How does the string of an acoustic guitar transfer energy to the guitar's body?

Question

I've learned through the answer to this post that the purpose of the soundboard of an acoustic guitar is to match the impedance of air, facilitating the energy transfer from the strings to the sound waves. The energy goes from the string to the guitar bridge to the soundboard to the air.

However, for this to occur, shouldn't the impedance of the string match that of the bridge, who in turn should match that of the soundboard? The whole purpose of the soundboard is to have a different impedance from the strings (closer to the air's) but through this reasoning they should all be equal.

How exactly does the string transfer energy to the soundboard? Is there something I am missing?

Answer 1

If the impedance of the string were truly matched to that of the soundboard, then a string vibration would be completely absorbed into the soundboard with no reflection and the string would almost immediately stop vibrating i.e., the guitar would have no sustain and the sound radiated off the soundboard into the air would sound loud, but dull and lifeless.

Some degree of mismatch is required so the vibrating string can "sing" for a while after being plucked. The guitar designer's job is then to balance projection (how loud it is) with sustain (how long a note lasts) by playing with the details of how the string is coupled to the soundboard and how flexible the soundboard is.

This is a complicated business because the impedances are all functions of frequency, and you do not want the system to exhibit uneven response where one string is much louder than the others, or where a note played at one particular position on the neck sounds sour.

Answer 2

Transverse soundboard vibrations are amplified most efficiently since the soundboard moves the most air with waves that vibrate normal to its surface.

A guitar string vibrates transversely as well but only string vibrations perpendicular to the soundboard will be amplified efficiently. The purpose of the bridge is to convert as much of the strings vibration into waves that are normal to the soundboard so the string is amplified most effectively.

Classical guitarists take advantage of this amplification using a finger stroke (rest stroke) that pushes vertically on the string rather than horizontally (free stroke) and the tone and volume are noticeably different for the same finger force.

Soundboard design (particularly struts and bracing) is responsible for assuring an equal response across all frequencies with as much volume as possible.

Answer 3

In the violin acoustics field (equivalent to guitar acoustics for our purposes here), it is common to study the body of the instrument and the air which surrounds it all at once. So one measures the response and impedance of the instrument/air system as felt by the strings at the bridge. This is achieved by striking the bridge with a calibrated hammer and recording the radiated sound and the movement of the bridge.

An illuminating edge case of this kind of analysis would be an incredibly stiff, nearly massless soundboard. It would actually perform quite well! The soundboard would serve as an increase in radiative surface area for the string. (This is why we luthiers use very light and stiff woods like spruce and cedar) In this case, the impedance of the soundboard in a vacuum would be absurdly mismatched to both the string and the air, but energy is transferred anyway, because the soundboard cannot move without moving the air around it.

There is definitely impedance matching going on in instrument design, but at the level of granularity you are using to model the system, I’m not sure it is a very useful metric. At the very least it is problematic to try to separate soundboard impedance from air impedance.

A very useful consideration at the instrument/air interface is that of the overall dimensions of the vibrating area. If the dimensions are much smaller than the wavelength of the sound in question, the air will just slosh around the instrument rather than causing meaningful radiative sound. So the strings themselves can’t radiate any but the highest pitches directly into the air because they are so thin. The lower the frequency range, the bigger the instrument, not only because resonators like strings get longer, but because a violin simply can’t radiate a cello’s fundamental frequencies, even if we tune the violin’s strings accordingly. The air has time to whoosh all the way around the violin before the pressure reverses.

This isn’t exactly a clear cut answer to your question, but those are surprisingly hard to find in instrument acoustics. I hope it sheds light on one of the many complexities at play here.