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I am told that I can model the vibration of a guitar string of length $l >0$ by the following Sturm-Liouville equation $$ -u'' = \lambda u \: \: \text{ on } [0,l],$$ with boundary conditions $$u(0)=0 \: \text{ and } \: u(l)=0.$$

And the vibrations of air in a pipe (with one end closed and the other open) by $$-u'' = \lambda u $$ with BCS $$ u(0)=0 \: \text{ and } u'(l)=0. $$

It is easy to see that the eigenvalues of the first and second problem are $\lambda_k = \left( \frac{k \pi} {l}\right)^2$ and $\lambda_k = \left( \frac{(2k-1) \pi} {2l}\right)^2$, respectively.

My question is how does this explain the difference in sound from a plucked guitar string and a pipe even when they are tunes to the same pitch?

Sorry, maybe this is obvious but I don't have the physical intuition and I am curious...

Thx!

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up vote 2 down vote accepted

Each eigenfunction behaves independently as a simple harmonic oscillator. From the eigenvalues, you can determine the frequencies of oscillation, which are $f_k^s = k/2l$ and $f_k^p = (2k-1)/4l$ for the string and the pipe respectively. These are the frequencies of the harmonics that sound when the instruments are played. As you can see, the harmonics of the string are all the multiples of a fundamental frequency $f_1^s$, while those of the pipe are only the odd multiples of $f_1^p$. So even if both instruments are tuned to the same pitch, that is, the same fundamental frequency $f_1^s = f_1^p$, the sound of the pipe will be missing the even-numbered harmonics.

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Although surely the precise distribution of the overtones is of significant consequence in timbre, the pipe-versus-string comparison, creating the odd-versus-all overtones, has a huge impact apart from the further niceties about precise harmonics. If one has any sort of electronic tone-generation device, the hearable impact of excluding even overtones is immediate and huge! Note that this issue is clear even with the one-dimensional model. –  paul garrett Nov 1 '12 at 22:55
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