Why do some things resonate more than others? I was holding a guitar next to a bass amplifier, when the bass hit the string at a certain note, the same note began to vibrate in the guitar. If I pressed certain strings in the guitar, the bass strings in the guitar, the chord vibrated too without having to pluck it. However, if I did the same for the thin strings I didn't see much vibration (maybe there was and I didn't perceive it). Why doesn't it vibrate the same if they have both the same frequency as the source?
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$\begingroup$ The question is clear and makes sense. $\endgroup$– Bill NCommented Nov 3, 2017 at 20:35
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1$\begingroup$ Also note that the guitar body is not intended to be a resonator, but a phase inverter (a bass reflex system) while the guitar top matches the impedance of the strings to the impedance of the air. Both the body and top ensure the maximum energy transfer not only from the strings to the air, but also from the air to the strings. This explains your guitar reacting to the external sound. A more detailed explanation is here: physics.stackexchange.com/questions/365557/… $\endgroup$– safesphereCommented Nov 4, 2017 at 5:05
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3 Answers
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Resonance is a phenomenon in which the frequency of some driving force closely matches a natural frequency of the system exposed to the force. The system will have low damping at that frequency and energy from the driving force source can build up and cause vibrations in the exposed system.
If a guitar string is exposed to a sound which has a frequency component, a la Fourier decomposition and analysis, then all those frequencies could be driving the guitar string. If the string has a natural frequency component which matches one of the frequencies in the sound, it will absorb energy begin to vibrate with that frequency.
In a stringed instrument, the natural frequencies are integer multiples of a lowest frequency, the fundamental, $f_0$. The fundamental frequency depends on the length ($L$), density ($\rho$), radius ($r$), and tension ($F_T$) of the string: $$f_0=\frac{1}{2L}\sqrt{\frac{F_T}{\rho\pi r^2}}.$$
For a given musical note on the bass (e.g., $A_2$), the 6 lowest natural frequencies will be close to $A_2$, $A_3$, $E_4$, $A_4$, $C^{\#}_5$, $E_5$. The strengths of these overtones will decrease as you go up in frquency. If a natural frequecy of the guitar string is $A_3$ or $E_4$, you might expect a good amount of interaction. This would be case for the lower $A$ and $E$ strings. The thinner strings will have a higher fundamentals, and although their frequencies might theoretically match an overtone (harmonic) of the bass string, the amplitude of that higher frequency will be too small to produce any noticeable resonant behavior.
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$\begingroup$ "The system will have low damping at that frequency" - for a typical constant coefficient 2nd order ODE, the damping is independent of frequency so I wonder if you're thinking of something else here. $\endgroup$ Commented Nov 3, 2017 at 21:07
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$\begingroup$ @AlfredCentauri Damping may not be the precise word. Maybe you can help me. When one plucks a guitar string, the input is multiple frequencies, based on the triangular shape giving multiple wavelengths. The frequencies from the waveform of the pluck which persist match the natural frequencies of the string. What happens to the others? $\endgroup$– Bill NCommented Nov 4, 2017 at 1:44
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The structure of some systems is such that admitted energy can flow internally between different forms at a regular frequency (potential and kinetic or electrical and magnetic as examples) with minor energy losses. Or else the rate of loss of energy is no greater than the rate at which it is driven. These are known as resonant, high-Q systems, systems whose structures do not include means for dissipating energy input say in the form of heat. Resonant systems can feed off external energy, particularly if the energy is supplied at the same natural frequency as the system itself.
Two identical mechanical systems are thus said to vibrate sympathetically with one another by the transfer of energy, say from your example through sound waves. But the same can happen between radio transmission and receiving circuits by means of electromagentic wave transmission.
When two high Q systems have different natural frequencies, one system that transmits energy between them by whatever means will likely not trap and excite resonance within the other. They are anti-sympathetic. This is why the cochlea, a structure within your ear cannot receive and interpret the ultrasound from a dog whistle for example.
answered Nov 3, 2017 at 21:31
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Trying to think the guitar string as a circuit analogy, not all resonating circuits have the same "Q factor", which implies that a bigger Q will be a more sharpen resonator centered in the resonating frequency with ampitudes increasing as the Q factor increases I do not know how to calculate the Q factor in a mechanical circuit but I guess that the Q factors of each string are different ant this cause of course different effects
