# Why do higher harmonics have a lower amplitude than the fundamental frequency?

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.

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.

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.