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I'm curious as to how many calories(or joules) are directly transferred to a guitar through a player's arm and pick when strumming the guitar. Let's say that the player has an average weight (for the weight of his arm) and he is strumming a regular six-string acoustic guitar that produces from 80 to 90 dB.

Would this be enough information to make a ball park estimate? Thanks!

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    $\begingroup$ A straightforward way to determine this might be to calculate how much energy per unit time is transported along each string, given the wavelength of each note played, the amplitude, and the frequency on each string. Some energy from the strings is transferred to the body of the guitar, and ultimately to the air as sound, but the upper limit on energy going into the guitar would be what is transported by the strings. hyperphysics.phy-astr.gsu.edu/hbase/Waves/powstr.html $\endgroup$ – Ernie Mar 19 '17 at 2:44
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Ultimately, you still need to know how the sound level changes over time.

Here's where we can get with what you have: first, convert dB to power in W using:

$L=10\log (P/P_0) $

where $L $ is loudness in dB, $P $ is intensity in W, and $P_0=10^{-12} $ W. So now we have energy per unit time. Assuming that the guitar just keeps ringing at 80 dB for the entire time it's playing and then abruptly stops after a while, you could get a very rough estimate by just multiplying the power by the note length.

But this is a pretty terrible way of describing reality, as string sound power decays over time. Typically, this is exponential, i.e.

$P(t)=Pe^{-kt}$

where $P $ is the initial power and for some decay constant $k $. In this case we integrate the power function from $t=0$ to infinity to get $E=P/k $.

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  • $\begingroup$ Of course, due to damping the string doesn't actually ring for an infinite time, but adding those effects in gives you only a minor correction to the value given above. $\endgroup$ – probably_someone Mar 18 '17 at 3:09
  • $\begingroup$ thanks for the help, but where is the I in the equation? I'm trying to plug in the values into the equation you gave me. $\endgroup$ – Chet Spalsky Mar 19 '17 at 0:05
  • $\begingroup$ @Chet_Spalsky Sorry, that should have been a P. $\endgroup$ – probably_someone Mar 19 '17 at 0:09
  • $\begingroup$ So I did the math and the answer comes out to be that 80 dB takes about 2.9^-9 watts. This seems strange, does this mean I'll have to strum a guitar hundreds of millions of times for the energy equivalent to 1 Watt to have been transferred through it? This doesn't seem intuitively accurate. $\endgroup$ – Chet Spalsky Mar 19 '17 at 0:28
  • $\begingroup$ There is an efficiency factor that you would have to estimate for the transfer of power from the arm to the strings. I suspect it's a very small number, meaning the player power is much larger than the sound power. Consider there is a lot of friction between the pick and strings, and a lot of the energy fails to make it into the air. $\endgroup$ – Bill N Mar 19 '17 at 1:39

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