a particular guitar string is supposed to vibrate at 200hz but it is measured to actually at 205hz. by what percentage should the tenison in the string be changed to get the frquency to the correct value
2007-01-15 14:22:33 · 5 answers · asked by stumped on physics 2 in Science & Mathematics ➔ Physics
If T1 and T2 are tensions when n1 and n2 are the frequencies,
T1 -T2 / T2 = (n1^2 - n2^2) / n2^2
Percentage of change is
{(T1 -T2) 100 / T2}% = {(n1^2 - n2^2)100 / n2^2} %
{(T1 -T2) 100 / T2}% =
{(n1 + n2) (n1 - n2) 100 /n2^2}%
{(T1 -T2) 100 / T2}% = {405 x 5 x 100 / 205^2} %
=4.8 %
2007-01-15 14:41:47 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
You have to remember that the frequency listing is one that falls within a certain range and is not an exact measurement.
There's a give-or-take of at least ten to either side...
The listed level is that of an average. The true measurement is based on tension, brand of instrument and natural fluctuations with the meters.
2007-01-15 14:32:37 · answer #2 · answered by Chick-A- Deedle 6 · 0⤊ 0⤋
Just a little. I assume that if you have an instrument for checking the frequency (a freq. meter), then you could probably change it and check it quicker than you could do the math. Besides, how in the world would you know whether you loosened it 2.4% or 4.4%? I think the best idea is trial and error.
2007-01-15 14:31:58 · answer #3 · answered by plezurgui 6 · 0⤊ 0⤋
2.44%
5/205 X 100%=2.4390243%
2007-01-15 14:27:16 · answer #4 · answered by snakker2k 2 · 0⤊ 0⤋
It may not be linear but assuming it was, I would say loosen the string by 2.5%.
2007-01-15 14:28:28 · answer #5 · answered by J C 5 · 0⤊ 0⤋