Physics headache: string tension and frequency


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Hi,

I have difficulties to get the string frequency from some easy formulas.

Here's the chart I use:

http://www.daddario.com/DADProductD...id=7&sid=50f400af-ca35-4823-aaea-a3a4b96ce6a4
Diameter Tension
Item #-----Note--Inches---mm------lbs---------kg--------lbs/inch
PL009-----E------0.0090---0.2280---13.140---5.960---.00001794
PL011-----B------0.0110---0.2790---11.020---5.000---.00002680
PL016-----G------0.0160---0.4060---14.680---6.660---.00005671
NW026---D------0.0260---0.6604---18.380---8.340---.00012671
NW036---A------0.0360---0.9144---19.040---8.640---.00023964
NW046---E------0.0460---1.1684---16.910---7.670---.00038216

lb * 0.45359237 = kg
inch * 25.4 = mm

But, if I apply the formula f=(k/L)*SQRT(T/u)
With f=freq(s-1), k=1/2, L=length(m), T=tension(kg.m.s-1), u=linear mass(kg.m-1)

whatever the unit conversion I use, I am unable to get an approaching 110Hz with a 648mm length, using the A string.

The best I did is to use the US units as is, and divide the result with 1.979, and I magically got an almost right value for all the strings.
 
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I think your u is wrong (in the video, it is rho = ρ). ρ is kg/m^3 and not kg/m
Thus, it is calculated using the volume of a cylinder (called ρ * A) A being the area of the diameter of the snare.

A cylinder with radius r units and height h units has a volume of V cubic units given by

cylind1.gif


However, we have the diameter (in millimeters, thus meters * 10^-3), not the radius, thus r^2 is (d/2)^2 and there is where you get that funky 4

Also, k/L is n/(2L), thus k=0.5 means that n=1 (and you still get 1/(2*L) ), so your k=1/2 is the same as n=1 from the video.

TL;DR: your k, which has relation to n, which is an integer which gives the same note at diferent octaves can be n=1/2, which means you divide once more by 2 (you have 1.979 ... almost 2)... so your calculation might be right... just had to take a different n
 
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So the Nickel plated steel called NickelWound 0.0360 inches in diameter aka NW036


Diameter = 0.9144 mm = 2*r = 0.0009144 m = D
Tension = 8.640 kg
Steel density = 7850 kg/m^3

STEELDENSITY:=7850
D:=0.0009144
if A = Area, is pi * r^2
u = ρ * A = STEELDENSITY * pi * (D/2)^2 = 0.00515503920935



freq_hz = n / (2L) * sqrt( T / u ) = 110 Hz (A2)

L= 648mm = 0.648 m
T= 8.640 kg (from the chart)

L:=0.648
Hz:=110
u:=0.00515503920935

T= 4*L^2*u*Hz^2 = 104.764594807 Newton = 10.6 Kg

from http://www.convertunits.com/from/kg/to/N we get that 1 kilogram is equal to 9.80665002864 N.

And the guitar site gives that Tension T should be 8.640 Kg (not Newtons)... so as Nickel is a bit heavier than Iron, it is denser, but there is a "high carbon steel core", and Carbon is a lot less dense...
so what is the advertized u?

T/(4*L^2*Hz^2)= u = 0.000425126687753 (directly to Kg, not Newtons)

And now:


L:=0.648
Hz:=110
u:= 0.000425126687753
T = 4*L^2*u*Hz^2 = 8.64 <--- We get the number back from the advertised table but it is in Kg, not in Newtons

So, now back to your calculations:

advertised Tension = 8.640 Kg = 8.640 * 9.80665002864 = 84.7294562474 Newton

T:=84.7294562474
L:=0.648
Hz:=110
D:=0.0009144
u = T/(4*L^2*Hz^2) = 0.00416906864463 = ρ * A = STEELNICKELDENSITY * pi * (D/2)^2

STEELNICKELDENSITY = 0.00416906864463 / (pi * (D/2)^2) = 6348.58194696


Thus,

STEELNICKELDENSITY:=6348.58194696
Nw:=9.80665002864 (from Kg to Nw)
D:=0.0009144
u = ρ * A = STEELNICKELDENSITY * pi * (D/2)^2 = 0.00416906864463

u:=0.00416906864463 for Newtons


Now you can use your formula with the re-calculated u


As you can see, the strings are less dense than pure steel.
[doublepost=1495233985,1495233682][/doublepost]Thus, with these numbers:

n:=1
L:=0.648
T:=84.7294562474
u:=0.00416906864463

Frequency = n / (2L) * sqrt( T / u ) = 110 hz
 
Thank you a lot for this detailed reply, I'm now able to have the good frequency for all strings, and also all frets.
It took me a while to figure out how you proceed, but I think I got it.

I owe you a bunch of beers, or whatever you'd prefer !
 
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