## Sound/accoustics - guitar string question

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.

 Quote by Rap 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.
 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.

Recognitions:
Gold Member
 Quote by 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. 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.

 Quote by sophiecentaur 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.
 Recognitions: Gold Member Science Advisor Excel will do most things if I'm determined enough! Cheers.
 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.
 Recognitions: Gold Member Science Advisor I think at least two of us have sorted our ideas out now. I am pretty omnivorous when it comes to eating fruit.

 Quote by 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.
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.

 Quote by rbj 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 $$x_1=\sum_{n=1}^2 \sum_{m=1}^2 A_{mn}e^{i \omega_{mn} t}$$$$x_2=\sum_{n=1}^2 \sum_{m=1}^2 B_{mn}e^{i \omega_{mn} t}$$ 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).
 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: $$A_{mn}=\frac{-i}{4\,\omega_{mn}}\left(1-\frac{(-1)^m\delta}{\sqrt{\delta^2+\epsilon^2}} \right)$$ and $$B_{mn}=\frac{-i}{4\,\omega_{mn}}\left(\frac{(-1)^m\epsilon}{\sqrt{\delta^2+\epsilon^2}}\right)$$
 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.