Oscillations - Part 2

How does $$e^{i \theta}$$ lead to oscillations?

Euler's Formula

\begin{equation} e^{i\theta} = \cos \theta + i \sin \theta \label{eq:eulersformula} \end{equation}

A number: $$z = a + b y$$ on the complex plane.

This can also be expressed in polar coordinates: $$z = r \left( \cos{\theta} + i \sin{\theta}\right)$$

Any complex number can be expressed using the exponential: $$z = |r| e^{ix}$$

Now, we have $$e^{i x} = r \left( \cos{\theta} + i \sin{\theta}\right)$$ We need to figre out what $r$ and $x$ equal.

Start by taking the derivative of both sides w.r.t $x$ $$ \frac{d}{dx} e^{ix} = ie^{ix} $$ and $$ \frac{d}{dx} \left[ r \left( \cos{\theta} + i \sin{\theta}\right) \right] = \frac{dr}{dx}\left( \cos \theta + i \sin \theta \right)+ r(-\sin \theta + i \cos\theta)\frac{d\theta}{dx}$$ Thus: $$ ie^{ix} = \frac{dr}{dx}\left( \cos \theta + i \sin \theta \right)+ r(-\sin \theta + i \cos\theta)\frac{d\theta}{dx}$$ Comparing reals to reals and imaginary to imaginary: $$i r \cos \theta = \frac{dr}{dx} i \sin \theta + r i \cos \theta \frac{d\theta}{dx}$$ and $$-r \sin \theta = \frac{dr}{dx} \cos \theta - r \sin \theta \frac{d\theta}{dx}$$ we can surmise that: $$\frac{dr}{dx} = 0$$ and $$\frac{d\theta}{dx} = 1 \Rightarrow \theta = x + C $$ Since $e^{i 0} = 1$, we then also can say that $$r(0) = 1$$ and $$\theta(0) = 0$$ thus $r = 1$ and $C = 0$ and $\theta = x$ Which means that $$e^{i\theta} = 1 \times \left( \cos{\theta} + i \sin{\theta}\right)$$

Underdamped, revisited

The damped eq. of motion was \begin{equation} \ddot{x} + 2 \beta \dot{x} + w_0^2 x = 0 \label{eq:dampedoscillatoreqofmotion} \end{equation} which was solved by: $$x \propto e^{\alpha t}$$ where \begin{equation} \alpha = -\beta \pm \sqrt{\beta^2 - \omega_0^2} \end{equation}

If $$\beta < \omega_0$$ then we see that $\sqrt{\beta^2 - \omega^2}$ will be imaginary.

Let's see how this affects our solution.

Noting that $$\sqrt{\beta^2 - \omega_0^2} = i \sqrt{\omega_0^2 - \beta^2}$$ we can recast the solution as: \begin{equation} x(t) = e^{-\beta t} \textrm{Re} \left(A_1 e^{i \omega_1 t} + A_2 e^{- i \omega_1 t}\right) \label{eq:dampedsolutionreals} \end{equation} with $$\omega_1 = \sqrt{\omega_0^2 - \beta^2}$$ (we only want the real part: $\textrm{Re}$)

We can use the Euler Formula \eqref{eq:eulersformula} to re-write \eqref{eq:dampedsolutionreals} as \begin{equation} x(t) = e^{-\beta t}\left(\bar{A}_1 \cos \omega_1 t + \bar{A}_2 \sin \omega_1 t \right) \label{eq:dampedsolutioncossin} \end{equation}

Using the trig identity: \begin{equation} \cos \left(\theta + \phi \right) = \cos \theta \cos \phi - \sin \theta \sin \phi \end{equation} \eqref{eq:dampedsolutioncossin} becomes: \begin{equation} x(t) = A e^{-\beta t} \cos \left(\omega_1 t + \phi \right) \label{eq:dampedsolutionphase} \end{equation} where $A = \sqrt{\bar{A}_1^2 + \bar{A}_2^2}$ and $\phi = \tan^{-1}\left(\frac{-\bar{A}_2}{\bar{A}_1} \right)$

One thing to note is that the frequency of oscillation has now changed...

Oscillations - Part 2

Euler's Formula

Underdamped, revisited

Applications