# Fourirer Analysis

1. Oct 26, 2008

### Fourier mn

1. The problem statement, all variables and given/known data

String of length L and string mass-density of$$\mu$$ is fixed at both ends. at t=0
y(x,t)=
4xh/L 0<x<L/4
2h-4xh/L L/4<x<L/2
0 L/2<x<0
Find the first four coefficients, is any one of them is zero? If so why is it?

/\
/ \
|/ \______|
the peak is L/4 and the right corner of the triangle is L/2.
2. Relevant equations

2/L{$$\int$$(4xh/L) sin(nx$$\pi$$/L) over 0<x<L/4 +
$$\int$$(2h-4xh/L)sin(nx$$\pi$$/L) over L/4<x<L/2 +
$$\int$$ 0*sin(nx$$\pi$$/L) over L/2<x<L}

3. The attempt at a solution

so the third integral is equal to zero, we are left with only two.
After integrating the first two I've got the following result for An=
(8h)/(n$$\pi$$)^2[2sin(n$$\pi$$/4)-sin(n$$\pi$$/2)]
is this result correct? I've never seen Fourier series described by two sine terms. should I change the height to (h+1) so the last integral wont be zero and then integrate???
any other ideas how to go about it?

2. Oct 26, 2008

### gabbagabbahey

Use the following trig identity: $sin(2x)=2sin(x)cos(x)$ to rewrite your second sine term as $2sin(\frac{n \pi}{4})cos(\frac{n \pi}{4})$...what does that make A1,A2,A3 and A4?

3. Oct 26, 2008

### Fourier mn

hmmmmm......I didnt think I can use cos as part of the Fourier series.
is it ok? but I'm still going to have two terms of sin and cos--
2sin(npi/4)(1-cos(npi/4)) it's still not uniformly zeros...

4. Oct 26, 2008

### Fourier mn

every fourth term is a zero

5. Oct 26, 2008

### gabbagabbahey

Yes, every fourth term is zero....is there any physical interpretation for this result?

6. Oct 26, 2008

### Fourier mn

I'm still trying to figure out how to graph the wave.
But, I think that it means that the wave is symmetric about the x-axis and that's why it's zero, correct?

7. Oct 26, 2008

### gabbagabbahey

Not really; it means that every fourth harmonic is zero...in other words the initial wave pulse doesn't contain any harmonic of the frequency n*pi

8. Oct 26, 2008

### Fourier mn

so the other half of the string never moves?
i agree that every multiple of 4 will cause the amplitude to be zero, but i thought that the main notion from Fourier is that one should always consider harmonic frequency when using his analysis (i.e the frequency is the same for all=>harmonic)

9. Oct 26, 2008

### Fourier mn

how would you go about graphing it? just choose arbitrary values?

10. Oct 26, 2008

### Fourier mn

A1,A2 and A3 do contain the harmonic n*pi

11. Oct 26, 2008

### gabbagabbahey

A1,A2,A3 contain harmonics of npi/4, npi/2 and 3npi/4 not npi....this means that the fourier series of the initial waveform does not contain any terms with frequency (0,pi,2pi,....).

To graph the function just do a 3D plot of y(x,t) with t as your z-axis.

12. Oct 26, 2008

### Fourier mn

ohhh....got you. thanks