This implies the $2\times 2$ system of equations
$$
\begin{array}{rcrcl}
-300A &+& 1000B &=& 0, \\
-1000A &-& 300B &=& 400.
\end{array}
$$
%solve({-300*A + 1000*B = 400,
%-1000*A - 300*B = 0},{A,B});
%
%                       /    -12      40 \
%                      { A = ---, B = --- }
%                       \    109      109/
%

The solution by Cramer's rule (the easiest method) is $A=-12/109,B=40/109$. Then the steady-state current is
$$I(t)=A\cos(20t)+B\sin(20t)=\frac{-12\cos(20t)+40\sin(20t)}{109}.$$
The {\bf steady-state current} is defined to be the sum of those terms in the general solution of the differential equation that remain after all terms that limit to zero at $t=\infty$ have been removed. The logic is that only these terms contribute to a graphic or to a numerical calculation after enough time has passed (as $t\to\infty$).

\bigskip

{\large\bf Part (d)} Let $\cos(\delta)=B/I_0, \sin(\delta)=-A/I_0, I_0=\sqrt{A^2+B^2}$. Use the trig identity
$$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$$
to rearrange the current formula as follows:
$$I(t)=A\cos(20t)+B\sin(20t)=I_0(\sin(20t)\cos(\delta)-\sin(\delta)\cos(20t))=I_0\sin(20t-\delta).$$
Compute $I_0=\sqrt{A^2+B^2}=\frac{4}{\sqrt{109}}$. Compute $\tan(\delta)=\frac{\sin\delta}{\cos\delta}=-A/B=12/40$. Then $\delta=\arctan(12/40)$ and finally $\delta/\omega=\arctan(3/10)/20=0.01457$.

{\bf Another method}, using Edwards-Penney Section 3.7: Compute the {\bf impedance} $Z=\sqrt{R^2+S^2}=\sqrt{250^2+75^2}=\sqrt{68125}=25\sqrt{109}$ and then $I_0=E_0/Z=4/\sqrt{109}$. The phase $\delta = \arctan(S/R)=\arctan(75/250)=\arctan(3/10)$. Then the time lag is $\delta/\omega=\frac{\arctan(0.3)}{20}=0.01457$.

% evalf(arctan(3/10)/20.);  0.01457283972

\rule{\textwidth}{0.5ex}

\end{document}
