Reedbeta

Derivation of the equation for simple harmonic motion

There must be an easier way to derive this.

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

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

$\ F=-kx$

Since force is mass times acceleration,

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

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:

${\frac {d}{dt}}{\begin{bmatrix}x\\v\end{bmatrix}}={\begin{bmatrix}0&&1\\-\omega ^{2}&&0\end{bmatrix}}{\begin{bmatrix}x\\v\end{bmatrix}}$

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:

${\mathcal {B}}=\left\{{\begin{bmatrix}-{\frac {i}{\omega }}\\1\end{bmatrix}},{\begin{bmatrix}{\frac {i}{\omega }}\\1\end{bmatrix}}\right\}$,

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 ${\mathcal {B}}$.

Applying Euler's formula:

$x(t)={\frac {-ic_{1}}{\omega }}\left(\cos \omega t+i\sin \omega t\right)+{\frac {ic_{2}}{\omega }}\left(\cos -\omega t+i\sin -\omega t\right)$

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

${\begin{bmatrix}x_{0}\\v_{0}\end{bmatrix}}={\begin{bmatrix}-{\frac {i}{\omega }}&&{\frac {i}{\omega }}\\1&&1\end{bmatrix}}{\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}$

Inverting the matrix, we have

${\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}\omega i&&1\\-\omega i&&1\end{bmatrix}}{\begin{bmatrix}x_{0}\\v_{0}\end{bmatrix}}$ 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

$A={\sqrt {x_{0}^{2}+{\frac {v_{0}^{2}}{\omega ^{2}}}}}={\sqrt {x_{0}^{2}+{\frac {mv_{0}^{2}}{k}}}}$ and

$\phi =\theta -{\frac {\pi }{2}}=\arctan {\frac {x_{0}\omega }{v_{0}}}-{\frac {\pi }{2}}$,

then we finally end up at the desired solution

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

Gradient of an ellipsoid

The basic ellipsoid equation is

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

So the gradient is

$\nabla f(x,y,z)=\left\langle {\frac {2x}{a^{2}}},{\frac {2y}{b^{2}}},{\frac {2z}{c^{2}}}\right\rangle$

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