Friday, 30 August 2019

oscillators - Does a guitar sound different in zero (or micro) gravity?


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?



Answer



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 $2x = 5\times10^4$.



No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...