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?

• The assumption in your question is wrong. The maximum amplitude is not necessarily the fundamental frequency. Look at the graphs here, showing the frequency response of some notes on a piano. acs.psu.edu/drussell/Piano/Dynamics.html – alephzero Jul 5 '20 at 21:22
• @alephzero good point, though I wager that at each point of the string itself, the fundamental does still have the strongest amplitude. It's just that at the low end of the piano's range, the sound board isn't able to transmit that accordingly and thus a microphone will pick up the fundamental weaker than some of the harmonics. – leftaroundabout Jul 5 '20 at 21:44
• I removed the note about just commenting with the link because that's not really something we want to encourage here. Sure someone might leave a comment with a useful link, but that's not good enough for an answer itself; the hope is that someone else would come along later and turn that link into an actual answer. (And anyway, you have answers now, so that part of the question isn't really needed anymore.) – David Z Jul 6 '20 at 0:41
• Does this answer your question? Is the first harmonic the prevalent frequency? – John Rennie Jul 6 '20 at 8:38
• It may be observed that this question has more upvotes, and more answers, than the already closed "duplicate" question to which it has been compared. Please do not close better questions and answers in favour of less good ones. – Charles Francis Jul 6 '20 at 18:50

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.

• What Im asking maybe stupid. But how are the energies distributed among the modes. Energy of a wave depends on both amplitude and frequency, right? So on higher harmonics, wont the increase in frequency makeup for the decrease in amplitude? And wont these higher modes take up more fraction of energy of the wave? – Rahul R Jul 5 '20 at 6:49
• The energy of mode $n$ ($n$'th harmonic) is $\frac{1}{4}M\omega^2b_n^2=\frac{1}{4}M\omega_0^2n^2b_n^2$ where $\omega_0$ is the angular frequency of the fundamental. If $b_n$ decreases as $1/n^2$ asymptotically, $b_n^2$ decreases as $1/n^4$ and the energy decreases as $1/n^2$. – Puk Jul 5 '20 at 7:39
• This is a very good answer, and I follow the math. But I'm wondering if anyone has an intuitive answer for why this happens? – gardenhead Jul 5 '20 at 15:14
• What you have done here is basically Fourier series decomposition of the standing wave! Let me then ask a more general question - why is the amplitude of higher harmonics in the Fourier series have lower amplitudes. You have given a specific example but what would be a more general answer? – abir Jul 5 '20 at 15:49
• @abir See my previous comment for an explanation that appeals to physics. Mathematically, Parseval's identity requires that the squares of the Fourier series coefficients add to a finite value. This requires that the absolute value of the coefficients be bounded by a sequence that decays sufficiently fast with $n$ as $n\to\infty$. – Puk Jul 5 '20 at 16:52

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.

• Also, any harmonic that happens to have a node at the point where you pluck the string will have an amplitude of zero. That's why plucking a string precisely in the middle has such a characteristic sound: All odd harmonics have a node there, and are thus suppressed. E-guitarrists frequently use this for solos, playing on the high end of the fret-board so that the pick and pick-up are located squarely in the middle of the vibrating string. – cmaster - reinstate monica Jul 6 '20 at 10:15
• @cmaster-reinstatemonica: I think you may have switched around even / odd. The even harmonics have a node in the middle and are suppressed. You get the fundamental, 3rd, 5th, etc. – Dietrich Epp Jul 6 '20 at 22:03
• @DietrichEpp Depends on whether you call the fundamental the 1st or 0th harmonic. Where I learned physics, the convention was to to call it 0th harmonic (or rather, to not call it an harmonic at all) so that the first harmonic is already the octave above the fundamental. I agree that your counting makes more sense, though. – cmaster - reinstate monica Jul 7 '20 at 7:39
• I believe (as a fan of Mr Zakk Wylde) that the phrase is a "pinched harmonic", not "squelched" .musicradar.com/tuition/guitars/… audio here: mos.musicradar.com/audio/tutorial-audio/guitar/play-guitar-like/… – ProfRob Jul 10 '20 at 15:58

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.

• Is this Energy equally distributed among the different modes? – Rahul R Jul 5 '20 at 5:45
• @RahulR No, it depends on how you pluck the string. – Puk Jul 5 '20 at 6:12
• I like this answer a lot, energy considerations quickly tell you that amplitudes of the harmonics must go down faster than $1/n$ asymptotically. – Puk Jul 5 '20 at 6:49
• @Rishab Navaneet According to your logic, it is true that the amplitude of the $n^{th}$ harmonic should be zero as $n->\infty$. But is there any guarantees that there won't be a maxima in one of the intermediate harmonics, say for example, in the $3^{rd}$ harmonic! – abir Jul 5 '20 at 15:46
• @abir that clearly depens on how you pluck the strings, for example you can mute the fundamental etc – fqq Jul 5 '20 at 18:55

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.

• This should be the top answer. Clear, concise, conclusive. Can't upvote this hard enough. – cmaster - reinstate monica Jul 6 '20 at 10:23

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.

• In an idealized string, do higher harmonics have the same amplitude? Or are you suggesting that the thickness of the string is a contributing factor to the pre-existing drop in amplitude? – awe lotta Jul 5 '20 at 13:21
• @awelotta There is a general theorem in mathematical physics (equipartiion theorem), that asserts: when a system has multiple degrees of freedom then over time the available energy will become evenly distributed over all degrees of freedom. Example: a gas consisting of diatomic molecules: total energy will be sum of translational kinetic energy and molecule rotational energy. In the case of an idealized string there is no upper limit to the number of harmonics. Not necessarily all the same amplitude; what the theorem asserts is that they will tend towards equal amounts of energy. – Cleonis Jul 5 '20 at 15:06
• @Cleonis does the equipartition theorem apply here? In the ideal string the modes are decoupled. In reality they probably interact through imperfect endpoints, nonlinearities etc. On the other hand, a guitar sounds different depending on where you pluck the strings, suggesting that initial conditions matter. – fqq Jul 5 '20 at 18:30
• @fqq I concur: in the case of an idealized string there is no coupling amongst the modes. To excite a great many modes simultaneously would require a complicated setup. My purpose was to emphasize that while the idealized string simplification is acceptable for the lower harmonics the higher up the worse the failure. (And yeah, as we know: with an actual guiter (and any string instrument) there is not much transfer of energy between modes.) – Cleonis Jul 5 '20 at 18:48
• Actually, the most notable effect of a string's stiffness is that the higher harmonics get off-pitch: The stiffness adds to the amount of force that's trying to straighten out the string, so the vibrations happen faster than they do on the idealized string. This effect gives piano tuners real headaches because the low/high strings in a piano exhibit so much stiffness that they noticeably skew the tuning. – cmaster - reinstate monica Jul 6 '20 at 10:21

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!

• This is a nice visualisation, but it's not clear why all modes should have the same energy (and it's probably not true). – fqq Jul 5 '20 at 18:26
• @fqq: in thermodynamic equilibrium, all the modes have the same energy. And if you pluck a string and wait, it should eventually go to thermodynamic equilibrium (although in real life, the vibrations damp out first). – Peter Shor Jul 6 '20 at 22:10
• Yes that's the point, in usual conditions vibrations damp out before reaching thermodynamic equilibrium, as the coupling is rather weak. Empirically, when playing e.g. a guitar the sound changes when plucking the string differently, so initial conditions matter a lot. – fqq Jul 7 '20 at 1:23

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.