Coefficients of Fourier series for periodically driven oscillators

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Homework Statement


An oscillator is driven by a triangular periodic force (if that makes sense), which has period [tex]\tau[/tex] = 2.

(a) Find the long-term motion x(t), assuming the following parameters: natural period [tex]\tau[/tex][naught] = 2 (that is, [tex]\omega[/tex][naught] = π), damping parameter ß = 0.1, and maximum drive strength fmax = 1. Find the coefficients in the Fourier series for x(t) and plot the sum of the first four terms in the series for 0 <= t <= 6.

Homework Equations





The Attempt at a Solution



For starters: the Fourier coefficients, An (A sub n), I see in my book the equation for. It has in it f sub n, omega naught, omega, beta, and n. n is easy, beta is given, omega naught is given, but omega and f sub n confuse me. I would have thought, that since omega = (2*pi)/(tau), that, with a period of 2, omega would equal pi. But that doesn't seem correct, and, also, I don't know how to find f sub n.

Further: if anyone knows the program Matlab well enough, could you share how I would be able to graph this? I only first began my relationship with the program last night (early this morning, really).

If what I'm asking doesn't make sense, sorry. Chalk it up to my sleeplessness.
 

Answers and Replies

  • #2
vela
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I think the fn's are the Fourier coefficients for the driving force.
 
  • #3
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Yes, the driving force f(t), which is f(t) = sum{fn*cos(wnt)}

But I'm really bad at this stuff and I don't know how to know the fn's.
 
  • #4
vela
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OK, you're only using cosine terms, so the triangle wave must be symmetric about t=0. (If it's not, you'll need to use both the sine and cosine terms.)

You need to calculate

[tex]f_n = \frac{2}{\tau}\int_{-\tau/2}^{\tau/2} f(t)\cos(n\omega t)dt[/tex]

where [itex]\omega=2\pi/\tau[/itex]. If you're having problems doing this integral, post what you have so far or a more specific question so we can see where you're getting stuck.
 
  • #5
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Hmmm..... so f(t) over that interval will be fmax, ehe? So:

fn = ((8*fmax*n*pi)/(tau^2))sin((2*n*pi*∆tau)/(tau))
So that in this case, where
tau = 2
fmax = 1
∆tau =? 1,
fn = (2*pi*n)sin(pi*n), so
fn = 0 for any n because of the sin function.

But they're not all zero.
 
  • #6
vela
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Hmmm..... so f(t) over that interval will be fmax, ehe?
No, that would be a constant function, not a triangle wave. You first need to find a function of t that describe the triangle wave.
 
  • #7
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Is that so? But that would be, like,
f(t) = t
for 2n+1 <= t < 2n+2, where n is an integer, and
f(t) = -t
for 2n <= t < 2n+1

So should I just work it out differently for different n's?
 
  • #8
vela
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You'll have to define it in a piecewise way, but you only have to define it from [itex]t=-\tau/2[/itex] to [itex]t=+\tau/2[/itex], because that's the interval over which you are integrating.

I suggest you sketch it so you get the correct equations for the two pieces.
 
  • #9
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I don't get it. The answer is that

fn = 4/((n*pi)^2),

but I can't get to that point from this.
 
  • #10
vela
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What do you have for f(t)?
 

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