User talk:Reedbeta

Derivation of the equation for simple harmonic motion

There must be an easier way to derive this.

The desired closed-form solution is: $x(t)=A\cos (\omega t+\varphi )$ for some constants $A$, $\varphi$.

A differential equation for a 1-dimensional harmonic oscillator is:

$F=-kx$

Since force is mass times acceleration,

$\frac{d^{2}x}{d{t}^2}=-\frac{k}{m}x$

To solve this second-order differential equation, we introduce a new variable v, reducing it to a system of linear first-order differential equations. Also writing ${\omega }^2=k/m$ for convenience:

$\frac{dx}{dt}=v$

$\frac{dv}{dt}=-{\omega }^{2}x$

Rewrite this system using matrices:

$\frac{d}{dt}\left[\begin{array}{c}x\\ v\end{array}\right]=\left[\begin{array}{ccc}0& & 1\\ -{\omega }^2& & 0\end{array}\right]\left[\begin{array}{c}x\\ v\end{array}\right]$

The coefficient matrix has characteristic polynomial

$\det (A-\lambda I)={\lambda }^2+{\omega }^2$

so its complex eigenvalues are $\pm \omega i$. A complex eigenbasis is then:

${ℬ}=\{\left[\begin{array}{c}-\frac{i}{\omega }\\ 1\end{array}\right],\left[\begin{array}{c}\frac{i}{\omega }\\ 1\end{array}\right]\}$,

which leads to the general solution of our system,

$x(t)=c_1{e}^{\omega it}\frac{-i}{\omega }+c_2{e}^{-\omega it}\frac{i}{\omega }$

Here, $c_1$ and $c_2$ are the (complex) coordinates of $[x_0,v_0{]}^T$ (that is, the initial state vector of the system) with respect to the complex eigenbasis ${ℬ}$.

Applying Euler's formula:

$x(t)=\frac{-i{c}_1}{\omega }(\cos \omega t+i\sin \omega t)+\frac{i{c}_2}{\omega }(\cos -\omega t+i\sin -\omega t)$

Using the trigonometric identities $\cos a=\cos -a$ and $\sin a=-\sin -a$, this simplifies to:

$x(t)=\frac{i(c_2-c_1)}{\omega }\cos \omega t+\frac{c_1+c_2}{\omega }\sin \omega t$

Now, let us return to the coordinates $c_1$ and $c_2$. We know that

$\left[\begin{array}{c}{x}_0\\ v_0\end{array}\right]=\left[\begin{array}{ccc}-\frac{i}{\omega }& & \frac{i}{\omega }\\ 1& & 1\end{array}\right]\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$

Inverting the matrix, we have

$\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc}\omega i& & 1\\ -\omega i& & 1\end{array}\right]\left[\begin{array}{c}{x}_0\\ v_0\end{array}\right]$ where $x_0$ and $v_0$ are both real.

Substituting back into the equation for $x(t)$, we have

$x(t)=x_0\cos \omega t+\frac{v_0}{\omega }\sin \omega t$

Another trig identity gives:

$x(t)=\sqrt{x_0^2+\frac{v_0^2}{{\omega }^2}}\sin (\omega t+\theta )$, where $\theta =\arctan \frac{x_0\omega }{v_0}$.

If we then let

$A=\sqrt{x_0^2+\frac{v_0^2}{{\omega }^2}}=\sqrt{x_0^2+\frac{m{v}_0^2}{k}}$ and

$\varphi =\theta -\frac{\pi }{2}=\arctan \frac{x_0\omega }{v_0}-\frac{\pi }{2}$,

then we finally end up at the desired solution

$x(t)=A\cos (\omega t+\varphi )$.

Gradient of an ellipsoid

The basic ellipsoid equation is

$f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1$

So the gradient is

$\mathrm{\nabla }f(x,y,z)=⟨\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2}⟩$

ArbCom elections are now open!

{{Wikipedia:Arbitration Committee Elections December 2015/MassMessage}} MediaWiki message delivery (talk) 13:01, 23 November 2015 (UTC)