Sound/accoustics - guitar string question

[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.

 

[sophiecentaur]

I think this is a reason for needing to include some damping because the impulse response of your oscillator would extend for all time. If you use a real oscillator (which is the only kind that you can construct in an engineering lab) there is damping and the initial settling time will allow all the natural resonance to dissipate itself. To eliminate it quickly, you need to 'sneak up on it quietly' with your test signal and make sure you don't turn it on at a max or min - or even at a zero, because the time derivative is not continuous.
Your ideal model is implying something that you really can't expect in practice. I have 'squeaked' so many resonant circuits in my time and I have never actually seen this component at ω0 you refer to and the reason is that I have always had damping in there and, with a Q of thousands ( even ) the 'natural' frequency has gone well before you get to measure anything.
Your analysis is quite correct, I think - it just doesn't fit the real world.

 

[Rap]

The real world I am trying to deal with is the guitar strings. If the damping time ( a few seconds) is much longer than the oscillation periods (milliseconds), then you can treat it as an undamped system over windows much shorter than the damping time - a few tenths of a second. In other words, over any window of a few tenths of a second, you can treat it as an undamped oscillator, even though the amplitudes for that undamped oscillator may change appreciably for such windows a second or two apart.

For the electronic system you are talking about, as you have said - the damping time is much shorter than the time between squeaking the system and making the measurement - too big a window by the above reasoning, and so not approximating the real world you are talking about.

 

[sophiecentaur]

I realize we are coming at this from different directions (and that the OP was about guitar strings -but wandering is what it's all about) Actually, I look upon electronic resonators as the 'real world'. I'd bet they're actually measured far more often than mechanical resonators. Most music is not measured but appreciated 'by ear'.
When you turn on a continuous excitation signal, with a Q of 1000, the transient response at the natural frequency will have dropped to half power after only 1000 cycles. The steady state response will have established itself and won't ever change.
My problem with assuming no damping is that the response of the sympathetic string to an off-frequency excitation must be affected by where it sits on the response curve. How can this not be relevant? It strikes me that your treatment would imply that all strings would resonate to all other strings because, for an undamped resonator, the response is the same (zero) for all frequencies but the centre frequency.

 

[Rap]

I realize we are coming at this from different directions (and that the OP was about guitar strings -but wandering is what it's all about) Actually, I look upon electronic resonators as the 'real world'. I'd bet they're actually measured far more often than mechanical resonators. Most music is not measured but appreciated 'by ear'.
When you turn on a continuous excitation signal, with a Q of 1000, the transient response at the natural frequency will have dropped to half power after only 1000 cycles. The steady state response will have established itself and won't ever change.
My problem with assuming no damping is that the response of the sympathetic string to an off-frequency excitation must be affected by where it sits on the response curve. How can this not be relevant? It strikes me that your treatment would imply that all strings would resonate to all other strings because, for an undamped resonator, the response is the same (zero) for all frequencies but the centre frequency.

Well, first of all, I erroneously described the Q-factor of an undamped oscillator as Q=1/(\omega^2-\omega_0^2), so scratch that statement. An undamped oscillator has infinite Q at all frequencies, and an oscillator with very low damping has a large Q>>1. Its the Bode magnitude that is 1/|\omega^2-\omega_0^2| - the absolute value of the ratio of the input magnitude to the output magnitude assuming zero resonant contribution. Assuming small damping, that's equivalent to "steady state". The important point is that the response is not zero for all frequencies except resonant, it is non-zero and finite for all frequencies except resonant, where it goes to infinity.

For two weakly coupled undamped harmonic oscillators, (hopefully corresponding to two guitar strings), if one is oscillating strongly at its resonant frequency only, the second oscillator will be weakly driven at that frequency, and its response will be 1/(\omega^2-\omega_0^2) times the driving amplitude delivered by the first oscillator through the weak coupling. (The first oscillator will also be very weakly driven by the second oscillator at that same frequency). If you stop the first oscillator, the second will immediately begin weakly oscillating at its resonant frequency.

If an oscillator has a Q of 1000, then looking at it with, say, 10-cycle wide windows, its behavior will be well approximated by an undamped oscillator.

 

[sophiecentaur]

I think I am beginning to sort this out in my mind. There are two scenarios involved in this issue and some of my problem has been to confuse the two. There is what happens with a single oscillator, excited from an energy source and there is what happens with isolated coupled oscillators. I can't see how, in either case, it is useful to ignore damping but, in the case of coupled oscillators, it may be OK because there is only a finite amount of energy in the system at the start but there is still energy flow in and out of the two oscillators so each one is either being damped or driven.
For coupled oscillators, one way of looking at it is that energy flows back and forth between the two at the beat frequency. The direction of energy flow is determined by the phase relationship between the two oscillators, which is constantly changing. The notion that you have any 'steady state' condition must be flawed because the whole process involves nothing staying the same - energy is either flowing one way or the other - even if you eliminate the damping. The situation at time t is the result of the past. I can't accept your statement about 10 cycles 'usefully' representing a small enough window to treat the system as undamped because energy is constantly being lost to or gained from the other oscillator and this is a vital part of the co- resonance phenomenon. (This is analogous to treating the Volts on a charging capacitor as being constant, if you take a short enough observation time. Slope is slope, however short an interval its' measured over.)

In the case of a driven oscillator, the phase difference of driver and oscillator at the driving frequency settles down to a constant value and energy continually flows into the oscillator (hence my obsession with the requirement for damping). In this case, whatever happens for the first few cycles of excitation, I don't see how there can be any power flowing into the ω0 mode because the phases are constantly changing wrt the drive and must surely integrate to zero.

 

[Rap]

I think I am beginning to sort this out in my mind. There are two scenarios involved in this issue and some of my problem has been to confuse the two. There is what happens with a single oscillator, excited from an energy source and there is what happens with isolated coupled oscillators. I can't see how, in either case, it is useful to ignore damping but, in the case of coupled oscillators, it may be OK because there is only a finite amount of energy in the system at the start but there is still energy flow in and out of the two oscillators so each one is either being damped or driven.
For coupled oscillators, one way of looking at it is that energy flows back and forth between the two at the beat frequency. The direction of energy flow is determined by the phase relationship between the two oscillators, which is constantly changing. The notion that you have any 'steady state' condition must be flawed because the whole process involves nothing staying the same - energy is either flowing one way or the other - even if you eliminate the damping. The situation at time t is the result of the past. I can't accept your statement about 10 cycles 'usefully' representing a small enough window to treat the system as undamped because energy is constantly being lost to or gained from the other oscillator and this is a vital part of the co- resonance phenomenon. (This is analogous to treating the Volts on a charging capacitor as being constant, if you take a short enough observation time. Slope is slope, however short an interval its' measured over.)

In the case of a driven oscillator, the phase difference of driver and oscillator at the driving frequency settles down to a constant value and energy continually flows into the oscillator (hence my obsession with the requirement for damping). In this case, whatever happens for the first few cycles of excitation, I don't see how there can be any power flowing into the ω0 mode because the phases are constantly changing wrt the drive and must surely integrate to zero.

You can ignore damping in the case of one driven oscillator over a short window, because negligible energy is lost in that short window (by the definition of "short"), but I see your point about the beat frequency. On certain points, I've been confusing the two situations myself. Your point about the beat frequency is having me scratch my head. I mean, you could say that there is no steady state for a single undamped oscillator, its coordinate is cyclic, changing in time. Calling two undamped coupled oscillators a steady state is ok with me, even though there is a beat frequency, its just another frequency that does not die out.

I think 10 cycles for a damping period of 1000 cycles is ok. You can represent the behavior of the oscillator very well by assuming it is undamped. You can represent it very well by a period of 1/10 of a cycle too. The fact that you haven't covered a number of cycles is not an issue. The fact that for 10 cycles you may not have covered a number of "beat cycles" is not an issue.

There's three time periods here, the average period of the two oscillators (a good number if they are separated by a frequency difference small compared to their frequencies), the beat period (corresponding to the difference in the two frequencies) and the damping period, or damping time. I think if the damping time is much longer than the average period, then the system can be well represented by two coupled, undamped oscillators, in a time window small compared to the damping period, no matter what the beat period. If the beat period is much longer than the window, you don't see much of it, if it is short, you see a lot of it. Slope is slope, but if you take a short enough window, the percent change in the function itself is negligible, and by assuming zero slope, your percent error will be low.

You last question I can't visualize, and my notes don't give me any help, so I will fire up Mathematica and try to solve two coupled undamped oscillators.

 

[sophiecentaur]

I still can't see how you can think that the answer is in a 'quasi steady state' approach.
It's true, of course, that the energy in an undamped, undriven system will remain the same. Clearly, the mass on spring model is trivial and we both agree that you can look at a small time window if we want. But why?

If there are two ideal oscillators, there will be some time period (lowest common denominator or whatever you'd call it and it could be a very long time if you choose the numbers right) over which the behaviour will repeat. But, again, how does this help with understanding of what goes on within that cycle? Energy is exchanged from one to the other and both natural modes will be in existence, with ever changing shares of the energy.

To deal with a driven oscillator, you absolutely have to involve some loss as there is no steady state outcome without it. I still cannot see how any analysis of a driven, damped system can lead to an oscillation that involves two frequencies. You will need to write it out or reference it for me before I can accept it. The idea goes against what I have always though to be an obvious bit of bookwork. I found this on the first hit of a Google search

 

[Rap]

I still can't see how you can think that the answer is in a 'quasi steady state' approach.
It's true, of course, that the energy in an undamped, undriven system will remain the same. Clearly, the mass on spring model is trivial and we both agree that you can look at a small time window if we want. But why?

If there are two ideal oscillators, there will be some time period (lowest common denominator or whatever you'd call it and it could be a very long time if you choose the numbers right) over which the behaviour will repeat. But, again, how does this help with understanding of what goes on within that cycle? Energy is exchanged from one to the other and both natural modes will be in existence, with ever changing shares of the energy.

To deal with a driven oscillator, you absolutely have to involve some loss as there is no steady state outcome without it. I still cannot see how any analysis of a driven, damped system can lead to an oscillation that involves two frequencies. You will need to write it out or reference it for me before I can accept it. The idea goes against what I have always though to be an obvious bit of bookwork. I found this on the first hit of a Google search

But you agree that there will be a transient response consisting of two frequencies, so if we are in time frame where the transient signal is appreciable, then we are not in steady state. The undamped oscillator does not necessarily approximate a steady state, it approximates the behavior in a window at any time, where the width of the window is much smaller than the damping time. It may contain both frequencies if the window is in the transient regime. If the phenomenon you are looking at can be profitably analyzed inside such a window, then consideration of undamped oscillators will give good results. For a single guitar string, the damping time is a matter of seconds and there is no steady state except zero. For the resonance phenomenon to occur, only a few, maybe tens of cycles are needed. So using an undamped oscillator as a model is profitable.

To get more mathematical, a damped, driven oscillator is characterized by the driving amplitude A, driving frequency \omega, resonant frequency of undamped oscillator \omega_0, damping constant \gamma, and two constants C_1 and C_2 which modify the phase and amplitude of the transient signal. For \gamma<1, the transient signal frequency is sinusoidal with a frequency of \omega_0 \sqrt{1-\gamma^2}. Let's call the signal of the damped oscillator X[A,\omega,\omega_0,\gamma,C_1,C_2], with \gamma=0 being an undamped oscillator. What I am saying is, if you look only in a window that is short compared to the damping time, you can choose A', C_1' and C_2' for an undamped oscillator such that X[A',\omega,\omega_0,0,C_1',C_2'] closely approximates X[A,\omega,\omega_0,\gamma,C_1,C_2] inside that window - the rms error is small compared to the signal.

This is useful because if you have a case where the damping is small (\gamma<<1) then you can have a window that is short with respect to the damping time, yet contains many cycles. For a single guitar string, the damping time is much longer than the vibration period and so you can model it as an undamped oscillator over many cycles (but not too many) and I would expect the mathematics is considerably easier.

Extending this to the two coupled damped oscillators, no driving, their response is characterized by X[A_1,\omega_1,\gamma1,C_{11},C_{12}, A_2,\omega_2,\gamma2,C_{21},C_{22}] where the driving frequencies have been eliminated and the undamped resonant frequencies of the two oscillators are \omega_1 and \omega_2. The case of two damped coupled oscillators with no other driving force is totally transient - there is no steady state. If the damping is small, then inside any window small with respect to the damping time of both oscillators, you can well approximate the situation by two coupled undamped oscillators. Here, I am saying the same thing - if f you look only in a window that is short compared to the damping time, you can choose A_1', C_{11}', C_{12}', A_2', C_{21}' and C_{22}' for a pair of coupled, undamped oscillators such that X[A_1',\omega_1,0,C_{11}',C_{12}', A_2',\omega_2,0,C_{21}',C_{22}'] closely approximates X[A_1,\omega_1,\gamma1,C_{11},C_{12}, A_2,\omega_2,\gamma2,C_{21},C_{22}] inside that window - the rms error is small compared to the signal. The beat frequency may or may not cover many cycles inside that window, but nevertheless, it will be a good approximation.

I see nothing in the link you provided that contradicts this. As far as references go, I don't have any, but I hopefully could find them. I say hopefully, because I am now beyond my notes, trying to figure things out with Mathematica and blurting out my present understanding of things, which may be in error. What I would find interesting is a reference that contradicts these conclusions. Even more interesting, a good argument against the conclusions. Your arguments have not changed my intuition, but have sharpened it considerably.

 

[sophiecentaur]

Well, first of all, I erroneously described the Q-factor of an undamped oscillator as Q=1/(\omega^2-\omega_0^2), so scratch that statement. An undamped oscillator has infinite Q at all frequencies, and an oscillator with very low damping has a large Q>>1. Its the Bode magnitude that is 1/|\omega^2-\omega_0^2| - the absolute value of the ratio of the input magnitude to the output magnitude assuming zero resonant contribution. Assuming small damping, that's equivalent to "steady state". The important point is that the response is not zero for all frequencies except resonant, it is non-zero and finite for all frequencies except resonant, where it goes to infinity.

I have to take issue with what you say is the Q of the oscillator. The Q has nothing to do with the frequency of excitation. The Q of an oscillator is 1/|\omega(half power)^2-\omega_0^2|, where omega(half power) is the frequency where the response would be half power. For an undamped oscillator this is of no use as a formula because Q is infinite. (Is there any doubt that the Q of an undamped oscillator is infinite and that {finite} Q is a function of the Oscillator and not of the excitation frequency?) I think you use this later, too, which brings the rest into doubt.

But you agree that there will be a transient response consisting of two frequencies, so if we are in time frame where the transient signal is appreciable, then we are not in steady state. The undamped oscillator does not necessarily approximate a steady state, it approximates the behavior in a window at any time, where the width of the window is much smaller than the damping time. It may contain both frequencies if the window is in the transient regime. If the phenomenon you are looking at can be profitably analyzed inside such a window, then consideration of undamped oscillators will give good results. For a single guitar string, the damping time is a matter of seconds and there is no steady state except zero. For the resonance phenomenon to occur, only a few, maybe tens of cycles are needed. So using an undamped oscillator as a model is profitable.

To get more mathematical, a damped, driven oscillator is characterized by the driving amplitude A, driving frequency \omega, resonant frequency of undamped oscillator \omega_0, damping constant \gamma, and two constants C_1 and C_2 which modify the phase and amplitude of the transient signal. For \gamma<1, the transient signal frequency is sinusoidal with a frequency of \omega_0 \sqrt{1-\gamma^2}. Let's call the signal of the damped oscillator X[A,\omega,\omega_0,\gamma,C_1,C_2], with \gamma=0 being an undamped oscillator. What I am saying is, if you look only in a window that is short compared to the damping time, you can choose A', C_1' and C_2' for an undamped oscillator such that X[A',\omega,\omega_0,0,C_1',C_2'] closely approximates X[A,\omega,\omega_0,\gamma,C_1,C_2] inside that window - the rms error is small compared to the signal.
etc

This is only true because the exciting waveform consists of the driving frequency ω, modulated by a step function (or whatever attack wavefrom you give it). which will excite the centre ω of the oscillator. If the excitation is turned on at a zero crossing, the power outside ω will be low. If it's turned on at a peak, it will be higher. But, yes, of course you will have a transient response and this will involve an exponential decay of the power at the centre frequency and an exponential growth at the exciting frequency. Now, if you agree about the 'exponentials' then you have to consider the time variation. The response is never quasi-static, any more than, as I said before, the volts on a charging capacitor are constant.
This function is unspecific and actually misses out the time factor, t. Surely that is a major part of it. You use the word "transient" which implies time variation very strongly. Over a short enough interval, a damped or growing oscillation can be treated as constant but how would that help if you are after the relative amplitudes of transient and long-term behaviour? The relative amplitudes of the two components depends entirely on how long since the drive was applied and it is unreasonable to assume that you can jump in half way through the process and expect to get an answer.
As far as I can see, your treatment would imply that the response would be independent of the separation of the two frequencies. How can that be? How would that describe why sympathetic strings work at near - harmonics and not just anywhere?

I am not sure what Mathematica will do for you (except as a help with hard integrals etc.) It will only give you an answer if you ask it the right question and I am not sure you are doing that. I think your initial problem with the definition of Q needs some attention. The fact that you cannot find a reference to your ideas could be giving us a clue here. There is plenty of stuff on coupled and driven oscillators but no sign of your particular approach. I appreciate you want to see this through but imho you are not going to get anywhere with it.

 

[Rap]

I have to take issue with what you say is the Q of the oscillator. The Q has nothing to do with the frequency of excitation. The Q of an oscillator is 1/|\omega(half power)^2-\omega_0^2|, where omega(half power) is the frequency where the response would be half power. For an undamped oscillator this is of no use as a formula because Q is infinite. (Is there any doubt that the Q of an undamped oscillator is infinite and that {finite} Q is a function of the Oscillator and not of the excitation frequency?) I think you use this later, too, which brings the rest into doubt.

I agree that it is not a function of frequency. My understanding of Q is obviously in a state of flux, but I don't worry about it, to me its a semantic problem because I am not (at this point) understanding the problem in terms of Q, but I should not be throwing it around in a conversation without understanding it. I hate getting bogged down in semantic arguments. If I look at wikipedia, I get various expressions for Q, depending on which page or paragraph I look at. I have your definition above, which must be wrong, because it is a function of frequency. If I multiply your expression by \omega_0^2 then it is dimensionless, and I have yet another definition. They are all slightly different for finite Q, all converge to infinity as the damping goes to zero. This kind of stuff drives me crazy. Q is some number that tells you how sharp the response curve is, higher Q means sharper. For an undamped oscillator it is infinitely sharp, Q=infinity. Can we leave it at that?

This is only true because the exciting waveform consists of the driving frequency ω, modulated by a step function (or whatever attack wavefrom you give it). which will excite the centre ω of the oscillator. If the excitation is turned on at a zero crossing, the power outside ω will be low. If it's turned on at a peak, it will be higher. But, yes, of course you will have a transient response and this will involve an exponential decay of the power at the centre frequency and an exponential growth at the exciting frequency. Now, if you agree about the 'exponentials' then you have to consider the time variation. The response is never quasi-static, any more than, as I said before, the volts on a charging capacitor are constant.
This function is unspecific and actually misses out the time factor, t. Surely that is a major part of it.

I agree on the exponentials. I did neglect to put in time t as a parameter of the X[...] functions above, but after inserting it, the statement still represents what I am saying. I wonder if this clears things up? In other words X[...,t] is the instantaneous position of the mass on a spring, or the instantaneous charge on the capacitor in a driven RLC circuit. By "steady state" I do not mean the voltage is constant, I mean that the function describing the voltage is not decaying, no e^{-\gamma \omega_0 t} term, where \gamma is the damping factor, zero for undamped. That means that even a function which has a beat frequency in it can be steady state.

You use the word "transient" which implies time variation very strongly. Over a short enough interval, a damped or growing oscillation can be treated as constant but how would that help if you are after the relative amplitudes of transient and long-term behaviour?

It would not, but in the case of the two guitar strings, we are not after the relative amplitudes of transient and long term behavior. Long term behavior is two strings at rest. We are interested in the resonance phenomenon. For this we need to look at a number of cycles. Not a period so short that the signal is constant, but so short that the oscillation approximates that of an undamped oscillator over a number of cycles. If damping is low, we can do this, we can examine the resonance phenomenon inside a time window which is large enough to contain many cycles, yet for which the difference between the damped and undamped oscillator is negligible. And for the two guitar strings, damping is low, the damping period covers many, many cycles of both strings.

The relative amplitudes of the two components depends entirely on how long since the drive was applied and it is unreasonable to assume that you can jump in half way through the process and expect to get an answer.

I don't mean that plucking an undamped string will give the same results as plucking a damped string. I mean that if I pluck a damped string, then between, say, 1 and 1.1 seconds, I can imagine an undamped oscillator that will practically match in detail the behavior of the damped oscillator over that time interval. This imaginary undamped oscillator, if I look outside the window, back to time zero, will have a different amplitude and phase from the damped oscillator, but inside that window, it is a good match. Since I am only interested in the resonance phenomenon, which can be well characterized in such a small window, I can analyze an undamped oscillator and be free of all the damping complications.

As far as I can see, your treatment would imply that the response would be independent of the separation of the two frequencies. How can that be? How would that describe why sympathetic strings work at near - harmonics and not just anywhere?

No, it wont. The behavior of two undamped coupled oscillators depends on the separation of the two frequencies. The entire problem is transient. For two damped coupled oscillators, the steady state is flat zero. If you want to examine the resonance phenomenon and the damping is low enough that you can profitably look inside a window that is small compared to the damping period, then you can profitably look at two undamped coupled oscillators. If the two frequencies are so close together that the beat period is comparable to the damping time, then you cannot model the beat effect over its entire period, but you can model the small beat effect that occurs inside that window. The bottom line is that for small damping, you can approximate two coupled damped oscillators as two undamped coupled oscillators multiplied by an exponential decay.. There will exist a window size that covers many cycles, yet for which the damping effect is essentially multiplication by a constant.

I am not sure what Mathematica will do for you (except as a help with hard integrals etc.) It will only give you an answer if you ask it the right question and I am not sure you are doing that.

The second sentence is very true. As for the first, Mathematica will solve systems of differential equations (e.g. two damped coupled oscillators) and you can examine the results to get a mathematical understanding, or plot the results for various parameters to get an intuitive feel.

 

[sophiecentaur]

Firstly, you really should sort out the basic definition of Q. There has never been any confusion in what I have been told. It is a dimensionless quantity and can either describe the number of cycles taken for the energy in an oscillator to halve or the fractional frequency difference for half power response. (Sorry - I didn't make all the changes I should have when I tried to correct y our definition).

The bottom line is that for small damping, you can approximate two coupled damped oscillators as two undamped coupled oscillators multiplied by an exponential decay..

but, at any given time, one is increasing and the other is decreasing; they are not both decaying exponentially.
If you are including a "damping factor" then that means energy loss and, for a single oscillator, that must mean finite Q.

 

[Rap]

but, at any given time, one is increasing and the other is decreasing; they are not both decaying exponentially.
If you are including a "damping factor" then that means energy loss and, for a single oscillator, that must mean finite Q.

For the two guitar strings, they are both decaying exponentially. Steady state is two non-vibrating guitar strings. Yes, you will have a finite Q but the question is - under what conditions will a large Q and and infinite Q be made to give practically the same results? Certainly not if you look at a guitar string over a period of a few seconds, but if you consider only time intervals on the order of a tenth of a second, it can.

I've seen pendulums that are 100 pound weights attached to a piano wire that is several stories high. Their oscillation period is tens of seconds, and they can go for weeks before they need to be re-energized. They demonstrate the rotation of the earth. You cannot say that it is hopeless to describe their behavior over a period of an hour without considering damping forces. If there were two pendulums coupled by a weak spring, you cannot say that its hopeless to analyze them without considering damping. You don't need to know the initial conditions of when they were energized a week ago, all you need to do is measure their simultaneous position and velocity now, and those will serve as your initial conditions and you can treat them as lossless pendulums and predict their behavior, for an hour, anyway.

 

[sophiecentaur]

For the two guitar strings, they are both decaying exponentially. Steady state is two non-vibrating guitar strings. Yes, you will have a finite Q but the question is - under what conditions will a large Q and and infinite Q be made to give practically the same results? Certainly not if you look at a guitar string over a period of a few seconds, but if you consider only time intervals on the order of a tenth of a second, it can.
et.

Hang on just one cotton picking second my friend. The string that is struck decays in amplitude but what about the sympathetic string? That only has one source of energy and that is from the struck string. It will gradually build up its energy (from zero) at the expense of the struck string. One decreases and one increases in energy. Given enough time and a high enough Q, the strings will cyclically exchange energy as coupled oscillators do. You still insist that they are both decaying at the same time?

Your long pendulum story is only a scaled up version of any two coupled oscillators. Of course a few oscillations will not be enough to describe what's going on and, yes, the amplitudes will not change very much. But, if one is increasing and the other is decreasing, can you ignore this?
You are right to say that you can describe the motion in a short window of time to a 'reasonable' accuracy (just as if you take a 1F capacitor and discharge it through a 10MΩ resistor, the Volts won't change much in a second- but that doesn't alter the fundamental exponential nature of the change). Your model misses out the basic mode of operation of the system. The simple model would never predict the long term alternation of energy from one to the other and back again. Isn't that a necessary requirement?

What was the frequency difference between these two super pendulums and what was their Q ( Possibly in the order of ten thousand). What was the period of the energy exchange?
But you are still not addressing my basic question about the driven oscillator which seem to be insisting will reach a steady state with a frequency component of ω₀ and you haven't commented on the definition of Q, either. This just has to be relevant.

 

[Rap]

The string that is struck decays in amplitude but what about the sympathetic string? That only has one source of energy and that is from the struck string. It will gradually build up its energy (from zero) at the expense of the struck string. One decreases and one increases in energy. Given enough time and a high enough Q, the strings will cyclically exchange energy as coupled oscillators do. You still insist that they are both decaying at the same time?

I guess I have to say it mathematically. For two coupled, damped harmonic oscillators, the differential equations are: x_1''+2\gamma x_1'+\omega_{01}^2=\epsilon(x_2-x_1) x_2''+2\gamma x_2'+\omega_{02}^2=\epsilon(x_1-x_2) This is a special case, the two oscillators have the same damping constants (\gamma) and coupling constants (\epsilon), but different resonant frequencies (\omega_{01} and \omega_{02}). Also, the coupling is through the zeroth derivative (x_1 and x_2) which corresponds to two masses on springs with a spring connecting the two masses. I think the coupling for guitar strings is through the first derivative, I have to think about that. Anyway, I expect the results will be qualitatively the same.

If we say \omega_{01}=1-\delta/2 and \omega_{02}=1+\delta/2 then we have an average resonant frequency of 1 and the difference between the resonant frequencies is \delta. That keeps things simple.

Crunching the thing through in Mathematica, the solution is: x_1=e^{-\gamma t} \left( A_{11} e^{i (\omega_1 t+\phi_{11})})+A_{12} e^{i (\omega_2 t+\phi_{12})}\right) x_2=e^{-\gamma t}\left(A_{21} e^{i (\omega_{21} t+\phi_{21})})+A_{22} e^{i (\omega_2 t+\phi_{22})} \right) Basically what we expect, they oscillate with two frequencies \omega_1 and \omega_2 which are essentially the two resonant frequencies, but thrown off by the coupling and the damping. The whole thing is multiplied by a damping factor e^{-\gamma t} which sends the whole thing to zero as time increases. The amplitudes (A_{ij}) and phases (\phi_{ij}) depend on initial conditions. The amplitudes, phases, and frequencies are functions of \gamma,\epsilon and \delta.

It can be seen that the two signals in each oscillator decay exponentially because of the e^{-\gamma t} term which multiplies the sum. That's what I meant when I said that they both decay exponentially. The beat phenomenon occurs because its the sum of two sinusoids at different frequencies. If, at time t=0, you excite the first oscillator while the second is motionless and at zero position, the increase in the second oscillators amplitudes will be part of the beat phenomenon. It can't be anything else, because the solution is just the sum of two decaying sinusoids.

Your long pendulum story is only a scaled up version of any two coupled oscillators. Of course a few oscillations will not be enough to describe what's going on and, yes, the amplitudes will not change very much. But, if one is increasing and the other is decreasing, can you ignore this?

I'm not ignoring this. What I am saying is that, for small damping (\gamma<<1), the solution is practically the same as when the damping is identically zero, as long as you don't look at it over time periods that are on the order of 1/\gamma - the damping period, which is long because \gamma is small. The beat phenomenon is still there for the undamped oscillators, because its still the sum of two sinusoidal signals at different frequencies. If the beat period is of the order of the damping period, then we cannot model multiple beat cycles using the undamped approximation, but we can certainly model the initial rise of the second string, which is part of the beat phenomenon.

But you are still not addressing my basic question about the driven oscillator which seem to be insisting will reach a steady state with a frequency component of ω₀ and you haven't commented on the definition of Q, either. This just has to be relevant.

With regard to the two damped, coupled oscillators (or guitar strings) there is no external driving force, so the steady state is two oscillators totally at rest. The problem is entirely one of transients. If you want to approximate things using two undamped, coupled oscillators, then you cannot talk about the steady state solution for the damped case, because that would require a window infinitely wide, which you cannot have if you want to use the undamped approximation. But that's ok, we are not interested in the steady state of two motionless strings, we are interested in the transient resonance phenomenon which we can analyze inside a window containing a bunch of cycles. IF the damping is low enough, we can find such a window that contains a bunch of cycles, but is narrow enough that we can ignore the damping.

And yes, I still have not addressed the precise Q issue, because it gives me a headache. We already understand it qualitatively. If you would like to examine how the Q factors change with various parameter changes, I will do it, but isn't Q only useful for steady state solutions? The steady state here is two unmoving guitar strings, and we are not interested in that.

 

[sophiecentaur]

Can we bring this to a close (soonish)?
There are two issues here - one about coupled oscillators and one about a driven oscillator.
That result from Mathematica looks pretty fair. With two pendulums you can use the same damping factors but for guitar strings you might need to use different values. (Shouldn't the exponential decay be outside the whole thing, though? Oh yes - you put in an extra bracket in there which confused me (I think you need to tidy up your suffixes too)
I am trying to make sense of those two expressions. You should be able to eliminate the phase of the struck oscillator and I should have thought that you could set the initial condition for one oscillator to zero amplitude at t=0. I suggest that there is a reduced version (preferably without complex notation) which would describe the motion in a more recognisable form.

The issue of the driven oscillator is another one which is the solution of one of those equations with the right hand side replaced with a sinusoidal force, rather than a force which is proportional to the difference in the displacements. (finite Q is essential here). The main reason I got involved with this thread was that you made a statement that the steady state situation involves two frequencies of oscillation (several pages ago). That is not as I remember from notes and not what I remember from measurements. Your definition of Q was not correct so your argument was not conclusive. The only frequency of oscillation under those conditions will end up as the excitation frequency. As you have the means for putting it into Mathematica, perhaps you could try it and prove it for yourself.

 

[Rap]

Well, if we agree on those two equations (without the extra bracket) then I disown any statement I may have made contrary to them, as well as any incorrect statements of Q. So my notes have definitely expanded as a result of this thread. Also, I agree that if one or both oscillators are driven, the steady state will consist only of the driving frequency.

I think the suffixes are correct, but maybe better choices could be made. I used the complex notation because its mathematically easier. For linear systems, you can just take the real part of the final answer and get (usually) a much messier equation.

You can get the specific case of a plucked oscillator by specifying initial conditions: x_1(0)=P, x_1'(0)=0, x_2(0)=0, x_2'(0)=0, where P indicates how much dispaced the plucked string was when it was let go. You can then solve for the four unknowns - the two amplitudes and two phases. I put that into mathematica and the results are not too messy and they all go to the expected limit as the damping or the coupling go to zero.

I think the answer given to the OP given previously is good - I would modify it to read:

The answer to the OP is that a guitar string can be quite accurately thought of as having a single lowest possible frequency, the fundamental frequency. The string also vibrates at integer multiples of the fundamental frequency at various, usually lower intensities than the fundamental frequency. These various frequencies that the string vibrates at are called harmonics, the first harmonic being the fundamental, etc. If any harmonic of a string is at or very near any harmonic of another string, then either string, when plucked, will "ring" the other - it will cause it to vibrate at that frequency they have in common (or at the two frequencies they have nearly in commn). For example the lowest guitar string is E1 at 82.407 cycles per second (cps). Its harmonics are 164.814, 247.221, 329.628 etc cps, each harmonic is weaker than the previous one. The fifth string is B3 at 246.942 cps, almost exactly three times the E1 frequency. So the third harmonic of the first string will "ring" or "drive" the fifth string at 247.221 hz also causing it to vibrate at its natural frequency of 246.942 hz. If you then damp the E1 string, it will then continue to vibrate at its natural frequency of 246.942 hz.

Unless you have an objection to that summary, I think that should do it.

 

[rbj]

the answer isn't bad, and i haven't read through the whole thread, but

The answer to the OP is that a guitar string can be quite accurately thought of as having a single lowest possible frequency, the fundamental frequency.

well, this also depends on initial conditions (determined at the time of the pluck).

by touching nodal points, the string can ring with a new fundamental that is at one of the harmonic frequencies. we do this often to tune (relatively) the guitar by ear. so it doesn't always have the lowest possible frequency as its fundamental. or, you can think of it as, when doing harmonic tuning (touching a nodal point when plucking) that only harmonics that are at multiples of 2 or at multiples of 3 or 4 are ringing. but that is equivalent to the fundamental taking on 2 times or 3 or 4 times the fundamental of the open string.

 

[Rap]

How about this:

A guitar string can be quite accurately thought of as vibrating at a number of distinct frequencies, each at possibly different amplitudes. The lowest frequency is called the fundamental frequenciy, and the higher frequencies are integer multiples of the fundamental frequency. These various frequencies that the string vibrates at are called harmonics, the first harmonic being the fundamental, the second harmonic being twice that, etc. If you pluck a string, all of its harmonics are excited, the lower harmonics being the strongest. If any harmonic of a string is at or very near any harmonic of another string, then either string, when plucked, will "ring" the other - it will cause it to vibrate at that frequency they have in common (or at the two frequencies they have nearly in common). For example, for a perfectly tuned guitar, the lowest guitar string is E1 at 82.407 cycles per second (cps). Its harmonics are 164.814, 247.221, 329.628 etc cps. The fifth string is B3 at 246.942 cps, almost exactly three times the E1 frequency. So the third harmonic of the first string will "ring" or "drive" the fifth string at 247.221 cps also causing it to vibrate at its natural frequency of 246.942 cps. If you then damp the E1 string, it will then continue to vibrate at its natural frequency of 246.942 cps.

 

[sophiecentaur]

Well, if we agree on those two equations (without the extra bracket) then I disown any statement I may have made contrary to them, as well as any incorrect statements of Q. So my notes have definitely expanded as a result of this thread. Also, I agree that if one or both oscillators are driven, the steady state will consist only of the driving frequency.

I think the suffixes are correct, but maybe better choices could be made. I used the complex notation because its mathematically easier. For linear systems, you can just take the real part of the final answer and get (usually) a much messier equation.
etc.

That's all quite reasonable (those suffixes look OK now - the version I saw first had a couple of missing digits). Respect to you for actually working it all out. That's worth an awful lot of arm waving - mea culpa.

What I'd like to know is what happens when the two frequencies are wide apart. This effect is most marked when the two oscillators have a small fractional frequency separation. The beat / amount of energy transfer will depend upon the coupling and the frequency separation. It is intuitive that oscillators with wide separation won't do this but it may just be because the beating effect is just not perceived as well when the two frequencies are close.

Years ago, I set up two series LC circuits, tuned to a few tens of kHz, iirc. I then coupled them with a small C and drove one with a string of well separated pulses (say at 1Hz), via another small C. Oscilloscope probes on the two circuits gave a very good picture of ten or so of the beat cycles, with the maximum amplitudes of the two modulated waveforms interleaving nicely and the overall amplitude decaying in a very convincing way. It was just like the coupled pendulum practical I had done many years previous to that at Uni - but at a higher rate and a lower Q. I seem to remember there was an optimum separation of frequencies, for the best picture.

 

[Rap]

That's all quite reasonable (those suffixes look OK now - the version I saw first had a couple of missing digits). Respect to you for actually working it all out. That's worth an awful lot of arm waving - mea culpa.

What I'd like to know is what happens when the two frequencies are wide apart. This effect is most marked when the two oscillators have a small fractional frequency separation. The beat / amount of energy transfer will depend upon the coupling and the frequency separation. It is intuitive that oscillators with wide separation won't do this but it may just be because the beating effect is just not perceived as well when the two frequencies are close.

Years ago, I set up two series LC circuits, tuned to a few tens of kHz, iirc. I then coupled them with a small C and drove one with a string of well separated pulses (say at 1Hz), via another small C. Oscilloscope probes on the two circuits gave a very good picture of ten or so of the beat cycles, with the maximum amplitudes of the two modulated waveforms interleaving nicely and the overall amplitude decaying in a very convincing way. It was just like the coupled pendulum practical I had done many years previous to that at Uni - but at a higher rate and a lower Q. I seem to remember there was an optimum separation of frequencies, for the best picture.

I could give you the expressions for the amplitudes and phases as functions of {\gamma,\epsilon,\delta} for the initial conditions I mentioned above (one oscillator displaced, but at zero velocity, the other at zero position and velocity, at time zero). You could plot the results. You could play around with the parameters and see what happens. Would that work? One problem is that some expressions which are real for small values of the parameters can turn complex for large values. If you stay with real parameters, then you have to make decisions at those points, but if your plotting package can take the real part of a complex expression, that won't be necessary.

 

[sophiecentaur]

Excel will do most things if I'm determined enough! Cheers.

 

[DragonPetter]

I had to jump out of this thread and pull my parachute string a while ago. There is too much of person A talks apples and person B talks oranges and person C tells person A he is wrong because his apple is not an orange or that he thought his apple looked like an orange but didn't see the difference, and it started to feel like I was eating an apple and it tasted like an orange. And in the middle of all of this, some people were talking mashed potatoes and claimed it tasted like a fruit salad.

 

[sophiecentaur]

I think at least two of us have sorted our ideas out now. I am pretty omnivorous when it comes to eating fruit.

 

[rbj]

How about this:
A guitar string can be quite accurately thought of as vibrating at a number of distinct frequencies, each at possibly different amplitudes. The lowest frequency is called the fundamental frequenciy, and the higher frequencies are integer multiples of the fundamental frequency. These various frequencies that the string vibrates at are called harmonics, the first harmonic being the fundamental, the second harmonic being twice that, etc. If you pluck a string, all of its harmonics are excited, the lower harmonics being the strongest.

that is not necessarily the case. if you pluck your guitar string in the very center of it, only the odd-numbered harmonics are excited. if you pluck the string while constraining the nodal point in the very center of the string, then only even-numbered harmonics will be excited. if you pluck the string while constraining the nodal point 1/3 of the string length from either side, only harmonics that are integer multiples of 3 will be excited.

do you understand that, while the wave equation of the string supports all of these harmonics possibly existing (with non-zero amplitude), there is more to the complete solution of a differential equation than just solving the diff. eq.: there are also the initial conditions that determine the complete result.

 

[Rap]

that is not necessarily the case. if you pluck your guitar string in the very center of it, only the odd-numbered harmonics are excited. if you pluck the string while constraining the nodal point in the very center of the string, then only even-numbered harmonics will be excited. if you pluck the string while constraining the nodal point 1/3 of the string length from either side, only harmonics that are integer multiples of 3 will be excited.

Well, I wouldn't call raising one point while constraining another "plucking". But even if you define plucking as simply raising one point and letting go, you are still correct - when the distance to the pluck point divided by the length of the string is a rational number, not all harmonics will be excited. Let's amend the statement to read "If you pluck a string, then various harmonics are excited ..."

Of course, the probability of plucking exactly a rational number is zero...

To SophieCentaur - I changed the problem to one where the first oscillator has a fixed velocity at time zero, rather than a fixed offset. The math seems simpler. The easiest way to express the solution is x_1=\sum_{n=1}^2 \sum_{m=1}^2 A_{mn}e^{i \omega_{mn} t}x_2=\sum_{n=1}^2 \sum_{m=1}^2 B_{mn}e^{i \omega_{mn} t} where we are using four terms instead of two terms with phases. The natural frequency of the first oscillator is 1-\delta/2, the natural frequency of the second is 1+\delta/2, \gamma is the damping constant, and \epsilon is the coupling constant (same for both oscillators). The four frequencies are: \omega_{mn}=\frac{(-1)^m}{2}\sqrt{1-\gamma^2+(\delta/2)^2+\epsilon+(-1)^n\sqrt{\delta^2+\epsilon^2}} You can see that when the damping and coupling go to zero, you get back the two resonant frequencies, and their negatives. The amplitudes are: A_{mn}=\frac{1}{2\omega_{mn}}\sqrt{i-\frac{(-1)^n i\delta}{\sqrt{\delta^2+\epsilon^2}}} and B_{mn}=\frac{-i\omega_{mn}(-1)^n}{\sqrt{\delta^2+\epsilon^2}}
Note I might have made a mistake, so I will keep going over this until I am sure its right, but this is a start. To get the real signals, just take the real part of the x_1 and x_2. I use \delta=0.1, \gamma=0.05, \epsilon=0.02 to get a nice plot from 0 to 50 seconds. (I use the direct solution without simplification, and the errors I might have made are in the simplification).

 

[Rap]

I kind of figured those numbers were wrong. The four frequencies are: \omega_{mn}=-(-1)^n\sqrt{1-\gamma^2+(\delta/2)^2+\epsilon+(-1)^m\sqrt{\delta^2+\epsilon^2}} and the amplitudes are: A_{mn}=\frac{-i}{4\,\omega_{mn}}\left(1-\frac{(-1)^m\delta}{\sqrt{\delta^2+\epsilon^2}}<br /> \right) and B_{mn}=\frac{-i}{4\,\omega_{mn}}\left(\frac{(-1)^m\epsilon}{\sqrt{\delta^2+\epsilon^2}}\right)

 

[PhilDSP]

Hi all,

Just looking at this thread for the first (and probably last) time. But I thought it would be worth mentioning that body resonances are very important for acoustic and hollow body electric guitars. And of course for other string instruments. If I remember correctly, all are pretty much configured so that the primary body resonance occurs near the second lowest open string frequency or one octave away. In Violins, especially, a mark of quality is how close the body resonance matches that characteristic. If the resonance becomes overly pronounced it is called a wolf tone and can be difficult for a player to control.