Fourier Analysis: String Vibrations with Fixed Ends and Varying Height

[Fourier mn]

Homework Statement

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 anyone of them is zero? If so why is it?

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

Homework 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}

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 won't be zero and then integrate?
any other ideas how to go about it?

 

[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?

 

[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...

 

[Fourier mn]

every fourth term is a zero

 

[gabbagabbahey]

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

 

[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?

 

[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

 

[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)

 

[Fourier mn]

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

 

[Fourier mn]

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

 

[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.

 

[Fourier mn]

ohhh...got you. thanks