- #36
Chrono G. Xay
- 92
- 3
It’s been at least a few years since I’ve added to this thread. Starting with the linearized equation you gave earlier for ##Δf## and substituting in the unconverted equation for initial tensile force $$F_0 = {μ\left(2Lf_0\right)^2}$$ as well as $$Δf = f - f_0$$ we can obtain $$f_0 = \frac {EAα} {2μL^2f_0} {ΔT} + f.$$ With a little rearranging we obtain the quadratic form $$f_0^2 - f f_0 - \frac {EAα} {2μL^2} {ΔT} = 0$$ $$\left(ax^2 + bx + c = 0\right).$$ By setting our variables asChestermiller said:You did a very impressive analysis for a guy who is doing music education. Dazzling in fact.
I have a few comments. It is not necessary to take into account the change in the cross sectional area because that is negligible. So this should simplify things considerably.
The G doesn't belong in the equation for the fundamental frequency (since gravity has nothing to do with this). The equation should read:
$$f=\frac{1}{2L}\sqrt{\frac{F}{μ}}$$
where L is the clamped length of the string, F is the tension in the string, and μ is the linear density of the string (mass/length).
The equation for the change in tension is:
$$ΔF = -αEAΔT$$
where α is the coefficient of thermal expansion, E is the Young's modulus, and T is the temperature.
So if f is the frequency at temperature T0 and tension F0 and f + Δf is the frequency at temperature T0+ΔT and tension F0+ΔF, then
$$Δf=\frac{1}{2L}\sqrt{\frac{F_0-(αEAΔT)}{μ}}-\frac{1}{2L}\sqrt{\frac{F_0}{μ}}$$
If we linearize this with respect to the temperature change ΔT, we obtain:
$$Δf=-\frac{EAα}{L\sqrt{F_0μ}}ΔT$$
The fractional change in frequency is:
$$\frac{Δf}{f}=-\frac{EAα}{2F_0}ΔT$$
According to these equations, when you increase the temperature, the frequency decreases (because the string expands and "slackens").
Chet
$$x = f_0 , a = 1 , b = -f , c = -\frac {EAα} {2μL^2} {ΔT}$$ we can obtain the equation $$f_0 = \frac f 2 + \sqrt{ \left( \frac f 2 \right)^2 + \frac {EAα} {2μL^2} {ΔT} }.$$ Now whenever ##ΔT = 0, f_0 = f##.
…How did I do?