*Any* periodic driving force [5.8]

Any periodic driving force

...can be written as (Thanks Fourier!): $$f(t) = a_0 +\sum_{n=1}^{\infty}[a_n \cos (n\omega t) + b_n \sin (n\omega t)]$$

We have found the solution, x(t) to the driven, damped oscillator equation of motion: $$\uv D x(t)=g_0\cos(\omega t)=g(t)$$ where $$\uv D\equiv\frac{d^2}{dx^2}+2\beta\frac{d}{dt}+\omega_0^2.$$

Imagine that we have solutions $x_1(t)$ and $x_2(t)$ for two different driving forces $g_1(t)$ and $g_2(t)$.

Because $\uv D$ is a linear operator, the solution to: $$\uv D x(t)=g_1(t)+g_2(t)$$ is $x(t)=x_1(t)+x_2(t)$.

Show a bit more...

Fourier components as "unit vectors"

A vector, $\myv v$, can be expressed in terms of its cartesian components as a sum of coefficients times unit vectors: $$\myv v=v_x \uv x+v_y \uv y+ v_z\uv z\equiv v_1\uv e_1+v_2\uv e_2+v_3\uv e_3,$$

The coefficients are given by: $$v_1=\myv v \cdot \uv e_1;\ \ v_2=\myv v\cdot\uv e_2; \text{ etc.}$$
The unit vectors are mutually orthogonal: $$\uv e_i \cdot \uv e_j= \left\{ \begin{matrix} 0 & \text{ if }i\neq j\\ 1 & \text{ if }i= j \end{matrix} \right. $$
Any possible 3-d vector, $\myv v$ can be written in this way, so we say that the set of unit vectore $\{\uv e_1, \uv e_2, \uv e_3\}$ spans the space of possible 3-d vectors.

A fourier series is a sum of coefficients times functions of $t$. $$ f(t) = a_0 +\sum_{n=1}^{\infty}[a_n \cos (n\omega t) + b_n \sin (n\omega t)] $$

Let's pretend that the functions are "unit vectors" according to this relationship...

$\uv{e}_n \equiv \sin(n\omega t)$ for positive integers $n>0$,
$c_n\equiv b_n$ when $n>0$.
$\uv{e}_n \equiv \cos(-n\omega t)=\cos(n\omega t)$ for negative integers $n<0$,
$c_{n}\equiv a_{-n}$ for $n<0$,
$\uv{e}_0 \equiv \frac{1}{2}$ for $n=0$,
$c_{0}\equiv a_0$.

Then Fourier's series can be written as: $$f(t) = \sum_{n=-\infty}^{+\infty} c_n \uv{e}_n(t),$$ remembering that each of the "unit vectors", $\uv{e}_n$ is a function of $t$.

Our recipe for finding the coefficients was $$a_n=\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} f(t) \cos (n\omega t) \,dt,$$ $$b_n=\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} f(t) \sin (n\omega t)\,dt,$$ and $$a_0 = \frac{1}{\tau} \int_{-\tau/2}^{\tau/2} f(t) \,dt = \langle f \rangle.$$

In terms of these "unit vectors", this translates to $$c_n=\frac{2}{\tau} \int_{-\tau/2}^{\tau/2} f(t) \uv e_n(t) \,dt.$$

Now, let the right hand side of the equation above serve as the definition of a "dot product of functions", and then we can write: $$c_n=f(t)\cdot\uv e_n(t).$$ Fourier's discovery that *any* $\tau$-periodic function can be written as a linear combination of these functions is like saying the set of unit vectors spans the (vector-like) space of all possible $\tau$-periodic functions.

As you can check, the "unit vector" functions are orthogonal to each other in the same way as the more familiar unit vectors according to:

$\uv{e}_n \cdot \uv{e}_m = 0$ if $m \neq n$

$\uv{e}_n \cdot \uv{e}_n = 1$

Here are desmos graphs of the products of sin cos functions and sin sin functions.

Homework

Chapter 5: 51 (and 47 from 5.7)