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Seeing a video of astronaut Chris Hadfield playing a guitar on the International Space Station made me wonder if a guitar or other stringed instrument played in zero-G would sound any different than on earth.

It seems that on earth, the downward pull of gravity could cause an asymmetrical oscillation of the guitar string, with a larger amplitude as the string moves downward due to the pull of gravity. (of course, it wouldn't take zero-G to test this since a guitar could be held vertically)

With the amount of tension that a guitar string is under, perhaps this effect is so minuscule so as to be unnoticeable?

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If gravitational effects were significant, guitars would sound different depending on their orientation in the earth's gravitational field. They don't. – Ben Crowell May 13 '13 at 3:11
I think that the relevant variables are the ones that have to be with the medium, because the gravitational force upon a string is really small. – Ana May 13 '13 at 3:38
oh. yes. yes i do... (•_•) ( •_•)>⌐■-■ (⌐■_■) – wim May 22 '13 at 1:40
up vote 18 down vote accepted

The effect of gravity is miniscule, and here's why:

The speed of sound in a string is basically $$ v = \sqrt{\frac{T}{\lambda}}, $$ where $T$ is the tension and $\lambda = M/L$ is the mass per unit length. The frequency of a plucked string will then be this sound divided by the length of the oscillator: $f = v/L$. Combining and rearranging tells us the tension keeping the string taught will be $$ T = MLf^2. $$

Gravity can alter this tension only by something of the order of $$ T_\mathrm{grav} = Mg, $$ give or take some factor of order unity depending on orientation. Thus the intrinsic tension overwhelms the effect of gravity by a factor of something like $$ x \equiv \frac{T}{T_\mathrm{grav}} = \frac{Lf^2}{g} \approx \frac{(1\ \mathrm{m})(500\ \mathrm{Hz})^2}{10\ \mathrm{m/s^2}} = 2.5\times10^4. $$

Since frequency goes as the square root of tension, moving the string from free fall to standing still in the Earth's gravitational field will change the frequency by something like one part in $2x = 5\times10^4$.

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It's worth noting that a frequency change of one part in $5\times 10^4$ is about $\log2\left( \frac{5\times 10^4+1}{5\times 10^4} \right) \approx 2.9\times 10^{-5}$ octaves, or less than one 2500th of a semitone - quite a bit too small to be detected by the human ear. – Nathaniel May 13 '13 at 5:47
Note that moving the string in a vacuum is going to produce no sound whatsoever... – Shadur May 13 '13 at 8:01
I like your answer, but the last sentence saying "…moving the string from vacuum to the Earth's gravitational field…" hurts very badly. Theoretically you can't leave the Earths gravitational field at all. You mean, going back from free fall… – tamasgal May 13 '13 at 8:25
@septi Corrected that unfortunate wording. Thanks. – Chris White May 13 '13 at 14:45

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