By David Haimes
The lowest note possible on a guitar is the open low E string. Using a guitar-tuning app
on my phone, I measured the frequencies to be about 80 Hz.
Frequency: 80 Hz
Length: About 25 inches from fixed point to fixed point
Mass: 10 grams (very rough estimate/guess)
The Force due to tension equation is
FT = ( m / L ) * v2
We need to find v.
The equation for v is: v = ƒλ
And λ = 2L = 2 (25 inches * (.0254 m / 1 inch) ) = 1.27 m
Now we can solve for v : v = ƒλ = 80 Hz (1.27 m) = 101.6 m/s
Now we can solve for FT= ( .010 kg / (25 inch (.0254m/1 inch) ) * (101.6 m/s)2 = 65 N
So the approximate tension on the string is 65 N
The website lists the recommended tension for the string on my guitar to be 27 lbs which
is about 120 N. Assuming all the calculations are correct, the one big assumption/
approximation I made was the mass of the string. Working backwards, a tension of 120 N
would require a mass of…
m = FT / (v2*L) = 120 N / (101.6 m/s * 1.27m) = .0092 kg
I plan to bring the string to lab on Thursday to see how close the mass of the string is to
this! However, the string is larger than length used in the equation, due to excess beyond
the fixed points.