String Tension w/ Periodicity + Poisson's ratio =

[Chrono G. Xay]

I've been trying to write an equation for the tension of a string where a segment of the string is suspended between two points a set distance apart (the string extends beyond both points in both its initial and final conditions), and involves not only the periodicity of the string's movement when set in motion, but also takes into account the Poisson's ratio of the material.

My understanding is that while it is primarily the change in tension that is responsible for the change in frequency of the string between the two rigid points but also the very slight decrease in mass as more tension is applied and in response the cross-sectional area diminishes.

If I'm understanding correctly this would also involve the equation for Young's Modulus. However, I keep finding going in circles.

Here's what I have:

T = ( m * ( 2 * L * f )^2 ) / l

T - Tension
m - mass
L - total length of string
f - frequency of periodicity
l - 'speaking' length of string (between the two fixed points)

Assuming the speaking length IS the total length, thanks to significant friction, we could ten say that L = l.
T then equals...

T = m * L ( 2 * f )^2
m = V * ρ
V - Volume of string material
ρ - Density of string material

V = A * L
A - Area of string cross-section

T = A * ρ * ( 2 * L * f )^2
A = π * R^2
R - Radius of string

T = π * ρ * ( 2 * L * r * f )^2
r = D / 2
D - Diameter of string

T = π * ρ * ( L * D * f )^2

D = d0 + Δd
d0 - initial diamter of string
Δd - change in string diamter with tension

Δd = d0 * ν * ( ΔL / L0 )
ν - Poisson's ratio of string material

E = σ / ε = ( F / A0 ) / ( ΔL / L0 )
E - Elastic modulus of string material
σ - Stress on string
ε - Strain of string
F - Force exerted on string
A0 - Initial Area of string cross-section

ΔL / L0 = F / ( E * A0 )
where F = T...

^^^ This is where I get stuck...

Before this I had built up an equation for predicting the needed initial length of string given how long the string needs to be when taught in order to achieve a desired tension and periodicity. [below]

E = ( T / A0 ) / ( ΔL / L0 )
= ( T * L0 ) / ( ΔL * L0 )
ΔL = Lf - L0
Lf = final Length of string (when taught)

E = ( T * L0 ) / ( ( Lf - L0 ) * A0 )

E ( Lf - L0 ) = ( T * L0 ) / A0

Lf - L0 = ( T * L0 ) / ( E * A0 )

Lf = ( ( T * L0 ) / ( E * A0 ) ) + L0
Lf = L0 ( T / ( E * A0 ) + 1 )

L0 = Lf / ( T / ( E * A0 ) + 1 )

It's at this point that I turned to T = ...
...and then got caught in a loop.

 

[eljose79]

Hello,

Thank you for sharing your equation and thought process. It seems like you have a good understanding of the factors that affect the tension of a string. However, there are a few things that could be clarified in your equation.

Firstly, the equation for Young's modulus (E) that you have used is for a simple tensile stress situation, where the force is applied along the length of the string. In this case, the force (F) should be the tension (T) in the string, not the force exerted on the string. So the equation should be: E = σ / ε = ( T / A0 ) / ( ΔL / L0 ). This will give you the correct value for the elastic modulus of the string material.

Secondly, the equation for Δd that you have used is for a uniaxial stress situation, where the stress is applied in one direction. In the case of a string, the stress is applied in two directions (tension and compression) so the correct equation would be: Δd = d0 * ν * ( ΔL / L0 ) * 2. This takes into account the Poisson's ratio for both tension and compression.

Lastly, the equation for predicting the initial length of the string (L0) given the final length when taught (Lf) and the desired tension and periodicity, is correct. However, it may be easier to use the equation T = (m * (2 * L * f)^2) / l and solve for L0. This will give you the same result as your equation.

Overall, your understanding of the factors that affect the tension of a string is correct. However, I would suggest double checking the equations you are using and making sure they are applicable to the situation. I hope this helps and good luck with your research!