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\header{8.03}{1}
\problem{1.1}
\part{a}
\begin{equation} 14 + 8\sqrt{2}j \end{equation}
\part{b}
\begin{equation} \pm\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\jmath \end{equation}
\part{c}
\begin{equation} -1, \frac{1}{2} \pm \frac{3}{\sqrt{2}}\jmath \end{equation}
\part{d}
\begin{equation} j^j = e^{-\jmath\frac{\pi}{2}\jmath} = e^{\frac{-\pi}{2}} \end{equation}
\problem{1.2}
\begin{equation} dz = A\jmath e^{\jmath\theta}d\theta = \jmath\vec{z}d\theta \end{equation}
\vspace{1.8in}
\problem{1.3}
\part{a}
\[ e^{\jmath\theta} = 1 + (\jmath\theta) + \frac{(\jmath\theta)^2}{2!} \ldots = \] \\
\begin{equation} e^{\jmath\theta} = \cos(\theta) + \jmath\sin(\theta)\end{equation} \\ \begin{equation}
e^{-\jmath\theta} = \cos(\theta) - \jmath\sin(\theta) \Longrightarrow \cos(\theta) = \frac{e^{\jmath\theta} + e^{-\jmath\theta}}{2}\end{equation} \\ \begin{equation}
\Longrightarrow \sin(\theta) = \frac{e^{\jmath\theta} - e^{-\jmath\theta}}{2\jmath} \end{equation}
\part{b}
\vspace{2in}
\problem{1.4}
\begin{equation} x(t) = Re(\cos(\omega t + \phi) + \jmath\sin(\omega t + \phi)) \end{equation}
\vspace{1.5in}
\problem{1.5}
\part{a}
\vspace{1in}
\begin{equation} x(t) = x_0(1 + 2\cos\phi)\cos(\omega t) \end{equation}
\part{b}
\begin{equation} x_1(t) = x_0e^{\jmath(\omega t - \phi)}\hspace{.4in}x_2(t) = x_0e^{\jmath\omega t}\hspace{.4in} x_3 = x_0e^{\jmath(\omega t + \phi)}\end{equation} \\ \begin{equation}
x_1 + x_2 + x_3 = x_0e^{\jmath\omega t}(e^{\jmath\phi} + e^{-\jmath\phi} + 1) = x_0(1 + 2\cos\phi)\cos(\omega t)\end{equation} \\ \begin{equation}
\end{equation}Max\begin{equation} \phi = 0\hspace{.4in}
\end{equation}Min\begin{equation} \phi = \frac{2\pi}{3} \end{equation}
\vspace{.7in}
\problem{1.6}
\part{a}
\begin{equation} \vec{\tau} = \dot{\vec{L}}\hspace{.4in}\vec{L} = I\dot{\theta}\end{equation} \\ \begin{equation}
-c\theta = I\ddot{\theta}\hspace{.4in}\ddot{\theta} = \frac{-c}{I}\theta
\omega_0 = \sqrt{\frac{c}{I}} \end{equation}
\part{b}
\begin{equation} m\ddot{x} = \frac{-2Tx}{\sqrt{x^2 + l^2}} \end{equation}
Small values of x, so...
\begin{equation} \frac{-2Tx}{ml} = \ddot{x}\hspace{.4in}\omega_0 = \sqrt{\frac{2T}{ml}} \end{equation}
\part{c}
\begin{equation} -g\rho Ax = m\ddot{x}\end{equation} \\ \begin{equation}
\ddot{x} = \frac{g\rho Ax}{m}\hspace{.4in}\omega_0 = \sqrt{\frac{g\rho A}{m}} \end{equation}
\part{d}
\begin{equation} L\ddot{q} = \frac{q}{C}\hspace{.4in}\omega_0 = \sqrt{\frac{1}{LC}} \end{equation}
\problem{1.7}
\begin{equation} F = -kx\hspace{.4in} 9.8N = k(.1 m)\hspace{.4in}k = 98 \frac{kg}{s^2}\end{equation} \\ \begin{equation}
.04 = \frac{mg_m}{k}\hspace{.4in}mg_m = 3.92\end{equation} \\ \begin{equation}
-kx -mg_m = m\ddot{x}\hspace{.4in}d = x + \frac{mg_m}{k}\end{equation} \\ \begin{equation}
-d = \frac{m\ddot{x}}{k}\hspace{.4in}\frac{-kd}{m} = \ddot{d}\end{equation} \\ \begin{equation}
\omega_0 = \sqrt{\frac{k}{m}} = 2\pi\hspace{.4in}m = 2.48 kg\end{equation} \\ \begin{equation}
g_m = 1.58 \frac{m}{s^2} \end{equation}
\problem{1.8}
\begin{equation}
I_0 = \frac{MR^2}{4} + M\frac{(l-d)^2}{12}\hspace{.4in}k^2 = \frac{R^2}{4} + \frac{(l-d)^2}{12}\hspace{.4in}h = \frac{(l+d)}{2}\end{equation} \\ \begin{equation}
\omega_0^2 = \frac{g\frac{(l+d)}{2}}{\frac{l^2}{4} + \frac{ld}{2} + \frac{d^2}{4} + \frac{R^2}{4} + \frac{l^2}{12} - \frac{ld}{6} + \frac{d^2}{12}}\end{equation} \\ \begin{equation}
4l^2 + 4ld - \frac{6g}{\omega_0^2}d + 4d^2 + 3R^2 - \frac{6gl}{\omega_0^2} = 0\end{equation} \\ \begin{equation}
4d^2 - 3.42d - 10.67 = 0\hspace{.4in}d = 2.11\end{equation} \\
The containter is therefore $\frac{11}{12}$ full.
\problem{1.9}
\begin{equation} -L\ddot{q} = \frac{q}{C} \end{equation}
\part{a}
\begin{equation} I = \dot{q} = -CV_1\sqrt{\frac{1}{LC}}\sin(\omega_0 t) \end{equation}
\part{b}
\begin{equation} q = CV_1\cos(\omega_0 t)\hspace{.4in}\omega_0 = \sqrt{\frac{1}{LC}} \end{equation}
\problem{1.10}
\part{a}
\begin{equation} E = \frac{1}{2}I\dot{\theta}^2 + \frac{1}{2}c\theta^2\end{equation} \\ \begin{equation}
\frac{dE}{dt} = I\ddot{\theta} + c\theta = I\ddot{\theta} + c\theta = 0\end{equation} \\ \begin{equation}
\omega_0 = \sqrt{\frac{c}{I}} \end{equation}
\part{b}
\begin{equation} E = \frac{1}{2}m\dot{x}^2 + \frac{Tx^2}{l}\end{equation} \\ \begin{equation}
\frac{dE}{dt} = m\ddot{x} + \frac{2Tx}{l} = 0\hspace{.4in}\omega_0 = \sqrt{\frac{c}{I}} \end{equation}
\part{c}
\begin{equation} E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}g\rho Ax^2\hspace{.4in}\frac{dE}{dt} = m\ddot{x} + g\rho Ax = 0\hspace{.4in}\omega_0 = \sqrt{\frac{g\rho A}{m}} \end{equation}
\part{d}
\begin{equation} E = \frac{1}{2}L\dot{q}^2 + \frac{1}{2}\frac{q^2}{C}\hspace{.4in}\frac{dE}{dt} = L\ddot{q} + \frac{q}{C} = 0\end{equation} \\ \begin{equation}
\omega_0 = \sqrt{\frac{1}{LC}} \end{equation}

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