It seems that the more tension on the string, the more slowly the sound would decay after being plucked. Is there a formula relating the two? How is it derived?
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Basically, no. The decay is a function of friction (air drag, loss at the bridge and nut), non-elasticity in the string material, energy transferred to the air or instrument body as the sound is generated, and similar factors.
For a very simple example: a string tuned to a highly resonant frequency of a guitar or violin body will produce a longer-lasting sound than one tuned to a non-resonant frequency. This is primarily because the entire system, when nonresonant, is essentially far less "elastic" in the sense of returning energy to the desired frequency.
let me also point out that the decay in the perceived sound level is a strong function of frequency because of the way our ears work.
answered Oct 14, 2015 at 15:44
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1$\begingroup$ Thanks. So is there a function of actual sound level? Or just perceived? $\endgroup$ Commented Oct 14, 2015 at 19:45
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1$\begingroup$ Could you elaborate more on the decay in our perception? $\endgroup$– KF GaussCommented Feb 13, 2019 at 13:02
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$\begingroup$ @KFGauss Simply that our spectral sensitivity is not "flat," meaning it takes different energy (or sound pressure) levels for us to perceive one pitch as "same loundness" as some other pitch. $\endgroup$ Commented Feb 13, 2019 at 14:32
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$\begingroup$ @Carl Witthoft I see... I thought you were implying that human spectral sensitivity is heavily dependent on loudness at higher frequencies. $\endgroup$– KF GaussCommented Feb 13, 2019 at 17:00