Why do higher harmonics have a lower amplitude than the fundamental frequency? When we pluck a string, it vibrates in all possible modes of vibrations. The lowest frequency possible is the fundamental frequency and it is the most significant part of sound.
But why do the amplitude of higher harmonics decrease? Which formula is responsible?
Also, how is the energy of wave distributed among different modes?
A Google search didn't give any explained answer.
 A: As is often the case in physics, when the properties of string vibration are described invariably the string is treated as an idealized string. Among these idealizations: the string is treated as infinitely bendable. For the lower harmonics the error introduced by that simplification is acceptably small.
That simplication fails for higher harmonics.
On, say, a guitar, the lowest harmonic vibration can go up to an amplitude of a couple of milimeters or so. Now image a section of guiter string, cut to, say, a 1/16th of the total length between the bridge and the nut. Such a short section of string is quite stiff, the elastic properties are more like those of a stick than those of an idealized string. While it is possible to excite the 16th harmonic, the amplitude you can excite is limited.
So: even if your string plucking is very close to the bridge not much energy is going into exciting higher harmonics; the string isn't bendable enough for that to happen.
A: As a musician, the answer seems obvious. I can observe it when I play a guitar.
When you pluck an open string, the total displacement looks like this.

When you pluck a second harmonic with equal energy, you have to displace both sides of the string. The total displacement stays approximately the same.


P.S. I now await the physicists tearing into me!
A: Why not calculate it?
Consider a string of length $L$, with its ends fixed at $x=\pm\frac{L}{2}$. Let's assume for convenience that at time $t=0$ the string is "plucked" at $x = 0$, so that the string displacement relative to its equilibrium position is given by
$$f(x)=A\left|1-\frac{2x}{L}\right|.$$
The standing wave solutions to the wave equation obeying the boundary conditions are
$$\psi_n(x)=\cos\left(\frac{(n-\frac{1}{2})2\pi x}{L}\right) $$
with $n\ge1$, $n=1$ corresponding to the fundamental, $n=2$ to the third harmonic, $n=3$ to the fifth harmonic and so on. Note that I haven't included the odd solutions (even harmonics) here, because these modes won't be excited since $f(x)$ is even.
It is a straightforward exercise to show that $\psi_n$ are orthogonal:
$$\int\limits_{-L/2}^{L/2}\psi_m(x)\psi_n(x)dx=\frac{L}{2}\delta_{mn}$$
where $\delta_{mn}$ is the Kronecker delta. If
$$f(x)=\sum\limits_{m=1}^\infty a_m\psi_m(x),$$
multiplying by $\psi_n$, integrating and using the orthogonality relation yields
$$a_n = \frac{2}{L}\int\limits_{-L/2}^{L/2}f(x)\psi_n(x)dx=\frac{4A}{L}\int\limits_{0}^{L/2}\left(1-\frac{2x}{L}\right)\cos\left(\frac{(n-\frac{1}{2})2\pi x}{L}\right)dx.$$
Evaluating the integral gives
$$a_n=\frac{2A}{\pi^2\left(n-\frac{1}{2}\right)^2}.\tag{1}$$
So the amplitude of the harmonics decreases roughly as $1/n^2$.
You find that if you pluck the string closer to the ends, the amplitude of the harmonics goes down slower, i.e. there are more "overtones". Specifically, if the string is plucked a distance $\ell$ from one of the ends, the amplitudes are
$$ b_n = \frac{2AL^2}{\pi^2\ell(L-\ell)n^2}\sin\left(\frac{n\pi\ell}{L}\right)\tag{2}$$
where the sine factor accounts for the slower decay of $b_n$ when $\ell$ is small. $(2)$ is more general than $(1)$ as it is also valid when the string is not plucked in the middle, and is also consistent with how a guitar string is normally picked.
Note: the meaning of $n$ in $b_n$ is different from before: here, $n=1$ is the fundamental, $n=2$ is the second harmonic, $n=3$ is the third harmonic and so on. The difference is because when the string is plucked in the middle, the even harmonics are not excited.
As for the energy distribution, the energy in the $n$'th harmonic is
$$ E_n = \frac{1}{4}M\omega_n^2b_n^2 = \frac{1}{4}M\omega_1^2n^2b_n^2$$
where $M$ is the total mass of the string and $\omega_n=n\omega_1$ is the angular frequency of the $n$'th harmonic.
A: The answer is actually very dependent on how you pluck the string.  If you pluck it closer to the center, you put more energy into the lower modes.  Pluck it near either end, and you have more higher harmonics.
And then there's the overtone techniques, which intentionally squelch lower harmonics, leaving only higher harmonics.
A: It's simple energy conservation. With an increase in harmonics, the frequency of vibration of the string increases. We know that each particle in the string is executing a simple harmonic motion with energy:
$e=\frac{1}{2}m{\omega}^2A^2$
We have a continuous distribution of such oscillating masses, each oscillating with different amplitudes. Integrating them would give the total energy and obviously, that too would be dependent on frequency.
Now since the device we use to oscillate the string supplies a fixed energy, as the harmonic increases, the amplitude should drop.
A: A simple answer: the total energy of the vibration has to be finite.
Given that we have an infinite number of possible modes of vibration (not only harmonics, but let's start with them), you need some distribution of the energy between few of them (in order to hear something at all) and you get less and less energy left for higher ones.
p.s. you don't always get maximal amplitude for the basic frequency, it depends on a lot of factors and there are techniques for changing the harmonic content of the tone for most string instruments. But you still get few vibration modes getting most of the energy.
A: Well, it is because the frequency of vibration is decided by the length of the string and the tension in the string and once you have your device you are pretty much guaranteed to have a major frequency range and the rest will all have minor components with lesser amplitudes because of the way the wave oscillation gets decomposed.
