How does a string thickness affect the frequency of its harmonics?

Question

The harmonics of a theoretically infinitely small diameter string are pure integer multiples of the fundamental frequency. However, a real string has a thickness, and when vibrating in a harmonic, the additional node or nodes should be taken into account when considering the string's length. This should mean that higher harmonics would have frequencies higher than the corresponding integer multiple of the fundamental frequency. What is the formula that allows for this calculation, and what effect might it have on the quality of the sound an instrument might produce?

Answer 1

Generally you try to excite many harmonics when you play an instrument. You can see this by the fact that a piano hammers the strings near the end. Likewise a guitar or violin.

The strings do not deviate very far from straight as they vibrate. A guitar string might be a couple feet long, and the maximum sideways motion might be $0.1$ inch. This means the string curves as it vibrates, but the curvature is very shallow.

A strongly curved string might vibrate in complex ways because the outside of the curve is more stretched than the inside. The thickness of the string would make this effect bigger. But a gently curved string is like an ideal spring. Very little extra harmonics would be expected because of the thickness.

Answer 2

The ideal string has:

$$ f_n = nf_0 $$

for the $n$-th harmonic.

Including a finite thickness modifies that to:

$$ f_n = nf_0\sqrt{1+Bn^2}\big(1+\frac 2{pi}\sqrt B + \frac 4{\pi} B\big) $$

with

$$B = \frac{EAK^2}{TL^2} $$

$E$ is the modulus of elasticity. $A=\pi r^2$ is cross section area, and $K=r/2$ is the radius of gyration. $r, T, L$ are the usual suspects.

I think this is responsible for deviations in the "ideal piano tuning" plot: