Let m, g, and L be positive constants. (a) Show that $V(\theta, \omega)=\frac{1}{2} m L^{2} \omega^{2}+m g L(1-\cos \theta)$ is positive definite in an open neighborhood containing $(\theta, \omega)=(0,0)$. (b) Show that $\dot{V}(\theta, \omega)$ is negative semidefinite on solution trajectories of

$$\begin{array}{l} \dot{\theta}=\omega \\ \dot{\omega}=-\frac{\mu}{m} \omega-\frac{g}{L} \sin \theta \end{array}$$

on an open neighborhood containing $(\theta, \omega)=(0,0)$.

Construct the Lissajous figures for the following motions: (a) $x=\cos 2 \omega t, y=\sin 2 \omega t$ (b) $x=\cos 2 \omega t, y=\cos (2 \omega t-\pi / 4)$ (c) $x=\cos 2 \omega t, y=\cos \omega t$.

Write $7 \sin \omega t+24 \cos \omega t$ in the form $A \sin (\omega t+\phi).$

Solve the equation. $t=t_0+\dfrac{k}{2}\omega,$ for $\omega$

Determine the current in each case: (a) $V=5 \mathrm{~V}, R=1.0 \Omega$ (b) $V=15 \mathrm{~V}, R=10 \Omega$ (c) $V=50 \mathrm{~V}, R=100 \Omega$ (d) $V=30 \mathrm{~V}, R=15 \mathrm{k} \Omega$ (e) $V=250 \mathrm{~V}, R=5.6 \mathrm{M} \Omega$

The formula $\omega=\frac{\theta}{t}$ can be rewritten as $\theta=\omega t$. Substituting $\omega t$ for $\theta$ changes $s=r \theta$ to $s=r \omega t$.

Use the formula $s=r \omega t$ to find the value of the missing variable.

r=6 centimeters, $\omega=\frac{\pi}{3}$ radians per second, t= 9 seconds

Determine the following: c$\int_{0}^{\pi / \omega} \sin \omega t \cos 2 \omega t d t$

The $N+1$ complex numbers $\omega_m$ are given by $\omega_m=\exp (2 \pi i m / N)$, for $m=$ $0,1,2, \ldots, N$. (a) Evaluate the following: (i) $\sum_{m=0}^N \omega_m$, (ii) $\sum_{m=0}^N \omega_m^2$, (iii) $\sum_{m=0}^N \omega_m x^m$. (b) Use these results to evaluate: (i) $\sum_{m=0}^N\left[\cos \left(\frac{2 \pi m}{N}\right)-\cos \left(\frac{4 \pi m}{N}\right)\right]$, (ii) $\sum_{m=0}^3 2^m \sin \left(\frac{2 \pi m}{3}\right)$.

The formula $\omega=\frac{\theta}{t}$ can be rewritten as $\theta=\omega t$. Using $\omega t$ for $\theta$ changes s=r$\theta$ to $s=r \omega t$. Use the formula $s=r \omega t$ to find the value of the missing variable. r=6 cm, $\omega=\frac{\pi}{3}$ radians per sec, t=9 sec

Show that. under appropriate conditions on $f$ and its derivatives. $\mathcal{F}_{S}\left[f^{(4)}(x)\right](\omega)=\omega^{4} \widehat{f}_{S}(\omega)-\omega^{3} f(0)+\omega f^{\prime \prime}(0).$

If $f(t)=e^{-a t} \sin \omega t$, show that $\textbf{F}(s)=\dfrac{\omega}{(s+a)^{2}+\omega^{2}}$.

Given that the complex number $\omega \neq-1$ is a solution of $z^3+1=0$, show that:

$\omega^2-\omega+\mathrm{I}=0$

The formula $\omega=\frac{\theta}{t}$ can be rewritten as $\theta=\omega t$. Substituting $\omega$ t for $\theta$ converts $s=r \theta$ to $s=r \omega t$. Use the formula $s=r \omega t$ to find the value of the missing variable. $r=6 \mathrm{~cm}, \omega=\frac{\pi}{3}$ radians per sec, $t=9 \mathrm{sec}$

Show that

$$\int_{-\infty}^{\infty}\left(\frac{\sin a \omega}{a \omega}\right)^{2} d \omega=\frac{\pi}{a}$$