Sound/accoustics - guitar string question

[DragonPetter]

I think, getting back to the OP, all of this is off-topic. The answer to the OP is that a guitar string can be quite accurately thought of as having a single lowest possible frequency. Let's call it the main frequency temporarily, to avoid further semantic thrashing. The string also contains integer multiples of that main frequency at various, usually lower intensities than the main frequency. These integers are 2,3,4,... Let's call them "non-main" frequencies. If any frequency (main or not) of a string is at or very near any frequency (main or not) of another string, then either string, when plucked, will "ring" the other - it will cause it to vibrate.

Then you can get nitpicky and say "what if a string is plucked in such a way that the frequency that rings the other string is absent?". Well, ok, then in that case it won't ring the other string.

I think you've added back in incorrect/inaccurate ideas that we've attempted to remove.

Specifically: "If any frequency (main or not) of a string is at or very near any frequency (main or not) of another string, then either string, when plucked, will "ring" the other - it will cause it to vibrate. "

There is no such thing as "very near". Either they share a common frequency component in their response, or they don't. If they share this component, then they share a harmonic and can transfer energy between each other. Also, I don't understand why the lowest frequency would be called a main frequency when it does not necessarily have any relation to the other frequencies nor does it have the largest amplitude necessarily.

I see no reason to invent your own terms when harmonic and fundamental frequencies are pretty clearly defined mathematically and explain the effect more accurately.

 

[someGorilla]

DragonPetter, you can stop it now. His description is simple, understandable, and correct.

 

[DragonPetter]

DragonPetter, you can stop it now. His description is simple, understandable, and correct.

No it isn't? Its not simple when there are trivial definitions like "main frequency" and "non main frequency" that don't mean anything to describe what's happening. Its not understandable or correct if I ask how can a frequency very near but not the same as the vibration modes of a string cause it to vibrate? Can you put "very near" in any mathematical context, as in +/- x hertz? I'm not trying to be argumentative, just objective and accurate.

 

[Rap]

I think you've added back in incorrect/inaccurate ideas that we've attempted to remove.

Specifically: "If any frequency (main or not) of a string is at or very near any frequency (main or not) of another string, then either string, when plucked, will "ring" the other - it will cause it to vibrate. "

There is no such thing as "very near". Either they share a common frequency component in their response, or they don't. If they share this component, then they can transfer energy between each other. Also, I don't understand why the lowest frequency would be called a main frequency.

I see no reason to invent your own terms when harmonic and fundamental frequencies are pretty clearly defined mathematically.

I read thru the previous posts and my eyes glazed over at the semantic arguments, so I just wanted to get a point across without staking a claim in that argument. With regard to "very near", strictly speaking, I should not have said that. If you assume that the guitar string frequencies are in a bandwidth of zero width (as I basically did), then you are right. I was just worried about the fact that e.g., the third harmonic of the lowest string (I call it E1) is at 247.221 hz while the next to highest string (I call it B3) is at 246.942 hz, yet they still ring each other, because the response curve is really not infinitely narrow, just very narrow.

 

[someGorilla]

"how can a frequency very near but not the same"

it can very well. Try to hit a D1 on a well tuned grand piano while you keep the F#3 pressed down, then release the D1. Does it ring? Yes. Is it the same exact frequency resonating? No. Point demonstrated. Period.

 

[DragonPetter]

"how can a frequency very near but not the same"

it can very well. Try to hit a D1 on a well tuned grand piano while you keep the F#3 pressed down, then release the D1. Does it ring? Yes. Is it the same exact frequency resonating? No. Point demonstrated. Period.

Here is a better Idea: isolate the D1 string, place a microphone near it and measure the audio spectrum from an impulse (piano hammer dropping and lifting). Do the same for the F#3 string. You will see that they both share a common frequency in their responses if one can cause the other to vibrate. There is no such thing as a linear system responding to a frequency that is not in its own response but "very near".

 

[someGorilla]

Here is a better Idea: isolate the D1 string, place a microphone near it and measure the audio spectrum from an impulse (piano hammer dropping and lifting). Do the same for the F#3 string. You will see that they both share a common frequency in their responses if one can cause the other to vibrate. There is no such thing as a linear system responding to a frequency that is not in its own response but "very near".

What do you use to measure the audio spectrum? The attached image is an analysis of a pure sine wave obtained with a simple software. It looks like a somewhat wide frequency range; but no, it's a pure sine wave at 360 Hz. So be careful you might get apparent ranges much wider than the actual spike should be.

And by the way, systems with no common frequency can interact and settle to a common intermediate frequency which is in neither's frequency response.

 

[DragonPetter]

What do you use to measure the audio spectrum? The attached image is an analysis of a pure sine wave obtained with a simple software. It looks like a somewhat wide frequency range; but no, it's a pure sine wave at 360 Hz. So be careful you might get apparent ranges much wider than the actual spike should be.

And by the way, systems with no common frequency can interact and settle to a common intermediate frequency which is in neither's frequency response.

Sorry, but that is not a pure sine wave in the frequency domain.

Anyway, This is the effect you're describing, no? http://en.wikipedia.org/wiki/Sympathetic_resonance

 

[DragonPetter]

And by the way, systems with no common frequency can interact and settle to a common intermediate frequency which is in neither's frequency response.

That would be the effect of a nonlinear system called mixing.

 

[someGorilla]

Sorry, but that is not a pure sine wave in the frequency domain.

That's my point. It doesn't look like a pure sine wave, but it is (I just made it), as pure as a computer-generated sine wave can be with 44100 Hz temporal resolution.

Anyway, This is the effect you're describing, no? http://en.wikipedia.org/wiki/Sympathetic_resonance

This is the effect the thread was about. What I said about no common frequency I referred to this kind of phenomenon: http://www.synthgear.com/2010/audio-gear/synchronization-of-metronomes/
It's not directly linked to what we're discussing, it was just to show that your statement that
There is no such thing as a linear system responding to a frequency that is not in its own response but "very near".

was a bit cocksure. And about linearity, what is linear in a real string's vibration? There are many non linear effects coming into play.

 

[DragonPetter]

This is the effect the thread was about. What I said about no common frequency I referred to this kind of phenomenon: http://www.synthgear.com/2010/audio-gear/synchronization-of-metronomes/
It's not directly linked to what we're discussing, it was just to show that your statement that
There is no such thing as a linear system responding to a frequency that is not in its own response but "very near".

was a bit cocksure. And about linearity, what is linear in a real string's vibration? There are many non linear effects coming into play.

No, the assumption that these 5 metronomes don't share a common resonant frequency peak, and the assumption that the table is not putting them in phase, but rather creating new non-common frequencies for each one is cocksure. P.S. read the article, he even references harmonics..

As far as the plot you gave, that only shows that you picked up noise or that your analysis software/transducer does not have perfect response for the sine wave. You cannot apply a sine wave to a system and get a frequency plot, that just doesn't make sense because that response will be just 1 data point, since a frequency plot is over a range of frequencies. Look in any signals and systems textbook and you will see a sine wave is a vertical line at its frequency in a frequency response.

 

[someGorilla]

No, the assumption that these 5 metronomes don't share a common resonant frequency peak, and the assumption that the table is not putting them in phase, but rather creating new non-common frequencies for each one is cocksure.

The metronomes, by themselves, don't share a common resonant frequency peak. You can try by yourself. Take a metronome and see if you can find its "additional" resonant frequencies. The whole system, with the metronomes connected in that way, does have a global resonant frequency.

As far as the plot you gave, that only shows that you picked up noise

No, it's computer-generated.

or that your analysis software does not have perfect response to the sine wave.

Of course. The same might be true for yours. That's why I asked what you use.

 

[DragonPetter]

The metronomes, by themselves, don't share a common resonant frequency peak.

Yes they do. They are damped oscillators with as close as possible masses, pendulum lengths, etc. Their peaks might not be exact, but their responses are centered around a common frequency.

 

[DragonPetter]

No, it's computer-generated.
Of course. The same might be true for yours. That's why I asked what you use.

I use MATLAB or pen and paper mostly when I perform frequency analysis. I use PSPICE for frequency response of circuits.

What your plot shows is an underdamped system response. A pure sine wave is undamped, and it is a signal - not a system.

 

[someGorilla]

Yes they do. They are damped oscillators with as close as possible masses, pendulum lengths, etc. Their peaks might not be exact, but their responses are centered around a common frequency.

Hmm I looked at the video again and you might be right, they seem to be "tuned" on similar or equal frequencies (pendulum lengths). I was just looking for something like that and I might have landed on the wrong video!
There are examples of the same thing with significantly different pendulum lengths, and of course no shared frequency, and they end up oscillating at the same frequency once they're connected.

 

[DragonPetter]

Hmm I looked at the video again and you might be right, they seem to be "tuned" on similar or equal frequencies (pendulum lengths). I was just looking for something like that and I might have landed on the wrong video!
There are examples of the same thing with significantly different pendulum lengths, and of course no shared frequency, and they end up oscillating at the same frequency once they're connected.

No shared frequency, but shared harmonic frequency. You really need to start looking at the math instead of youtube if you want to convince me.

 

[someGorilla]

I use MATLAB or pen and paper mostly when I perform frequency analysis. I use PSPICE for frequency response of circuits.

What your plot shows is an underdamped system response. A pure sine wave is undamped, and it is a signal - not a system.

Good. When you have a minute, can you post a spectral analysis of a pure sine wave, done with MATLAB?

 

[DragonPetter]

Good. When you have a minute, can you post a spectral analysis of a pure sine wave, done with MATLAB?

Someone already did it for you:
http://4.bp.blogspot.com/_4Q2hMGW5A...fy1Tebk/s1600/sineWave_widened_window_FFT.png

The datapoints on the sides of the peak at the base are from windowing effects of the FFT.

 

[Rap]

Here is a better Idea: isolate the D1 string, place a microphone near it and measure the audio spectrum from an impulse (piano hammer dropping and lifting). Do the same for the F#3 string. You will see that they both share a common frequency in their responses if one can cause the other to vibrate. There is no such thing as a linear system responding to a frequency that is not in its own response but "very near".

I don't think that's true. Consider the simpler case of an undamped 1-dimensional harmonic oscillator. It has a single natural frequency, but it can be "driven" at any frequency. The amount of energy needed to drive it to get a particular amplitude is zero when the driving frequency is the natural frequency, and grows larger the further the driving frequency is away from the natural frequency.

If you couple two such oscillators and get one going at its natural frequency, and they both have the same natural frequency then very quickly they will both oscillate at that frequency, with equal energy. If their natural frequencies are slightly different, they will still drive each other, so the motion is more complicated, but it is still there and eventually they will, on average, share the energy equally. If the frequencies are very different, they still drive each other, but the time it takes for them to finally share their energy equally will be very long.

I bet the same holds true for "linear" guitar strings with each frequency spike being infinitely narrow (a delta function in frequency). Two such linear strings that have their harmonics line up somewhere will drive each other rather efficiently. If there is a slight mismatch, they will drive each other a little less efficiently. If there is a large mismatch in frequencies, they will still drive each other, but the time it takes for the effect to occur is much longer than the damping time, so it never really happens in a practical sense. (yes, I introduced damping into the previously undamped model, just to show why it "doesn't happen" in the real world.)

 

[sophiecentaur]

If you couple two such oscillators and get one going at its natural frequency, and they both have the same natural frequency then very quickly they will both oscillate at that frequency, with equal energy. If their natural frequencies are slightly different, they will still drive each other, so the motion is more complicated, but it is still there and eventually they will, on average, share the energy equally. If the frequencies are very different, they still drive each other, but the time it takes for them to finally share their energy equally will be very long.

There is a fascinating and almost unbelievable 'party trick' experiment you can do with two lightly coupled pendulums to demonstrate this. Fix a horizontal string about 50cm long, tightly between two supports. Hang two equal length pendulums from the string (fine thread with identical masses hung on them - classically you use potatoes!). Their periods need to be be slightly different. With one pendulum stationary, start the other. Eventually, the stationary pendulum will start to move and the first will slow down and stop: energy has transferred from one to the other. Then the energy will transfer back to the first pendulum...and so on, until the energy dissipates. The time for the transfer is related to the difference in period of the two pendulums. This is a 'perfectly' linear system.
It works best with a clamped wire at the top and two steel rods as the pendulums; it will go on for ages as energy is dissipated so slowly.

 

[Rap]

There is a fascinating and almost unbelievable 'party trick' experiment you can do with two lightly coupled pendulums to demonstrate this. Fix a horizontal string about 50cm long, tightly between two supports. Hang two equal length pendulums from the string (fine thread with identical masses hung on them - classically you use potatoes!). Their periods need to be be slightly different. With one pendulum stationary, start the other. Eventually, the stationary pendulum will start to move and the first will slow down and stop: energy has transferred from one to the other. Then the energy will transfer back to the first pendulum...and so on, until the energy dissipates. The time for the transfer is related to the difference in period of the two pendulums. This is a 'perfectly' linear system.
It works best with a clamped wire at the top and two steel rods as the pendulums; it will go on for ages as energy is dissipated so slowly.

Excellent. It sounds like the frequency spectrum of each potato contains a difference or "beat" frequency. I bet the same type of thing happens with two guitar strings at nearly the same frequency. If dissipation is very low, they probably transfer energy back and forth between each other. In a real situation, dissipation is much too great for that to occur before the vibration dies out, but the plucked string will still ring the other string.

Thinking about it, I realized that frictional damping of the strings is probably not the most important mechanism for dissipation of energy of the guitar strings, its probably sound radiation.

 

[AlephZero]

There is a fascinating and almost unbelievable 'party trick'
experiment you can do with two lightly coupled pendulums to demonstrate this.

A different (and perhaps better) way to look at this is to see it as one vibrating system with two degrrees of freedom. The two natural frequences are close (but not identical). In one of the modes, the masses vibrate in phase with each other. In the other mode they move are 180 degrees out of phase.

When you start one pendulum swinging, you actually start a linear combination of both modes. Because the frequencies are slightly different, the amplitude of each pendulum shows "beats" as its ampltude increases and decreases.

If you look at it that way, there is no "magic" about transferring energy between different vibration modes at different frequencies required, and it is clear the system can work if it is completely linear.

Of course this also applies to the guitar, whcih is vibrating as ONE connected system consisting of all the strings, the guitar body, and the air inside (the effect of the air is not neglibible - the position and size of the sound hole on an acoustic guitar is a critical part of the design). In particular, there are two vibration modes with close frequencies which inolve BOTH the 1st and 5th strings. When you pluck one string, you excite both of them. End of story.

Actually there are 4 vibration modes not 2, because each string can vibrate in two different planes. The flexibility of the guitar bridge is different parallel and perpendicular to the body of the guitar, so the two modes have slightly different frequencies, and very different damping factors. All this is ignored in a first physics course on vibrations, but it is crucial to the way a real acoustic guitar actually sounds (and important for the playing technique as well).

@DragonPetter, I don't have time to go through all your posts in this thread trying to sort out the mistakes, but you seem to be very confused about much of this - for example the difference between the steady state response of a linear system and its transient response (for a guitar there IS no steady state response!) Note that some of the Wiki pages describing "harmonics" etc don't make this distinction clear either.

AS for the semantic debate about harmonics, overtones, partials, etc - IMO all of that terminology is more or less obsolete. Some of it only applies to simplfied models of real instruments and "musical sounds" (and let's not even try to define what is "muscial" and what is not!). If we are trying to do physics or engineering, I think it's better just to stick to "vibration frequencies", and number them 1,2,3 ... from the lowest upwards, if that is useful.

 

[DragonPetter]

@DragonPetter, I don't have time to go through all your posts in this thread trying to sort out the mistakes, but you seem to be very confused about much of this - for example the difference between the steady state response of a linear system and its transient response (for a guitar there IS no steady state response!) Note that some of the Wiki pages describing "harmonics" etc don't make this distinction clear either

I don't have time to go through all of my posts again either, but I'm not sure what you mean that I'm confused about the difference between steady state response and transient response. Where did I talk about this ever? Also, what do you mean by saying that a guitar has no steady state response? If I pluck a string, that vibration dies down to 0 after a long enough time. If what I just said is wrong, then I have to admit that I am very confused.

My point has been, if you look at the frequency response of an oscillator, and you see that it has 'gain' or 'transmission' or whatever you want to call it in context, at specific frequencies, or a range of frequencies, like you see a curve around a resonance for an oscillator, where that curve is sharper or wider depending on the Q factor, then the oscillator will respond to any frequency in that curve range. To say that you can apply a frequency where the response of the oscillator has no transmission/gain (I've been referring to this as "no response to a frequency") and make it oscillate is what I said does not make sense. In all of my posts about that, I have said "its response" in reference to its frequency response. Someone was saying that an oscillator is capable of oscillating to a frequency that is not in its frequency response . . that is almost an obvious contradiction that I have been trying to discount.

I'm not sure what steady state and transient has to do with anything I've said, as I've almost always been talking in the frequency domain this entire discussion. Steady state and transient response are both components of the frequency response.

 

[bahamagreen]

Long time guitar player to the rescue...

When you play the open high E 1st string you are playing the same pitch as the one sounded by fretting the A string at the 19th fret. The A string may be divided into three equal lengths - nut to 7th fret, 7th fret to 19th fret, and 19th fret to saddle/bridge. The 7th and 19th frets are the two nodes for the three vibrating sections of the open A string. The pitch of the open high E 1st string matches the pitch of the A string fretted at the 19th, or touched at the 7th fret node (what guitar players call playing a harmonic). You can play the same pitch as a harmonic at the 19th by touching it rather than fretting it. The E string drives the open A string to sound the same pitch on those 1/3 length sections... sympathetic vibration through the air and through the body of the instrument.

As far as terminology, it can be confusing. In music:

Frequency of the fundamental (f) is called the first harmonic.
The first overtone is f*2 and called the second harmonic.
The second overtone is f*3 and called the third harmonic.

So the order of the harmonic takes the same order as the "n" of f*n.
And the order of the overtone takes the order of n-1 unless n=1, then it is called the fundamental...

In the cited example, the open E 1st string is playing its own fundamental or first harmonic. That same note pitch is the second overtone or third harmonic of the open A string.

This same relationship is present between the open B string and the open low E 6th string... the third harmonic of the open low E is the same pitch as the open B.

 

[sophiecentaur]

I'm not sure what steady state and transient has to do with anything I've said, as I've almost always been talking in the frequency domain this entire discussion. Steady state and transient response are both components of the frequency response.

There is still some confusion here. There is a distinct difference between the two responses you mention:
The term 'frequency response' describes how the system responds to a continuously applied (steady state) excitation. This is the 'bell shaped' response we are familiar with. If you apply a steady frequency (on or off the resonance peak) and then remove the exciting signal, the frequency of oscillation will not change but stay at the value of the applied frequency.

The term 'transient response' describes the behaviour after an impulse is applied. This will result in an oscillation that is spot on the natural resonant frequency.
Same oscillator but two different frequencies result. So a guitar string will 'go off' at its natural frequency, when plucked (along with the overtones, of course) but will resonate to many other frequencies that are near to the fundamental (of higher order modes).

 

[DragonPetter]

There is still some confusion here. There is a distinct difference between the two responses you mention:
The term 'frequency response' describes how the system responds to a continuously applied (steady state) excitation.

I can't agree with that. The frequency response will tell me how a system responds to any type of excitation, including impulse, step, or continuous. The point you're getting at, I believe, is that Fourier transform is for continuous, periodic signals (integrate from t = -infinity to + infinity). Laplace transform, for which an oscillator can have a transfer function, is integrated from time = 0. When I talk about frequency response, I am not talking about a Fourier transform of a periodic signal.

If transient and steady state behavior are not in the frequency response, then you could not switch back and forth through the laplace transform and get valid results for analyzing systems.

When you want to know the steady state response of a system in the frequency domain, you look at the transfer function as frequency -> 0. That alone tells me that steady state is included in the frequency response. Given complete magnitude and phase plots of the frequency response of a system, you can derive both the transient and the steady state response for the time domain . . we did it all the time in controls class!

 

[sophiecentaur]

Your terms are muddled here. Frequency response is frequency response (as defined above) and impulse response is impulse response. Whilst one response can yield the other, if you don't use the terms appropriately then it is all too easy to jump to the wrong conclusions about the actual results of an experiment. For a simple oscillator, you can transform between the two responses but you can't do the same for something even as simple as a string because the actual point and direction of the applied impulse will affect the relative levels of the frequency components of the final note. There is no way that an applied impulse can result in any frequencies other than those of the natural modes - this is nothing like the response to a continuous applied signal.

 

[DragonPetter]

Your terms are muddled here. Frequency response is frequency response (as defined above) and impulse response is impulse response. Whilst one response can yield the other, if you don't use the terms appropriately then it is all too easy to jump to the wrong conclusions about the actual results of an experiment. For a simple oscillator, you can transform between the two responses but you can't do the same for something even as simple as a string because the actual point and direction of the applied impulse will affect the relative levels of the frequency components of the final note. There is no way that an applied impulse can result in any frequencies other than those of the natural modes - this is nothing like the response to a continuous applied signal.

If we can switch to electronics temporarily, frequency response to me is a Bode plot, i.e the response of a system over a spectrum. It is not the response of a system from an applied single continuous frequency signal.
http://en.wikipedia.org/wiki/Frequency_response

If you guys don't believe me, look here:
http://people.exeter.ac.uk/mmaziz/ecm2105/ecm2105_n4.pdf

Mathematical examples of deriving time domain steady state response from a frequency domain description.

 

[sophiecentaur]

You can have your own ideas but you can't ignore the fact that you get different results in practice from continuous and impulse excitation of a string. If you go to 99% of informed people, the amplitude / frequency response we know and love is the first thing we think of - followed by the phase response. You can't generate frequencies out of nowhere in a linear system. All frequencies exist in an impulse but only those corresponding to the normal modes survive after the initial settling time of a string.

PS Where do the frequencies on a Bode Plot come from, if they don't represent the response for each spot (continuously applied) frequency?

 

[DragonPetter]

You can have your own ideas but you can't ignore the fact that you get different results in practice from continuous and impulse excitation of a string. If you go to 99% of informed people, the amplitude / frequency response we know and love is the first thing we think of - followed by the phase response. You can't generate frequencies out of nowhere in a linear system. All frequencies exist in an impulse but only those corresponding to the normal modes survive after the initial settling time of a string.

These aren't my own ideas. I presented two links of other people's ideas, and these ideas have been around long before I was born and they are in my textbooks and lecture notes. Does the frequency response tell me what the steady state and transient response of a system will be from a single sine wave OR from an impulse applied? Yes, according to http://people.exeter.ac.uk/mmaziz/ecm2105/ecm2105_n4.pdf because he shows examples of doing it.

Of course you get different results in practice from continuous and impulse excitations. They represent different frequencies with different magnitudes/phases being applied to the system, but both responses stem from the string's frequency response (the term used in the way I've been using it and the way the professor in the link uses it). An impulse signal is broken down into components and applied just the same as a single frequency component can be applied. I am really unsure of what you are saying that I have gotten wrong or confused.

PS Where do the frequencies on a Bode Plot come from, if they don't represent the response for each spot (continuously applied) frequency?

Those points do represent single frequencies, the frequency response/Bode however does not. The frequency response represents the spectrum of these points and the entire frequency response is what determines steady state and transient behavior, not a single frequency.From page 15-16 of my link:

Advantages of frequency domain analysis:
- The transfer functions of complicated components can be determined experimentally by
frequency response tests (without deriving their mathematical models) using available signal
generators and precise measurement equipment (e.g. spectrum analysers).

- The amplitude and phase of the frequency response can be used to predict both time-domain transient and steady-state system performances.

- Systems may be designed to achieve transient and steady-state requirements using frequency
response analysis, and such analysis and design may be extended to certain nonlinear control
systems.

Now, tell me that this professor is wrong and that a frequency response does not contain the time domain steady state and transient response information.

 

[sophiecentaur]

I don't think I am disagreeing substantially with what your "professor" is saying. But the title 'professor' doesn't imply total infallibility any more than the title 'pope'; believe me, I have had many arguments with 'professors' and they are not always right - (how could they be if they are famous for disagreeing amongst themselves).
The link says nothing to disagree with what I am saying. It makes a distinction between the time response at the beginning and the final time response, which may consist of long-term 'ringing' - but the examples don't deal with that, I think.
That last quote of yours shows that he is ignoring the behaviour of many of the more complex systems - for instance, loudspeaker drive units.
There are two issues here. Of course a signal can be characterised in either time or frequency domain. When you are dealing with a system, though, its behaviour needs to be described in 'modes', which are the possible free vibrations.

I don't think that you can be disagreeing with what the 'frequency response' of a system really means, i.e. the response to an infinitely slow frequency sweep, where the response to each frequency has time to establish itself. For a string (with energy losses), this will not consist of a series of infinitely narrow peaks but a series of 'humps'.
If you pluck a string (impulse / time response) you have to agree that the 'spectrum' of the resulting sound will not be the same as the frequency response but a series of single frequencies - which is not a series of humps.
Do you say there is no difference? It certainly reads as if you do.

 

[Rap]

Ok, I looked at my notes for harmonic oscillators. They are not guitar strings, but their behavior is linear, and I think you can get a feel for the guitar string situation. An undamped harmonic oscillator has a natural frequency. It can be driven by a force which is oscillating at another frequency. When it is driven, its frequency spectrum is two spikes, one at its resonant frequency, one at the driving frequency. The strength of the resonant component depends on initial conditions, it can be totally absent. The strength of the driving component depends only on the amplitude of the driving force and the difference in the two frequencies. The larger the difference, the stronger the driving amplitude must be in order to maintain a given amplitude in the oscillator. When the driving frequency equals the natural frequency of the oscillator, the amplitude of the oscillator frequency is driven to infinity (or alternatively, the amplitude of the driving frequency must be zero in order to maintain a given amplitude in the oscillator).

If you have two harmonic oscillators that are weakly coupled (with no other forces driving the system), and their natural frequencies are different, the frequency spectra of each oscillator will have two spikes, one at its natural frequency, one at the other oscillator's natural frequency. (This makes sense, the second oscillator is driving the first and vice versa). If one oscillator is oscillating at its natural frequency only, then the other oscillator will be oscillating at that same frequency. Neither will have a component of the second oscillators natural frequency but this is a very specific case. For example, in the case where one oscillator is for the most part oscillating at its natural frequency while the other oscillator is momentarily at rest, in its zero position, that other oscillator will oscillate weakly at both frequencies, rather evenly distributed.

I'm guessing that if you have two guitar strings tuned to slightly different frequencies and you pluck one, the other will vibrate weakly at both its own natural frequency spectrum and the natural frequency spectrum of the plucked string, rather evenly distributed between the two. The plucked string will have a very, very weak spectrum of the other string also. If you damp the plucked string, the other will no longer be driven at an un-natural frequency and that frequency component will immediately disappear, and the string will weakly oscillate in a natural frequency spectrum. If you think of a guitar string as having a frequency spectrum that is a series of narrow spikes, then plucking any string will to some extent cause every other string to vibrate at both its own frequency spectrum and that of the plucked string. It is only when a harmonic of one is at or very near a harmonic of the other that this excitation may be strong enough to be heard.

 

[someGorilla]

I haven't been following the discussion lately nor I have the time to read it now.
So maybe this doesn't apply to the latest posts. However let me just note that we can easily disprove one of DragonPetter's claims: it's possibe to pluck a string at a certain frequency A and a second string at a slightly different frequency B, different enough to be discernable in analysis and by ear, and close enough for the first string's vibration to excite the second string. If you pluck the first and then damp it, the second will vibrate at frequency B, having been set into motion by frequency A which is not in its natural frequency response.
I can address the question experimentally and post an audio clip of plucked strings with these characteristics. Can DragonPetter suggest an experiment to test his claims (which I haven't entirely understood)?

 

[sophiecentaur]

Ok, I looked at my notes for harmonic oscillators. They are not guitar strings, but their behavior is linear, and I think you can get a feel for the guitar string situation. An undamped harmonic oscillator has a natural frequency. It can be driven by a force which is oscillating at another frequency. When it is driven, its frequency spectrum is two spikes, one at its resonant frequency, one at the driving frequency. The strength of the resonant component depends on initial conditions, it can be totally absent. The strength of the driving component depends only on the amplitude of the driving force and the difference in the two frequencies. The larger the difference, the stronger the driving amplitude must be in order to maintain a given amplitude in the oscillator. When the driving frequency equals the natural frequency of the oscillator, the amplitude of the oscillator frequency is driven to infinity (or alternatively, the amplitude of the driving frequency must be zero in order to maintain a given amplitude in the oscillator).

It is not clear what system you are describing here. If it is truly undamped, linear and 'simple' (mass on spring or LC circuit) then it has a single resonance 'spike' and you will not be able to make it oscillate at any other frequency. The truly undamped condition is not really worth considering as it doesn't represent any real situation. If you accept some damping (a finite Q) then the response to an excitation with frequency will give the well known Lorenz Bell shape. I would be interested to know of a mechanism that can excite the 'natural' frequency, assuming the system is truly linear and has only a single mode. The exciting signal will constantly be changing its phase wrt the notional 'natural' frequency oscillation - resulting in no net energy transfer after a period corresponding to what would be the 'beat' frequency.

 

[Rap]

Yes, its a simple mass on a spring, no damping. I think you can see that it can be made to exhibit oscillatory motion at any frequency. Suppose the mass on a spring has a natural frequency of one cycle per second. Just grab the mass and move it at one cycle every 5 seconds, and there you go, its motion is oscillatory at 1/5 hertz. The force you have to apply to make this happen will generally not be a pure sinusoidal force at 1/5 hz, but under the right initial conditions, it can be.

The equation for a forced harmonic oscillator is $$m \ddot{x}=-k x+F e^{i \omega t}$$ where $m$ is the mass, $k$ is the spring constant, $F$ is the amplitude of the driving force, and $\omega$ is the frequency of the driving force. The natural frequency is $\omega_0=\sqrt{k/m}$. The solution is $$x=-\frac{A e^{i \omega t}}{(\omega^2-\omega_0^2)}+C_1 e^{i \omega_0 t}+C_2 e^{-i \omega_0 t}$$ where $A$ is the acceleration due to the external force (=$F/m$) and $C_i$ are complex constants which depend on initial conditions. With the right initial conditions, you can have the $C_i$ both be zero, but if the initial conditions are not right, then I think you have to add a force at frequency $\omega_0$ to compensate to make the oscillator oscillate at a pure $\omega$ frequency. The Q is $1/(\omega^2-\omega_0^2)$ which is infinite at $\omega=\omega_0$, but adding damping turns it into the Lorentzian. As long as the periods corresponding to both frequencies are considerably shorter than the damping time, I think adding damping is an unnecessary complication.