0
$\begingroup$

I have been studying vibrations and I stumbled upon the overlapping of simple harmonic motions. Consider the case where the number of oscillators $n$ is $n \gg 1$, all of them have the same angular frequency $\omega$, and all have consecutive different phases, that is oscillator 1 is $\delta$ radians in front of oscillator 2 and so on. In this case, the total motion can be written as $$ R' = \text{Re} \left[ \sum_k A_k e^{i(\omega t + j\delta)} \right] $$ where $A_k$ is the amplitude of the $k^\text{th}$ oscillator and $j$ is a mute variable, from the sum.

Why do we take only the real part?

In a way I understand, because the imaginary part is called imaginary for a reason, but nonetheless it is a mathematical part of the solution. How should I guide my intuition? I would easily understand it if somehow we transformed the imaginary part into real, but merely ignoring it seems just so unusual that I can't get my head around it.

$\endgroup$
9
  • $\begingroup$ What's a "mute integral"? $\endgroup$
    – DanielSank
    Sep 26 '18 at 23:03
  • $\begingroup$ I didn't know the correct term but it felt that it suited the case. I call it mute because, it's just from the sum, such that $j$ is between $0$ and $n$, you know, for the first term of the sum $j=0$, the second $j=1$ and so on until $j=n$ $\endgroup$
    – Bidon
    Sep 26 '18 at 23:06
  • $\begingroup$ Ok I've tried to clean up the question a bit. Please take a look. I still don't understand what $j$ means. $\endgroup$
    – DanielSank
    Sep 26 '18 at 23:08
  • $\begingroup$ I noticed, but it is not what I really meant. In this case all SHM's have the same amplitude,$A$. For the $j$, suppose this:$$\sum{j}$$ from 0 to 3, that sum is equal to $0+1+2+3$, it has the same role in the sum on the post $\endgroup$
    – Bidon
    Sep 26 '18 at 23:13
  • 3
    $\begingroup$ @Duartejfs we call those dummy variables $\endgroup$ Sep 26 '18 at 23:18
2
$\begingroup$

We take the real part because the differential equations governing the dynamics of the system were real to begin with (and so were the initial conditions). To put it simply, there is 'no way to add apples with apples and get oranges'.

To explain why taking the real part is okay, let's start with the (hopefully) familiar differential equation for the damped driven harmonic oscillator: $$\underbrace{m\ddot{x}}_{\text{inertia}}=-\underbrace{m\omega_0^2x}_{\text{spring force}}-\underbrace{m\gamma\dot{x}}_{\text{damping}}+\underbrace{F\cos\omega_{\text{d}} t}_{\text{drive term}}.$$ Notice that this is a linear differential equation, meaning that only linear powers of $x$ and its time derivatives show up (no terms like $\sqrt{x},~x^2$, et cetera are present). The steady state solution to this differential equation is of the form $$x(t)=x_0 \cos(\omega_{\text{d}} + \phi)$$ where $x_0$ and $\phi$ are the steady state amplitude and phase shift, respectively. Now you can plug this into our differential equation to arrive at equations for $x_0$ and $\phi$, but there is another approach based on complex exponentials which can greatly simplify the mathematics when the systems become more complicated (e.g. three coupled oscillators). Things like differentiation, multiplication, and division are much simpler with exponentials than with $\sin$ and $\cos$. To see how this works for the problem at hand, let's create the complex variable $$\begin{align}z(t)&= x_0 \cos(\omega_{\text{d}} t + \phi) + i x_0 \sin(\omega_{\text{d}} + \phi) \\ & = x_0 e^{i(\omega_d t+\phi)}\end{align}$$ where the last line is by Euler's formula. Notice that $x(t) = \text{Re}[z(t)]$ and that $F\cos\omega_{\text{d}} t = \text{Re}[F e^{i\omega_{\text{d}} t}]$, and therefore our differential equation is simply the real part of the complex differential equation $$m\ddot{z(t)}=-m\omega_0^2 z(t)-m\gamma\dot{z(t)}+Fe^{i\omega_{\text{d}} t}.$$ This is actually two differential equations: our original differential equation plus an imaginary one that we added. Because the differential equation for this system is linear, we don't have to worry about products between real (one we care about) and imaginary (one we invented) terms. Let me know if you are still confused about this. It trips up everyone when they first see it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.