My body has various cavities, such as my throat, mouth, chest, and nose. This cavities have resonant frequencies. I also have a voice box, which creates sound. How do I create sounds at the resonant frequencies of those cavities? It is hard to directly measure the resonant frequencies of those cavities, so how can you calibrate your voice to their resonances?
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$\begingroup$ Although at heart this is a physics problem, I doubt there is much in the way of practical advice that can come from considering the physics from first principles. What you need are singing lessons - I imagine there are no end of appropriate tutorial videos on YouTube. $\endgroup$– N. VirgoApr 13, 2015 at 2:09
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$\begingroup$ @Nathaniel That seemed boring. I just think resonance is a cool phenomenon. $\endgroup$– PyRulezApr 13, 2015 at 10:51
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2 Answers
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The resonances are quite broad: each cavity will amplify a broad range of frequencies, spanning most of or more than an octave.
Driving those resonances isn't as simple as choosing a pitch. You have to do some work to efficiently couple the different cavities to your vocal apparatus, and to maintain the resonance while you're singing. The people who are very good at explaining how to do this are voice teachers.
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To the point of Is it hard to measure the resonant frequencies directly: it's tricky and careful discussion of the measuring procedures is needed. Some of the main problems:
Destruction of the open-end behavior: If you place the speaker and microphone in front of the vocal tract to measure the response, you may have just switched open end behavior of your lips to closed end.
Influence of walls and bends of the vocal tract: there are simple and elegant ways to model the waveguide of the vocal tract using e.g. the Webster equation, but the exact effect of the boundary layer on the walls is not clear. More to that, the Webster equation assumes propagation of the one dimensional plane wave, which is still good enough e.g. for some types of flaring horns, but the bends of the vocal tract might be too big.
Webster equation for the velocity potential:
$$ \frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial (\ln S)}{\partial x}\frac{\partial \Phi}{\partial x}=\frac{1}{c_0^2}\frac{\partial^2 \Phi}{\partial t^2}. $$
Clearly a generalization of 1-D wave equation for waveguides with changing cross-section S(x) along the x-axis.
answered Apr 14, 2015 at 14:08