11
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • 11
    $\begingroup$ If gravitational effects were significant, guitars would sound different depending on their orientation in the earth's gravitational field. They don't. $\endgroup$
    – user4552
    Commented May 13, 2013 at 3:11
  • 1
    $\begingroup$ 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. $\endgroup$
    – Ana S. H.
    Commented May 13, 2013 at 3:38
  • $\begingroup$ oh. yes. yes i do... (•_•) ( •_•)>⌐■-■ (⌐■_■) $\endgroup$
    – wim
    Commented May 22, 2013 at 1:40

1 Answer 1

Reset to default
19
$\begingroup$

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 $2x = 5\times10^4$.

Improve this answer
$\endgroup$
4
  • 3
    $\begingroup$ 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. $\endgroup$
    – N. Virgo
    Commented May 13, 2013 at 5:47
  • 4
    $\begingroup$ Note that moving the string in a vacuum is going to produce no sound whatsoever... $\endgroup$
    – Shadur-don't-feed-the-AI
    Commented May 13, 2013 at 8:01
  • $\begingroup$ 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… $\endgroup$
    – tamasgal
    Commented May 13, 2013 at 8:25
  • $\begingroup$ @septi Corrected that unfortunate wording. Thanks. $\endgroup$
    – user10851
    Commented May 13, 2013 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.