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If standing waves are established in musical instruments like the guitar, then how do we hear the sound because standing waves don't transfer energy from one place to another. At least that's what I have learned.

My guess is that standing waves are established in the string itself but when the string vibrates, it also vibrates the air around it

Qmechanic
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Spluesh
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    Standing waves don't transfer energy across the stationary nodes. But they can transfer energy to the surrounding air. – SmarthBansal Jan 13 '24 at 01:27
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    Does this answer your question? How does the string of an acoustic guitar transfer energy to the guitar's body? – Stevan V. Saban Jan 13 '24 at 01:42
  • This may be helpful: https://physics.stackexchange.com/a/615386/123208 – PM 2Ring Jan 13 '24 at 02:11
  • In physical systems, there are no perfect standing waves due to the termination of the string. From a mathematical point of view, the boundary conditions (which most probably, are frequency-dependent) do not correspond to either rigid or free termination and there is some loss of energy. When the strings are coupled to another system (instrument’s body) there is energy transfer from the string to the body which radiates. As Poisson Aerohead very well mentioned there is coupling between the transmitted sound to the strings but this can be small, like the radiation from the strings (cont.) – ZaellixA Jan 13 '24 at 11:36
  • (cont.’ed) themselves. The impedance mismatch between the vibrating strings and the surrounding air is quite large and the former do not radiate efficiently. If you treat the system as linear then that also means that they are not affected that much from the radiated sound from the body. – ZaellixA Jan 13 '24 at 11:38

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The short answer is that the energy carried away as music is like a "loss" or "friction" term. It can be complicated though. For an electric guitar it arises from Lorentz forces with the pickups / coils. An acoustic guitar is very complicated, because the strings resonate with the sound hole. I have been interested in "rigorous physics of music" questions and it is hard to find analyses that aren't overly hand wavy.

As the comment above says, standing waves don't transfer energy across stationary nodes. Obviously, there are damped standing wave solutions with loss or friction terms that decay over time, and it is generally ignored but those loss terms never actually destroy energy.

Poisson Aerohead
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  • This comment is addressed to you, as in: related to your answer, but not related to the question. About physics of music. I am an avid singer; I sing in a choir. In singing there is the phenomenon of singer's formant The singer's formant is usually associated with singing very loud, but at moderate volume it is already possible to sing with a lot of singer's formant quality, which I do. Given my physics background I have a keen interest in the physics of how the higher harmonics become so energetic. – Cleonis Jan 14 '24 at 12:57
  • Interesting. Out of curiosity, does that include male opera singers? I ask because 3 kHz is pretty high overall, but especially for a man. What little physics of music I have looked at has been mostly guitars and the electronics that are now used. I would expect vocals to be an extremely difficult analysis. I don't even really understand how vocal chords work (obviously they don't make sound every time we breathe). – Poisson Aerohead Jan 14 '24 at 16:15
  • Especially male singers. For instance, when singing at 220 Hz the 2nd harmonic is at 440Hz, 4th at 880Hz, 8th at 1760Hz, 16th at 3520Hz. So: in the range around 3kHz you get a closely spaced cluster of harmonics. (For instance, from the 15th to the 16th harmonic is a semitone) So when a singer has the ability to energize the harmonics all the way to the range of four octaves above the fundamental, that is when the audience experiences a big sound. For female singers it can get tricky, because when their fundamental is up high the harmonics of the fundamental are more widely spaced. – Cleonis Jan 14 '24 at 17:28