The diameter of the A string of a guitar is 0.229mm and 440Hz. If the tension of the strings are the same, what is the diameter of the D string if it is 293Hz? The length of the strings is 58cm.
**I tried using the V= the square root of [tension/(m/L)] formula, but I don't know mass or how to solve for diameter
You are referring to this equation:
. . . . .\(\displaystyle v_w\, =\, \sqrt{\strut \dfrac{T}{\left(m/L\right)}\,}\)
...where:
. . . . .\(\displaystyle v_w:\, \mbox{ wave velocity}\)
. . . . .\(\displaystyle T:\, \mbox{ tension in the string}\)
. . . . .\(\displaystyle m:\, \mbox{ mass of the string}\)
. . . . .\(\displaystyle L:\, \mbox{ length of the string}\)
In other words, you're using what is outlined (
here) by "Physics Lab".
But, for "frequency" questions such as the one you've posted, one might prefer to use the "resonance" equation:
. . . . .\(\displaystyle v_w\, =\, f\lambda\)
...where:
. . . . .\(\displaystyle v_w:\, \mbox{ wave velocity}\)
. . . . .\(\displaystyle f: \mbox{ resonance frequency of the string (in Hertz)}\)
. . . . .\(\displaystyle \lambda:\, \mbox{ string wavelength (being twice string length)}\)
In other words, one might prefer to use what is outlined (
here), also by "Physics Lab".
Combining the two, we get:
. . . . .\(\displaystyle \sqrt{\strut \dfrac{T}{\left(m/L\right)}\,}\, =\, 2fL\)
. . . . .\(\displaystyle \dfrac{T}{\left(m/L\right)}\, =\, 4\,f^2\,L^2\)
. . . . .\(\displaystyle \dfrac{TL}{m}\, =\, 4\,f^2\,L^2\)
. . . . .\(\displaystyle \dfrac{T}{4mL}\, =\, f^2\)
. . . . .\(\displaystyle \sqrt{\strut \dfrac{T}{4mL}\,}\, =\, f\)
We need to assume that the A string and the D strings are made from the same material and has same length.
Okay, that makes sense. But I still don't know how to go about solving this
What were your thoughts? What did you
try?
For instance, I don't think I've ever worked with this particular topic before but, starting with the previous helper's hint, I tried the following:
I noted that the expression "mass / length" gives one cross-sectional area, since the total mass of the string is the mass of the cross-sectional area (being a circle) multiplied by the length of the string. By dividing out the length, I was left with the mass of the cross-sectional area of each string. I know that the cross-sectional areas are different, because I know that the diameters are different. If the lengths were different, or if the materials were different, then I'd have an unsolvable problem. However, by assuming that the lengths are the same and that the strings have the same per-volume mass because of being made of the same material, I can say:
. . . . .\(\displaystyle d\, =\, 2r\)
. . . . .\(\displaystyle A_{circ}\, =\, \pi\, r^2\, =\, \pi\, \left(\dfrac{d}{2}\right)^2\, =\, \dfrac{\pi\, d^2}{4}\)
...so:
. . . . .\(\displaystyle 4mL\, =\, 4\, \left(\dfrac{\pi\, d^2}{4}\right)\, L\, =\, \pi\, d^2\, L\)
Then:
. . . . .\(\displaystyle f\, =\, \sqrt{\strut \dfrac{T}{4mL}\,}\, =\, \sqrt{\strut \dfrac{T}{\pi\, d^2\, L}\,}\)
Because I'm assuming that the lengths are the same, then the two different strings' information look like:
. . . . .\(\displaystyle \mbox{A-string: }\, 440\, =\, \sqrt{\strut \dfrac{T}{\pi\, d_A^2\, L}\,}\)
. . . . .\(\displaystyle \mbox{D-string: }\, 293\, =\, \sqrt{\strut \dfrac{T}{\pi\, d_D^2\, L}\,}\)
I have some equal values in these equations: the T and the L. So I'll solve for one and set the results equal. I'll be able to plug in the one given diameter, and then solve for the other.
Now I've shown my thoughts and (most of) my work. What did
you do?
Please be complete. Thank you!
