Coefficients of Fourier series for periodically driven oscillators

[Oijl]

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.

 

[vela]

I think the f_n's are the Fourier coefficients for the driving force.

 

[Oijl]

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.

 

[vela]

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.

 

[Oijl]

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.

 

[vela]

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.

 

[Oijl]

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?

 

[vela]

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.

 

[Oijl]

I don't get it. The answer is that

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

but I can't get to that point from this.

 

[vela]

What do you have for f(t)?