Sound/accoustics - guitar string question

[sophiecentaur]

I don't exactly understand what you mean, and I still think you're somewhat confused.

For example you wrote:
Overtone gets further away from the original question of how one string can cause another to vibrate, while a harmonic specifically addresses how 2 strings can share a resonance - they share at least one frequency mode of vibration.


This is not true. If you have string #1 made of a peculiar material, so that its fundamental is 100 Hz and its first overtones 230 Hz, 370 Hz... and then you have string #2 with its fundamental at 370 Hz, the first string will cause the second to vibrate, through an overtone which is not a harmonic.

There is no reason why the 370Hz resonance of either string should, in fact, be a harmonic of either 'fundamental'. All that is necessary is for the two systems to share (fairly closely) a natural mode with frequency of 370Hz for one to resonate with the other.

Ref to my earlier mention of circular drums, the first overtone of one membrane and the second overtone of the other membrane would definitely not have a common sub-harmonic in sight.

 

[someGorilla]

I'm not trying to argue, its simply a mathematical definition that no one can argue with:
http://en.wikipedia.org/wiki/Harmonic

I mean the 370Hz is not a harmonic of the first fundamental.
If you mean every note is a harmonic of itself, oh well guess what, I agree.

 

[DragonPetter]

I mean the 370Hz is not a harmonic of the first fundamental.
If you mean every note is a harmonic of itself, oh well guess what, I agree.

Yes, it does seem silly, but my point the entire time has been that harmonic is more accurate than overtone to describe why one string will vibrate from another. Why? Because a non-fundamental harmonic OR an overtone (still a harmonic technically) can BOTH cause a second string to vibrate. Perhaps overtones are the main contribution in guitars, I don't know, but saying harmonic is not as accurate was my contention.

 

[Rap]

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

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

 

[someGorilla]

Yes, apologies to the thread poster, we got carried away... Rap summed it up pretty well.

 

[DragonPetter]

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

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

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

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

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

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

 

[someGorilla]

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

 

[DragonPetter]

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

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

 

[Rap]

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

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

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

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

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

 

[someGorilla]

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

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

 

[DragonPetter]

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

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

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

 

[someGorilla]

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

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

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

 

[DragonPetter]

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

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

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

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

 

[DragonPetter]

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

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

 

[someGorilla]

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

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

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

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

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

 

[DragonPetter]

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

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

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

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

 

[someGorilla]

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

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

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

No, it's computer-generated.

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

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

 

[DragonPetter]

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

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

 

[DragonPetter]

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

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

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

 

[someGorilla]

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

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

 

[DragonPetter]

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

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

 

[someGorilla]

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

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

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

 

[DragonPetter]

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

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

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

 

[Rap]

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

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

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

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

 

[sophiecentaur]

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

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

 

[Rap]

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

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

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

 

[AlephZero]

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

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

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

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

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

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

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

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

 

[DragonPetter]

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

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

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

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

 

[bahamagreen]

Long time guitar player to the rescue...

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

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

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

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

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

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

 

[sophiecentaur]

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

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

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