Sound/accoustics - guitar string question

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  • #76
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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.
 
  • #77
sophiecentaur
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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
 
  • #78
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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 [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.
 
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  • #79
sophiecentaur
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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.
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.


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.
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.
 
  • #80
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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.
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?

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 [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.

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.
 
  • #81
sophiecentaur
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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.
 
  • #82
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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.
 
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  • #83
sophiecentaur
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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.
 
  • #84
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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: [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.

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 ([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.

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.
 
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  • #85
sophiecentaur
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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.
 
  • #86
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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: [itex]x_1(0)=P[/itex], [itex]x_1'(0)=0[/itex], [itex]x_2(0)=0[/itex], [itex]x_2'(0)=0[/itex], 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.
 
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  • #87
rbj
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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.
 
  • #88
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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.
 
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  • #89
sophiecentaur
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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.
 
  • #90
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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 [itex]{\gamma,\epsilon,\delta}[/itex] 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.
 
  • #91
sophiecentaur
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Excel will do most things if I'm determined enough! Cheers.
 
  • #92
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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.
 
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  • #93
sophiecentaur
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I think at least two of us have sorted our ideas out now. I am pretty omnivorous when it comes to eating fruit.
 
  • #94
rbj
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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.
 
  • #95
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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. Lets 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 [tex]x_1=\sum_{n=1}^2 \sum_{m=1}^2 A_{mn}e^{i \omega_{mn} t}[/tex][tex]x_2=\sum_{n=1}^2 \sum_{m=1}^2 B_{mn}e^{i \omega_{mn} t}[/tex] where we are using four terms instead of two terms with phases. The natural frequency of the first oscillator is [itex]1-\delta/2[/itex], the natural frequency of the second is [itex]1+\delta/2[/itex], [itex]\gamma[/itex] is the damping constant, and [itex]\epsilon[/itex] is the coupling constant (same for both oscillators). The four frequencies are: [tex]\omega_{mn}=\frac{(-1)^m}{2}\sqrt{1-\gamma^2+(\delta/2)^2+\epsilon+(-1)^n\sqrt{\delta^2+\epsilon^2}}[/tex] 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: [tex]A_{mn}=\frac{1}{2\omega_{mn}}\sqrt{i-\frac{(-1)^n i\delta}{\sqrt{\delta^2+\epsilon^2}}}[/tex] and [tex]B_{mn}=\frac{-i\omega_{mn}(-1)^n}{\sqrt{\delta^2+\epsilon^2}}[/tex]
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 [itex]x_1[/itex] and [itex]x_2[/itex]. I use [itex]\delta=0.1, \gamma=0.05, \epsilon=0.02[/itex] 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).
 
  • #96
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I kind of figured those numbers were wrong. The four frequencies are: [tex]\omega_{mn}=-(-1)^n\sqrt{1-\gamma^2+(\delta/2)^2+\epsilon+(-1)^m\sqrt{\delta^2+\epsilon^2}}[/tex] and the amplitudes are: [tex]A_{mn}=\frac{-i}{4\,\omega_{mn}}\left(1-\frac{(-1)^m\delta}{\sqrt{\delta^2+\epsilon^2}}
\right)[/tex] and [tex]B_{mn}=\frac{-i}{4\,\omega_{mn}}\left(\frac{(-1)^m\epsilon}{\sqrt{\delta^2+\epsilon^2}}\right)[/tex]
 
  • #97
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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.
 

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