Why are the harmonics of a piano tone not multiples of the base frequency?

This effect is known as inharmonicity, and it is important for precision piano tuning.

Ideally, waves on a string satisfy the wave equation $$v^2 \frac{\partial^2 y}{\partial x^2} = \frac{\partial^2 y}{\partial t^2}.$$ The left-hand side is from the tension in the string acting as a restoring force.

The solutions are of the form $\sin(kx - \omega t)$, where $\omega = kv$. Applying fixed boundary conditions, the allowed values of the wavenumber $k$ are integer multiples of the lowest possible wavenumber, which implies that the allowed frequencies are integer multiplies of the fundamental frequency. This predicts evenly spaced harmonics.

However, piano strings are made of thick wire. If you bend a thick wire, there's an extra restoring force in addition to the wire's tension, because the inside of the bend is compressed while the outside is stretched. One can show that this modifies the wave equation to $$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^2}.$$ Upon taking a Fourier transform, we have the nonlinear dispersion relation $$\omega = kv \sqrt{1 + (A/v^2)k^2}$$ which "stretches" evenly spaced values of $k$ into nonuniformly spaced values of $\omega$. Higher harmonics are further apart. We can write this equation in terms of the harmonic frequencies $f_n$ as $$f_n \propto n \sqrt{1+Bn^2}$$ which should yield a good fit to your data. Note that the frequencies have no dependence on the amplitude, as you noted, and this is because our modified wave equation is still linear in $y$.

This effect must be taken into account when tuning a piano, since we perceive two notes to be in tune when their harmonics overlap. This results in stretched tuning, where the intervals between the fundamental frequencies of different keys are slightly larger than one would expect. That is, a piano whose fundamental frequencies really were tuned to simple ratios would sound out of tune!

In plain English, there is stiffness at the ends of the strings where they are fixed in place, which makes the string's frequency of vibration slightly higher (sharper)—effectively shortening the length of the string slightly, for all practical purposes. And the resistance to bending is dependent on the frequency. It behaves more “stiffly” with regard to higher frequencies, reducing the string’s effective lengths more and more for each higher and higher frequency harmonic (“partial”). Therefore the upper partials go effectively sharper in pitch, the higher they get.

All stringed instruments exhibit this effect to some extent, and it’s part of the reason realistic piano sounds were hard to synthesize electronically on early synthesizers… when all the harmonics are in tune, the timbre sounds “deader” and less rich.

A real-life example of this would be guitar strings, which eventually go “dead” as you use them, and don’t sound as good as new strings. This is because the cumulative effect of playing them makes them more and more soft and flexible, reducing the stiffness at the fixed ends and making the upper partials more in tune with the fundamental. This sounds “duller”, much less rich in tone then new, much stiffer strings.

It also means that pianos have to be “stretch tuned”… low notes are tuned slightly flat, and high notes are tuned slightly sharp. Otherwise, playing a C1 and a C6 at the same time would sound slightly out of tune, as the partial of C1 closest to C6 would be slightly sharper than C6's mathematically precise frequency of 2^5 times the frequency of C1*, resulting in an audible, out-of-tune-sounding “beat” of the partial against the C6’s root, just like two strings whose fundamentals are close but not exactly in tune. Tuning C6 slightly sharp by the same amount that C1’s 6-octave partial is sharp, ironically, makes the piano sound more in tune.

EDIT: *A comment below drew my attention to the fact that I had originally made a mistake here. The frequency of C6 is 2^5 times the frequency of C1, because each octave is a doubling of frequency. In terms of frequency:

• C2 = 2*C1 (=2^1*C1)
• C3 = 2*2*C1 (=2^2*C1)
• C4 = 2^3*C1
• C5 = 2^4*C1
• C6 = 2^5*C1

So that's where the 2^5 comes from.]

Carrying out the Fourier transform, I get a slightly different result for the frequency spectrum than 'knzouh'. I used $u$ instead of $y$ and $c$ instead of $v$, so the PDE becomes:

$$u_{tt}=c^2u_{xx}-Au_{xxxx}$$ Fourier transforming the equation: $$F\{u_{tt}\}=F\{c^2u_{xx}\}-F\{Au_{xxxx}\}$$ Transforming $x$ to $k$: $$\hat{u}(k,t)=\int_{-\infty}^{+\infty}u(x,t)e^{-ikx}dx$$ $$\int_{-\infty}^{+\infty}u_{tt}(x,t)e^{-ikx}dx=\frac{\partial^2}{\partial t^2}\hat{u}(k,t)$$ $$F\{c^2u_{xx}\}=c^2(ki)^2\hat{u}(k,t)=-c^2k^2\hat{u}(k,t)$$ $$F\{Au_{xxxx}\}=A(ki)^4\hat{u}(k,t)=Ak^4\hat{u}(k,t)$$ Inserting we get:

$$\frac{\partial^2}{\partial t^2}\hat{u}(k,t)=-c^2k^2\hat{u}(k,t)-Ak^4\hat{u}(k,t)$$ $$\frac{\partial^2}{\partial t^2}\hat{u}(k,t)=-k^2(Ak^2+c^2)\hat{u}(k,t)$$ Call: $$\hat{u}(k,t)=U(t)$$ So $$U''(t)+k^2(Ak^2+c^2)U(t)=0$$ $$U(t)=c_1\sin(k\sqrt{Ak^2+c^2}t)+c_2\cos(k\sqrt{Ak^2+c^2}t)$$ $$\omega=2\pi f=k\sqrt{Ak^2+c^2}=kc\sqrt{\Big(\frac{A}{c}\Big)^2k^2+1}$$ $$f_n \propto n \sqrt{Bn^2+1}$$