- #1
jaumzaum
- 434
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Guitar strings behave like a spring when tuned:
F = k.x is the tension in the string, where k is the contant of the string and x the displacement (when tuned). So by the equation :
[itex] v = \sqrt{ F/u} [/itex]
where u is the linear density of the string.[itex] v = \lambda.f -> f = \sqrt{ F/u}/\lambda [/itex]
The first string of a guitar is E and has a frequency f1, when tuned, that is proportional to the square root of the string tension.
But when we play the first string weakly, we seem to hear the same E note (sure, more weak) and when we play strongly, we seem to hear this same E stronger. So the force we apply in the string does not seem to change the frequency, only the amplitude. But in a spring, when we make a vertical displacement
[PLAIN]http://img716.imageshack.us/img716/6427/sgfhdf.jpg
The tension do change
So why don't we have a change in the frequency?
F = k.x is the tension in the string, where k is the contant of the string and x the displacement (when tuned). So by the equation :
[itex] v = \sqrt{ F/u} [/itex]
where u is the linear density of the string.[itex] v = \lambda.f -> f = \sqrt{ F/u}/\lambda [/itex]
The first string of a guitar is E and has a frequency f1, when tuned, that is proportional to the square root of the string tension.
But when we play the first string weakly, we seem to hear the same E note (sure, more weak) and when we play strongly, we seem to hear this same E stronger. So the force we apply in the string does not seem to change the frequency, only the amplitude. But in a spring, when we make a vertical displacement
[PLAIN]http://img716.imageshack.us/img716/6427/sgfhdf.jpg
The tension do change
So why don't we have a change in the frequency?
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