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It's not just where you pluck the string: most electric guitars have a "neck" pickup and a "bridge" pickup (sometimes a third in the middle). The neck pickup is

It's not just where you pluck the string: most electric guitars have a "neck" pickup and a "bridge" pickup (sometimes a third in the middle). The neck pickup is closer to the middle of the string, and the bridge pickup is close to the end of the string. Regardless of where you pluck, the bridge pickup has a significantly more prominent high-end, to the point of being a bit shrill when played in isolation. Typically rock guitarists play rhythm with the neck pickup so they don't overpower the vocalist, then lead with the bridge pickup so they cut through the mix without needing to amp the volume too loudly. Why is this the case? It is funny that my guitarist's intuition seems very clear about it - "the string is tougher and clickier at the bridge compared to the neck, of course the tone is more shrill" - but in terms of actual analytical evidence I just have to say "something something Fourier coefficients" :) Refining the physical intuition a bit: I believe the boundary at the end of the string dampens lower-frequency (i.e. lower-energy) vibrations faster than higher-frequency vibrations, so the lower harmonics die off more quickly than the higher "nasal" harmonics.

mrob 2y ago
Isn't it just the geometry of the guitar constraining the ends of the string to have zero amplitude? The fundamental has peak amplitude only at the center of the vibrating part of the string. Higher harmonics have peaks in amplitude at multiple places along the string, and the higher the harmonic the closer one of those maxima is to the bridge.
- thrtythreeforty 2y ago
  Bingo! It's all about harmonics' nodes. For a visualization, Cycfi Research has a great series on how pickup position affects tones. He also sells a "modeling pickup" based on this theory. https://www.cycfi.com/2014/07/virtual-pickup-placement-part-1/ https://www.cycfi.com/2014/07/virtual-pickup-placement-part-...
- nicklecompte 2y ago
  The fundamental result of Fourier analysis is that we are saying the same thing :) Though I should have clarified that the kinetic energy is zero at the "boundary" (ie bridge). IMO which answer you prefer depends on perspective: - if you assume a wave can be broken down into sinusoidal overtones then your geometric approach is much more immediate and intuitive: sinusoidal overtones => higher overtones clearly have more kinetic energy near the boundary, just draw a picture. - if you assume that higher-pitched overtones have more kinetic energy then the physics approach explains why they are sinusoidal. Not the specific shape unless you do the math, but the "gist" of the slope. If the overtones were more like square waves, with no real difference in shape between frequencies beyond the length of the rectangle, then the pickup position wouldn't matter. But they can't be, the overtones have to be more "trapezoidal." And in particular, the lower overtones must have a more gradual slope than the higher overtones. The geometric approach makes a big (but correct) physical assumption for an easy analytical argument; the physical approach goes the other way, only depending on Newton's laws + a lot of elbow grease.

mrob 2y ago

Isn't it just the geometry of the guitar constraining the ends of the string to have zero amplitude? The fundamental has peak amplitude only at the center of the vibrating part of the string. Higher harmonics have peaks in amplitude at multiple places along the string, and the higher the harmonic the closer one of those maxima is to the bridge.

thrtythreeforty 2y ago
Bingo! It's all about harmonics' nodes. For a visualization, Cycfi Research has a great series on how pickup position affects tones. He also sells a "modeling pickup" based on this theory. https://www.cycfi.com/2014/07/virtual-pickup-placement-part-1/ https://www.cycfi.com/2014/07/virtual-pickup-placement-part-...
nicklecompte 2y ago
The fundamental result of Fourier analysis is that we are saying the same thing :) Though I should have clarified that the kinetic energy is zero at the "boundary" (ie bridge). IMO which answer you prefer depends on perspective: - if you assume a wave can be broken down into sinusoidal overtones then your geometric approach is much more immediate and intuitive: sinusoidal overtones => higher overtones clearly have more kinetic energy near the boundary, just draw a picture. - if you assume that higher-pitched overtones have more kinetic energy then the physics approach explains why they are sinusoidal. Not the specific shape unless you do the math, but the "gist" of the slope. If the overtones were more like square waves, with no real difference in shape between frequencies beyond the length of the rectangle, then the pickup position wouldn't matter. But they can't be, the overtones have to be more "trapezoidal." And in particular, the lower overtones must have a more gradual slope than the higher overtones. The geometric approach makes a big (but correct) physical assumption for an easy analytical argument; the physical approach goes the other way, only depending on Newton's laws + a lot of elbow grease.