Sound/accoustics - guitar string question

sophiecentaur

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 [tex]m \ddot{x}=-k x+F e^{i \omega t}[/tex] where [itex]m[/itex] is the mass, [itex]k[/itex] is the spring constant, [itex]F[/itex] is the amplitude of the driving force, and [itex]\omega[/itex] is the frequency of the driving force. The natural frequency is [itex]\omega_0=\sqrt{k/m}[/itex]. The solution is [tex]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}[/tex] where [itex]A[/itex] is the acceleration due to the external force (=[itex]F/m[/itex]) and [itex]C_i[/itex] are complex constants which depend on initial conditions. With the right initial conditions, you can have the [itex]C_i[/itex] both be zero, but if the initial conditions are not right, then I think you have to add a force at frequency [itex]\omega_0[/itex] to compensate to make the oscillator oscillate at a pure [itex]\omega[/itex] frequency. The Q is [itex]1/(\omega^2-\omega_0^2)[/itex] which is infinite at [itex]\omega=\omega_0[/itex], 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 realise 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

Well, first of all, I erroneously described the Q-factor of an undamped oscillator as [itex]Q=1/(\omega^2-\omega_0^2)[/itex], 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 [itex]1/|\omega^2-\omega_0^2|[/itex] - 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 [itex]1/(\omega^2-\omega_0^2)[/itex] 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

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

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 [itex]A[/itex], driving frequency [itex]\omega[/itex], resonant frequency of undamped oscillator [itex]\omega_0[/itex], damping constant [itex]\gamma[/itex], and two constants [itex]C_1[/itex] and [itex]C_2[/itex] which modify the phase and amplitude of the transient signal. For [itex]\gamma<1[/itex], the transient signal frequency is sinusoidal with a frequency of [itex]\omega_0 \sqrt{1-\gamma^2}[/itex]. Lets call the signal of the damped oscillator [itex]X[A,\omega,\omega_0,\gamma,C_1,C_2][/itex], with [itex]\gamma=0[/itex] 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 [itex]A', C_1'[/itex] and [itex]C_2'[/itex] for an undamped oscillator such that [itex]X[A',\omega,\omega_0,0,C_1',C_2'][/itex] closely approximates [itex]X[A,\omega,\omega_0,\gamma,C_1,C_2][/itex] 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 ([itex]\gamma<<1[/itex]) 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 [itex]X[A_1,\omega_1,\gamma1,C_{11},C_{12}, A_2,\omega_2,\gamma2,C_{21},C_{22}][/itex] where the driving frequencies have been eliminated and the undamped resonant frequencies of the two oscillators are [itex]\omega_1[/itex] and [itex]\omega_2[/itex]. 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 [itex]A_1', C_{11}', C_{12}', A_2', C_{21}'[/itex] and [itex]C_{22}'[/itex] for a pair of coupled, undamped oscillators such that [itex]X[A_1',\omega_1,0,C_{11}',C_{12}', A_2',\omega_2,0,C_{21}',C_{22}'][/itex] closely approximates [itex]X[A_1,\omega_1,\gamma1,C_{11},C_{12}, A_2,\omega_2,\gamma2,C_{21},C_{22}][/itex] 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

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 [itex]1/|\omega(half power)^2-\omega_0^2|[/itex], 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.


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 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 [itex]\omega_0^2[/itex] 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?

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 [itex]e^{-\gamma \omega_0 t}[/itex] term, where [itex]\gamma[/itex] is the damping factor, zero for undamped. That means that even a function which has a beat frequency in it can be steady state.

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.

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.

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.

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).
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

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

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

I guess I have to say it mathematically. For two coupled, damped harmonic oscillators, the differential equations are: [tex]x_1''+2\gamma x_1'+\omega_{01}^2=\epsilon(x_2-x_1)[/tex] [tex]x_2''+2\gamma x_2'+\omega_{02}^2=\epsilon(x_1-x_2)[/tex] This is a special case, the two oscillators have the same damping constants ([itex]\gamma[/itex]) and coupling constants ([itex]\epsilon[/itex]), but different resonant frequencies ([itex]\omega_{01}[/itex] and [itex]\omega_{02}[/itex]). Also, the coupling is through the zeroth derivative ([itex]x_1[/itex] and [itex]x_2[/itex]) 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 [itex]\omega_{01}=1-\delta/2[/itex] and [itex]\omega_{02}=1+\delta/2[/itex] then we have an average resonant frequency of 1 and the difference between the resonant frequencies is [itex]\delta[/itex]. That keeps things simple.

Crunching the thing through in Mathematica, the solution is: [tex]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)[/tex] [tex]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) [/tex] Basically what we expect, they oscillate with two frequencies [itex]\omega_1[/itex] and [itex]\omega_2[/itex] 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 [itex]e^{-\gamma t}[/itex] which sends the whole thing to zero as time increases. The amplitudes ([itex]A_{ij}[/itex]) and phases ([itex]\phi_{ij}[/itex]) depend on initial conditions. The amplitudes, phases, and frequencies are functions of [itex]\gamma[/itex],[itex]\epsilon[/itex] and [itex]\delta[/itex].

It can be seen that the two signals in each oscillator decay exponentially because of the [itex]e^{-\gamma t}[/itex] 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.

I'm not ignoring this. What I am saying is that, for small damping ([itex]\gamma<<1[/itex]), 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 [itex]1/\gamma[/itex] - the damping period, which is long because [itex]\gamma[/itex] 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.

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.