Physics of guitar strings

Guitarists normally press down hard on the frets and then pluck a string to obtain a note. However, one can also create notes by just touching the string above a particular fret and plucking.

For example, if I just press on a string about two thirds of the way up the guitar (from the bridge), a note of one octave higher in pitch is created than when I press down firmly at the same place. I know that an increase of one octave is the same as doubling the pitch.

How does this happen? What's the physics behind it? I assumed it was something to do with nodes, but I wasn't sure.

• Hint: the note you describe, one octave higher, is obtained exactly at half the length (12th fret), not 2/3.
– fqq
Commented Jan 20, 2015 at 21:26
• @fqq "one octave higher in pitch than when I press down firmly at the same place", i.e. the 3rd harmonic of the open string, compared to fretting the string to 2/3 of its open length. Commented Jan 20, 2015 at 21:35
• And to the OP, the phenomenon is called guitar harmonics and Wikipedia has a good explanation. Commented Jan 20, 2015 at 21:38
• @NathanReed you are right, I didn't read well, sorry about that. That's correct.
– fqq
Commented Jan 20, 2015 at 22:49

How does this happen? What's the physics behind it?

If you fret the string and pluck it, the string will vibrate strongest at the middle point (between the bridge and the fret). This is the fundamental vibrational mode of the string.

By placing your finger lightly at, e.g., the middle of the string and plucking (while quickly removing your finger), you force the string to vibrate such that the middle is a node.

In such a case, the string is vibrating in a higher order (2nd, 3rd, etc.) mode where there are one or more nodes along the string instead of just at the ends.

Image credit

The physics requires that, in the higher order modes, the frequency of the amplitude variations are higher which corresponds to higher pitch.

For example, in the 2nd mode, the pitch is twice as high as in the fundamental mode.

Animated gif credit

• Nice visuals! Did you do the animation in $\LaTeX$? Commented Jan 20, 2015 at 22:36
• @BillN, I was adding the credits just as your comment came in. Commented Jan 20, 2015 at 22:39
• The actual waveshape differs from what is shown above. If you look at slow-mo videos of guitar strings the waveforms look more like trapezoids than sine waves. Commented Jun 6, 2017 at 18:54
• @ja72, if you pluck the guitar string near the bridge, far from the anti-node, it will be 'bright' sounding indicating the presence of several harmonics as would be expected of say, a trapezoidal wave (however, the higher harmonics will decay faster than the fundamental and eventually, the string will vibrate more or less in the fundamental mode). If you pluck the guitar string at the anti-node, it will vibrate nearly sinusoidally. If, as the OP asks about, you lightly touch the string above say, the 12 fret and let go while plucking, you'll excited the string into the 2nd harmonic. Commented Jun 6, 2017 at 23:41
• @ja72, see for example, this Commented Jun 6, 2017 at 23:46

A string fixed at both ends, as on a guitar, can vibrate in a "standing wave" mode at several different frequencies. The lowest frequency, the fundamental, is such that the length of the string matches half of a wavelength, $L=\lambda/2$. The middle of the string has the maximum displacement from rest and the two ends don't move. In this case, the mid-point of the string is called an antinode and the ends are (always) nodes. The frequency of the string, and hence the frequency of the sound wave in the air generated by the string vibrations is $$f_1 = \frac{v}{\lambda} = \frac{v}{2L}$$ where $v$ is the speed of the wave in the string. This speed depends on the string material, thickness, and the tension (tightness) of the string. Thicker and more dense string means slower speed and tighter means higher speed. That's why higher pitched strings are generally thin and lower pitched are thick. It helps for the guitar strings to have a nearly uniform tension so that the neck isn't pulled sideways.

This fundamental isn't the only frequency present when a string is plucked. Another standing wave has a node in the middle and antinodes at 1/4 and 3/4. This means that the length of the string equals the wavelength of this wave, $L=\lambda_2$. So the frequency of this wave is $$f_2=\frac{v}{L} = 2f_1.$$ Aha! This second wave is one octave above the fundamental. It's called the 2nd partial, the 2nd (musical) harmonic, and the 1st overtone of a string (fixed at both ends). You can get this note to sound strongly in two ways: 1) press at the 12th fret, which forces the length of the string to $L_{new}=L/2$ so now the new fundamental is $v/L$ or 2) touch the string just over the 12th fret (at $L/2$) which forces a node where there would normally be an antinode of the fundamental. This prevents the fundamental from sounding, so you can hear the overtone.

You see, when you pluck a guitar string, you add energy to the string at hundreds of different frequencies. This is due to something called the Fourier Theorem. Only the frequencies in this pluck which match possible standing waves in the string will persist longer than a few milliseconds. All the others a quickly dissipated into the internal energy of the string and guitar components, never to be heard again. So a pluck will result in the string having the fundamental, the 1st overtone, the 2nd, the 3rd and so forth.

Let's look at the 2nd overtone. This requires another node, with the nodes evenly spaced on the string, so now we have 3 antinodes and the length of the string is split into 3 half wavelengths, $L=3\frac{\lambda_3}{2}$. The resulting frequency is $$f_3=\frac{v}{L(2/3)}=\frac{3v}{2L} = 3f_1$$ This frequency is also $3/2f_2$. A frequency ratio of 3/2 is a musical fifth, for example, G up to D.

If you consider the position of playing D on the G string, that's at the 7th fret, and it's also 1/3 of the way from the nut to the bridge (your 2/3 position). If you press there, the new length is the string is 2/3L, so the fundamental of this fretted string would be $$f_{fret7}=\frac{v}{2(L)(2/3)} = \frac{3v}{4L}$$. This is exactly half of the frequency of the 2nd overtone (also called the 3rd harmonic). Merely touching the string just over the 7th fret kills the original fundamental and 1st overtone, allowing the 2nd overtone to be heard with a frequency an octave higher than the 7th fret pressed note.

And that's how it happens. If you touch just above the 5th fret you will get a double octave because now you have killed the vibrations of the first three harmonics and forced a node at the nodal location for the 4th harmonic/3rd overtone.

• Small nitpick. The Fourier Theorem doesn't cause the multiple frequencies ("due to"), it helps describe them. Commented Jan 21, 2015 at 0:19
• @Floris Fourier Theorem allows you to show that the pluck waveform is equivalent to multiple frequency sine/cosine inputs. "Due to" is about "how we explain" the multiple f's. Never meant to imply cause. Commented Jan 21, 2015 at 4:45

Harmonic Response

Consider an elastic string stretched between two fixed points a distance $L$ apart. The harmonic response is a standing wave with speed $c=\sqrt{E/\rho}$

\begin{align} y(x,t) &=\sum_{i=1}^{\infty}\sin\left(\frac{i\pi x}{L}\right)\left(A_{i}\sin\left(\frac{i\pi ct}{L}\right)+B_{i}\cos\left(\frac{i\pi ct}{L}\right)\right) \\ \dot{y}(x,t) &=\sum_{i=1}^{\infty}\sin\left(\frac{i\pi x}{L}\right)\frac{i\pi c}{L}\left(A_{i}\cos\left(\frac{i\pi ct}{L}\right)-B_{i}\sin\left(\frac{i\pi ct}{L}\right)\right) \end{align}

with unknown coefficients $A_{i}$ and $B_{i}$ depending on the initial conditions. This is a result of the equation of motion

$$T\,\frac{\partial^{2}y(x,t)}{\partial x^{2}}-\rho\,S\,\frac{\partial^{2}y(x,t)}{\partial t^{2}}=0$$

where $T$ is the elastic tension in $\rm{[N]}$, $E$ the elastic modulus in $[{\rm N/m^2}]$, $S$ is the string area in $[{\rm m^2}]$ and $\rho$ the mass density in $[\rm{kg}/{m^{3}}]$.

Initial Conditions

Now consider at $t=0$ the shape and speed of the string to be $y(x,0)=U(x)$ and $\dot{y}(x,o)=V(x)$.

For example, for a slow pluck at point $x_p$ the initial speed is zero $V(x)=0$ and the shape is $$U(x)=\begin{cases} U_{0}\left(\frac{x}{x_{p}}\right) & 0\geq x\geq x_{p}\\ U_{0}\left(1-\frac{x-x_{p}}{L-x_{p}}\right) & x_{p}>x\geq L \end{cases}$$ where $U_0$ is the amplitude of the pluck

Fourier Analysis

By expanding out the following frequency decomposition we get the coefficients $A_i$ and $B_i$.

\left. \begin{aligned} \int\limits _{0}^{L}\sin\left(\frac{j\pi x}{L}\right)U(x)\,{\rm d}x&=\frac{L}{2}\sum_{i=1}^{\infty}\left[B_{i}\delta_{ij}\right]=\frac{L}{2}B_{j} \\ \int\limits _{0}^{L}\sin\left(\frac{j\pi x}{L}\right)V(x)\,{\rm d}x&=\frac{L}{2}\sum_{i=1}^{\infty}\left[\frac{i\pi c}{L}A_{i}\delta_{ij}\right]=\frac{i\pi c}{2}A_{j} \end{aligned} \right\} \boxed{ \begin{aligned} A_{i}&=\frac{2}{i\pi c}\int\limits _{0}^{L}\sin\left(\frac{i\pi x}{L}\right)V(x)\,{\rm d}x \\ B_{i}&=\frac{2}{L}\int\limits _{0}^{L}\sin\left(\frac{i\pi x}{L}\right)U(x)\,{\rm d}x \end{aligned} }

Again for the slow plucked string we get \begin{aligned} A_{i}&=0 \\ B_{i}&=U_{0}\frac{2L^{2}\sin\left(\frac{\pi ix_{p}}{L}\right)}{\pi^{2}i^{2}x_{p}\left(L-x_{p}\right)} \end{aligned}

Example

A slow plucked string in the middle $x_p = \frac{1}{2} L$ has solution

$$y(x,t) = \sum_{i=1}^{\infty} \frac{8 U_0}{i^2 \pi^2} \sin\left( \frac{i \pi x}{L} \right) \sin\left( \frac{i \pi}{2} \right) \cos\left( \frac{\pi c i t}{L} \right)$$ $$y(x,t) \approx \frac{8 U_0}{\pi^2} \left[ \cos(\theta) \sin\left( \frac{\pi x}{L} \right) - \frac{1}{9} \cos(3\theta) \sin\left( \frac{3 \pi x}{L} \right) +\frac{1}{25} \cos(5\theta) \sin\left( \frac{5 \pi x}{L} \right) + \ldots \right]$$

where $\theta = \frac{\pi c t}{L}$ is a new independent variable replacing time.

Additionally to the "usual" frequency of the string, it always produces tones with integer multiple frequencies with decreasing volume due to an effect called standing waves. When you damp say in the middle of the string only multiples of 2 remain because only those standing waves have a node there. What you hear will be an octave because the lowest of the frequencies will be the loudest. If you damp at a third of the string, only multiples of 3 times the ground frequency remain etc.