What happens when you pluck a guitar string?

[sophiecentaur]

So one would only expect resonances corresponding to overtones from a plucked string?
(Which was my original thought about resonant systems in general).

 

[thegreenlaser]

I'm actually working on a project related to this right now, so I'm hoping I can help here.

I read the link too, but I still have questions. What causes the harmonics? Does the string actually divide itself into nodes? If the harmonics are sharp, does that mean that the string tries to divide itself for example into two nodes, but since it doesn't bend perfectly at the middle section due to stiffness, the nodes are slightly shorter than theoretical and therefore sharp. Is this a correct understanding?

No, it doesn't actually divide itself into nodes like that. Ideally, it will form into a sum of those different waves. You won't actually observe most, if any, of the nodes present, though in a way they're there, because they're contributing to the sum. If you haven't taken any Fourier analysis, at least read a basic intro to Fourier series, because that's essentially what this is. You can take nicely spaced frequencies like the ones that show up on a guitar string, multiply them by various amounts and add them together and end up with a really messy-looking signal.

Does the observation that a plucked string has sharp triangular kinks mean that it also creates pressure waves that are sharp and contain harmonics? Why does the string's shape matter, as long as it is moving back and forth at a consistent speed? If a sharp kink is consistently moving at me and then away, shouldn't it also create a consistent pressure increase and decrease? I mean that for example a loudspeaker can have triangular, conical or all sorts of shapes, what matters is how it moves back and forth and what pressure waves it creates.

The initial shape of the string does indeed affect the frequency output, but you can't just find the Fourier series of the initial shape to find the Fourier series of the output. i.e. A triangle-shaped string does not guarantee a triangle-shaped output. For example, if you start an ideal string as a triangular wave and let it go, at any given time, the string will be in some sort of triangular wave shape. You can watch the triangular wave move back and forth. However, if you sample the displacement at one point (say, a third of the way along the string), you'll actually find that the output is more like a rectangular wave, and the proportions of that rectangular wave will actually change depending on where you sample along the string. Closer to the edges, there's more emphasis on high frequencies, which is why the bridge pickup on an electric guitar sounds brighter than the neck pickup.

There is no moving wave. How can the wave possibly move when there are always two nodes either end?

When you hold a string ready to be plucked, you have a piecewise continuous function (displacement vs distance from bridge). That function can be expressed using a Fourier series (imagining it is periodic beyond either node). Each frequency has an amplitude.

When you let go, the amplitudes will evolve. Within milliseconds (depending on the length of the string and the speed of sound within the material) most of the frequencies will decay, leaving the harmonics. They decay far slower, creating the sound.

Try picking up a guitar and plucking it at an extreme point. You can hear a "noise" before you hear the note, and the note is tinny. This is because the amplitudes of the non-resonant sinusoids making up the initial displacement are high, so a lot of energy is dissipated instantly. And then, a lot more energy is put into the 5th,6th,... harmonics which have a higher frequency, making it sound tinny.

If you pluck the string precisely in the centre, you don't hear that noise, and you hear a much purer sound. That's because more of the energy in the pluck is put into the fundamental frequency and not "wasted" standing waves (not harmonics) nor higher harmonics (in fact the 2nd, 4th, etc. harmonics will contain no energy).

I'm not so sure that this is completely accurate. In any linear string model, the only possible solutions will be linear combinations of frequencies, all of which are multiples of the fundamental frequency. I think what you're describing actually comes from frequency-dependent loss: A guitar string will dissipate high frequencies faster than low frequencies. When you first pluck a string, you're getting a much more even spread of frequencies, but they're still all multiples of the fundamental frequency of the string. Very quickly, though, the higher frequencies die off and the ones closer to the fundamental start to dominate, making the tone sound more pure. The reason that plucking in the middle rather than on the edges sounds 'purer' is similar: there's more emphasis on the frequencies close to the fundamental frequency. Again, you don't actually need to have non-fundamental frequencies in order to explain this behavior. The more emphasis you have on high frequencies far away from the fundamental, the more tinny and messy your sound is going to be. In a real string, there will obviously be some non-linearities, but the frequency-dependent loss is mostly what contributes to that initial distorted 'twang.' Non-linearities tend to mostly come into play when you pluck a string really hard or something.

 

[mikeph]

A Fourier analysis of the Shape of the string is not a frequency analysis nor a description of the initial waves on the string. The string is static at the instant of release. You are assuming that you have injected a particular set of waves onto the string (Static and Dynamic conditions), which is an entirely different situation. It is not valid to proceed any further with that argument.

I really don't understand why not.

Any wave at maximum displacement has zero velocity at the instant of release. The shape of the string when I pluck it is at maximum displacement, so the fact that it's static is not a problem.

 

[chingel]

No, it doesn't actually divide itself into nodes like that. Ideally, it will form into a sum of those different waves. You won't actually observe most, if any, of the nodes present, though in a way they're there, because they're contributing to the sum. If you haven't taken any Fourier analysis, at least read a basic intro to Fourier series, because that's essentially what this is. You can take nicely spaced frequencies like the ones that show up on a guitar string, multiply them by various amounts and add them together and end up with a really messy-looking signal.

The initial shape of the string does indeed affect the frequency output, but you can't just find the Fourier series of the initial shape to find the Fourier series of the output. i.e. A triangle-shaped string does not guarantee a triangle-shaped output. For example, if you start an ideal string as a triangular wave and let it go, at any given time, the string will be in some sort of triangular wave shape. You can watch the triangular wave move back and forth. However, if you sample the displacement at one point (say, a third of the way along the string), you'll actually find that the output is more like a rectangular wave, and the proportions of that rectangular wave will actually change depending on where you sample along the string. Closer to the edges, there's more emphasis on high frequencies, which is why the bridge pickup on an electric guitar sounds brighter than the neck pickup.

Do I understand it correctly, if I say that the string does divide itself into two parts, three parts, four parts etc, and these parts vibrate and the sum of these movements of the string (since they are vibrating at the same time and in some places add, in some places interfere) creates the overall triangular kinky wave seen on the string, that moves back and forth on the string and was discussed in the pdf linked beforehand?

Does the shape of displacement on a particular point on the string correspond to the shape of sound waves created?

 

[sophiecentaur]

I really don't understand why not.

Any wave at maximum displacement has zero velocity at the instant of release. The shape of the string when I pluck it is at maximum displacement, so the fact that it's static is not a problem.

I see what you are saying but why should one assume that the whole length of the string would be stationary at the same time? For instance, the different overtones are not at rest at the same time or place - the phases of any two could (would?) not be the same - hence one would be moving whilst the other is stationary. The string being stationary would require all the overtones to be to be at a suitable relative phase to each other to achieve a net KE of zero. I find that hard to believe - despite the justification given earlier. I appreciate that what I say would apply to the majority of cases but that the condition of zero KE need only be met for one particular situation.

 

[olivermsun]

...why should one assume that the whole length of the string would be stationary at the same time?...The string being stationary would require all the overtones to be to be at a suitable relative phase to each other to achieve a net KE of zero.

At the moment of initial release, the string is not moving and hence KE = 0. All the energy in the system resides in PE (the stretch of the string). Immediately after t = 0, the displacement at every point on the string decreases toward equilibrium, so that PE decreases and KE increases. The phases of all Fourier modes are necessarily such that the initial condition is fulfilled.

 

[sophiecentaur]

That makes sense. You actually impress those initial conditions so that defines the relative levels and phases of the normal modes. Fair enough.

 

[thegreenlaser]

Do I understand it correctly, if I say that the string does divide itself into two parts, three parts, four parts etc, and these parts vibrate and the sum of these movements of the string (since they are vibrating at the same time and in some places add, in some places interfere) creates the overall triangular kinky wave seen on the string, that moves back and forth on the string and was discussed in the pdf linked beforehand?

You are correct, however that's not really a good way of looking at it. It's true that the spatial 'modes' (i.e. the dividing the string into even parts and making sinusoids) affect the output, but it's not nearly as direct as it would seem. Like you said, thanks to Fourier analysis, we can indeed take the shape of the string at any time and turn it into a sum of sinusoidal terms based on some frequency omega, which is directly related to dividing the length of the string into even parts. So if f(x,t) is the displacement of the string at position x at time t, then we can say, for any t:
$$f(x,t)=\sum_{k=0}^{\infty} A_k(t) \cdot \sin (k\omega_0 x)$$
Where A_k are some functions of time, which we can find with Fourier analysis. In fact, the solution to the simple 1D wave equation comes exactly in a form like this. However, notice that the sin term is a function of x, and not t. This means that any time-variance has to come from the A_k(t) term. It is these terms that will determine the frequency output at any point x, not the sine terms, yet it's the sine terms that generate the shape profile. From what I've said here, there's not even a guarantee that the A_k(t) oscillate at predictable frequencies, yet the sine terms have predictable frequencies. It so happens that they do in most cases, but just looking at the shape of the string doesn't guarantee that. It's a much more finicky relationship. So yes, you can always break the string shape down into a sum of sinusoid-shaped strings, but this isn't as closely related to the frequency output as it might seem. What determines the frequency output is the way in which these 'sinusoidal strings' oscillate back and forth, which, while related, is a different story.

In response to your question about the displacement of the string at one point being the output: that is actually a fairly reasonable assumption, especially in the case of an electric guitar. In the case of acoustic instruments like an acoustic guitar or piano, such an assumption can't be made as easily, but even then, just summing the displacements of a couple points on the string can be a pretty good approximation. In that case, actually, a Fourier analysis of the shape can be useful because you can look at the sinusoids in the spatial domain to see where each frequency component is strongest (peaks of the spatial sinusoid) and weakest (nodes of the spatial sinusoid). Analysis like that explains why if you listen at the edge of a string, it sounds more 'tinny' than if you listen at the middle. However, this is only helping you find information about how those frequencies change as you move along the string, and not about the frequencies themselves.

At the moment of initial release, the string is not moving and hence KE = 0. All the energy in the system resides in PE (the stretch of the string). Immediately after t = 0, the displacement at every point on the string decreases toward equilibrium, so that PE decreases and KE increases. The phases of all Fourier modes are necessarily such that the initial condition is fulfilled.

(I'm kind of responding to the conversation you've been having with sophiecentaur rather than this particular post) The statement that a Fourier analysis of the string shape does not correspond to an analysis of the frequency output is true, as I've explained above. When you turn the shape at time t into a Fourier series, you're finding one particular set of A_k values. You can make arguments based on energy analysis and analysis of how the string should respond to draw conclusions about the time-dependence of the A_k(t), however, just from turning the shape profile at anyone time into a Fourier series, you cannot glean any information about the actual frequency output. It's true that the initial shape/velocity will uniquely determine the output, but that's less to do with the spatial frequency characteristic of the shape and more to do with the fact that those are boundary conditions for the PDE model being used. You must consider other factors in order to get information about the frequency output, as it's not a direct relationship. Particularly if you add non-linearities to the model, Fourier analysis of the shape can become almost meaningless. In that case, you can start with a triangular shape, which can easily be expressed as a sum of sine terms, and yet you might find that the frequency content of the output shifts over time, and the string may go in and out of tune. There would be no reason to believe that just from the shape profile, and yet if you pluck a guitar string really hard, you can hear this frequency distortion.

 

[olivermsun]

...if f(x,t) is the displacement of the string at position x at time t, then we can say, for any t:
$$f(x,t)=\sum_{k=0}^{\infty} A_k(t) \cdot \sin (k\omega_0 x)$$
Where A_k are some functions of time, which we can find with Fourier analysis. In fact, the solution to the simple 1D wave equation comes exactly in a form like this. However, notice that the sin term is a function of x, and not t. This means that any time-variance has to come from the A_k(t) term. It is these terms that will determine the frequency output at any point x, not the sine terms, yet it's the sine terms that generate the shape profile.

However, the 1-d wave equation has traveling wave solutions along the characteristics x+ct, x-ct, (c.f. d'Alembert's formula) so that the time dependence is related to the space dependence through the wave speed c. Put another way, the signal detected at a fixed point over time should be exactly the 'incoming' spatial wave shape(s). A similar principle applies for 'analogue' recording media such as LPs and magnetic tapes.

In response to your question about the displacement of the string at one point being the output: that is actually a fairly reasonable assumption, especially in the case of an electric guitar. In the case of acoustic instruments like an acoustic guitar or piano, such an assumption can't be made as easily, but even then, just summing the displacements of a couple points on the string can be a pretty good approximation.

For an acoustic instrument, I would think that most of the 'output' is actually whatever goes through the bridge to the soundboard(s) of the instrument.

You must consider other factors in order to get information about the frequency output, as it's not a direct relationship. Particularly if you add non-linearities to the model, Fourier analysis of the shape can become almost meaningless.

Most of the preceding discussion has been assuming a linear model of the string, but I'd argue that most real-world stringed instruments have behavior which is at least recognizably close to linear -- otherwise, they would be terribly hard to play.

 

[thegreenlaser]

However, the 1-d wave equation has traveling wave solutions along the characteristics x+ct, x-ct, (c.f. d'Alembert's formula) so that the time dependence is related to the space dependence through the wave speed c. Put another way, the signal detected at a fixed point over time should be exactly the 'incoming' spatial wave shape(s). A similar principle applies for 'analogue' recording media such as LPs and magnetic tapes.

I agree, in the case of a perfect string. Does a similar relationship still hold in a linear model of a string with stiffness and loss terms? I'll admit, I don't know enough about PDE's to know whether it would or not. My feeling is that stiffness and frequency dependent loss terms would deform the shape as time progressed (i.e. Simple triangle wave wouldn't remain a triangle wave for very long), which would mean a Fourier analysis of the initial shape would yield much less information about the frequency output of the system overall. You would have to know how the Fourier transform of the shape changes with time, and my gut feeling is that in the case of a more complicated PDE model, you wouldn't be able to predict that with very much ease at all. That would be in contrast with the ideal string case, where the manner in which the string moves is indeed quite easy to predict based on the initial position.

For an acoustic instrument, I would think that most of the 'output' is actually whatever goes through the bridge to the soundboard(s) of the instrument.

Point taken. Though I think it's still reasonable to talk about the displacement of the string at a single point as being the 'output.' It's crude, but it's not so different from the real sound that it's useless.

Most of the preceding discussion has been assuming a linear model of the string, but I'd argue that most real-world stringed instruments have behavior which is at least recognizably close to linear -- otherwise, they would be terribly hard to play.

Fair enough.

 

[olivermsun]

My feeling is that stiffness and frequency dependent loss terms would deform the shape as time progressed (i.e. Simple triangle wave wouldn't remain a triangle wave for very long), which would mean a Fourier analysis of the initial shape would yield much less information about the frequency output of the system overall.

You're absolutely right about stiffness and damping in a real string (especially one fitted on an instrument). You might model this with a frequency-dependent time decay (e.g., a complex impedence for wave transmission down the string), in which case the Fourier decomposition would be quite helpful. For weak damping, this should give the frequency shift due to damping as well. On the other hand, the true behavior really is nonlinear, so this approach wouldn't work perfectly as you've already pointed out.

Then again, the complex impedence might also be a useful way to allow partial transmission through the bridge...

 

[chingel]

You are correct, however that's not really a good way of looking at it. It's true that the spatial 'modes' (i.e. the dividing the string into even parts and making sinusoids) affect the output, but it's not nearly as direct as it would seem. Like you said, thanks to Fourier analysis, we can indeed take the shape of the string at any time and turn it into a sum of sinusoidal terms based on some frequency omega, which is directly related to dividing the length of the string into even parts.

Thank you for your reply. I still don't understand why does a string divide itself nodes (or parts), why doesn't it just oscillate at the fundamental, why does it also start oscillating at the partials? I also didn't understand why the harmonics last longer and the other frequencies die out quickly, if the other frequencies are even produced when plucking a string which I am not sure about either? There is supposedly some sort of an impedance for non-harmonic frequencies, but how it works I didn't understand. It had something to do with the nodes not fitting on the strings length, but then the videos were posted showing moving nodes on a plucked string.

 

[mikeph]

However, the 1-d wave equation has traveling wave solutions along the characteristics x+ct, x-ct, (c.f. d'Alembert's formula) so that the time dependence is related to the space dependence through the wave speed c.

A good way of showing there are no traveling waves on a guitar string,
f(x - ct) = 0, f(x + ct) = 0 at the nodes, say the nodes are x = 0,1, (wlog)
x = 0; f(-ct) = 0 and f(+ct) = 0, for all t
so f = 0.

 

[Pythagorean]

why does a string divide itself nodes (or parts), why doesn't it just oscillate at the fundamental, why does it also start oscillating at the partials?

The standing wave (with the nodes) is not the whole picture. When you pluck the string, you actually send out traveling waves. The disturbance goes out from your pluck, encounters both bounded ends, and travels back, hits the opposite side, etc, etc, superimposing on each other in such a way that gives rise to the pattern of standing wave harmonics due to the two fixed boundaries. As you change the tension on the string (with the keys) you change the speed of the wave, as you fret different strings, you change the length (shortening/lengthening the path of the traveling waves).

I also didn't understand why the harmonics last longer and the other frequencies die out quickly, if the other frequencies are even produced when plucking a string which I am not sure about either?

The string isn't massless, the tension isn't uniform, thermodynamics applies, etc, etc. So yeah, there's impurities in the harmonics, but they die out rather quickly because they don't "propel" each other like harmonics do (dosey do!) They're not synchronized in time like harmonics are with the fundamental (and the fundamental is the strongest force present, the "dominant" tone.)

 

[chingel]

I don't understand, are there nodes or not, is there a traveling wave or not? Reading this thread the answers seem to me very conflicting. Just looking at the last two posts, one says there are no traveling waves, the other says there are, which is correct?

The standing wave (with the nodes) is not the whole picture. When you pluck the string, you actually send out traveling waves. The disturbance goes out from your pluck, encounters both bounded ends, and travels back, hits the opposite side, etc, etc, superimposing on each other in such a way that gives rise to the pattern of standing wave harmonics due to the two fixed boundaries. As you change the tension on the string (with the keys) you change the speed of the wave, as you fret different strings, you change the length (shortening/lengthening the path of the traveling waves).



The string isn't massless, the tension isn't uniform, thermodynamics applies, etc, etc. So yeah, there's impurities in the harmonics, but they die out rather quickly because they don't "propel" each other like harmonics do (dosey do!) They're not synchronized in time like harmonics are with the fundamental (and the fundamental is the strongest force present, the "dominant" tone.)

Could you please elaborate on how they superimpose to give rise to the wave harmonics? I am having difficulty understanding. Also how do the harmonics propel each other? If the string divides itself into two parts/nodes, one is going up, the other is going down, is that correct? Doesn't that mean that one is necessarily not synchronized with the fundamental, but rather working against it? How does the synchronization with the fundamental work?

 

[Pythagorean]

I don't understand, are there nodes or not, is there a traveling wave or not? Reading this thread the answers seem to me very conflicting. Just looking at the last two posts, one says there are no traveling waves, the other says there are, which is correct?

Initially, they are traveling waves. As the waves meet each other and superimpose, they produce a standing wave.

Also how do the harmonics propel each other? If the string divides itself into two parts/nodes, one is going up, the other is going down, is that correct?

yes

Doesn't that mean that one is necessarily not synchronized with the fundamental, but rather working against it? How does the synchronization with the fundamental work?

You could call it "anti-synchronized" if you like, (just like anti-parallel vectors are still parallel lines). The work "against" the opposing bend sling shots it to the other side with more force than if the string would have been flat, due to tension.

Anyway, they're synchronized in that they reach their maxima at the same time (albeit in opposite directions).

 

[mikeph]

Please correct me if I'm wrong, this is a hunch,

If you pluck the string to create an initial shape as a V shape, for example, there will only be traveling waves if you use your finger.

If you have a V-shaped mould that you push against the string, and suddenly remove it, there will be no traveling waves, even at the start. The traveling waves only arise because it takes a finite time for the signal to move from the place where the finger was to the edges. If the entire string is supported and then instantaneously let free, there are no traveling waves.

I think that's they key.

 

[olivermsun]

A good way of showing there are no traveling waves on a guitar string,
f(x - ct) = 0, f(x + ct) = 0 at the nodes, say the nodes are x = 0,1, (wlog)
x = 0; f(-ct) = 0 and f(+ct) = 0, for all t
so f = 0.

This is incorrect. The wave reflects from the (clamped) boundary with a phase reversal. This allows the boundary condition to be satisfied for all time while conserving energy and momentum.

 

[olivermsun]

I don't understand, are there nodes or not, is there a traveling wave or not? Reading this thread the answers seem to me very conflicting. Just looking at the last two posts, one says there are no traveling waves, the other says there are, which is correct?

You should look at the videos linked earlier or the paper which discusses waves on a string and decide for yourself which one is correct.

Initially, they are traveling waves. As the waves meet each other and superimpose, they produce a standing wave.

Hint: for general initial conditions, a standing wave will not be created.

 

[sophiecentaur]

Remember that it takes time for a standing wave to establish itself. The wave has to travel the full length of the string and be reflected before a standing wave is formed.

Also, (@chingel) there is no conflict in first saying traveling waves and then saying standing waves. A standing wave is only the resultant interference pattern of a set of traveling waves.

 

[chingel]

I found a bunch of high speed videos showing a plucked string

http://physics.doane.edu/physicsvideolibrary/string/StringDirectory.html

Clicking on the links under the headline Videos will open them, some strings are plucked towards the end, some in the middle, they use different strings and weights etc.

It seems like initially there are traveling waves, but they seem to turn into standing waves.

Also, (@chingel) there is no conflict in first saying traveling waves and then saying standing waves. A standing wave is only the resultant interference pattern of a set of traveling waves.

How does that work? How do the interference patterns of traveling waves create standing waves?

 

[olivermsun]

I found a bunch of high speed videos showing a plucked string

http://physics.doane.edu/physicsvideolibrary/string/StringDirectory.html

Clicking on the links under the headline Videos will open them, some strings are plucked towards the end, some in the middle, they use different strings and weights etc.

It seems like initially there are traveling waves, but they seem to turn into standing waves.

Just looking at a couple of the examples where the strings are plucked near one end, I don't see any that turn into standing waves.

 

[sophiecentaur]

I found a bunch of high speed videos showing a plucked string

http://physics.doane.edu/physicsvideolibrary/string/StringDirectory.html

It seems like initially there are traveling waves, but they seem to turn into standing waves.
How does that work? How do the interference patterns of traveling waves create standing waves?

That's what standing waves are. They are the sum of traveling waves moving in different directions. You only get a standing wave resonance on a string (etc) if the length between the ends causes the constructive and destructive interference to coincide for all reflections from both ends. If you wiggle one end of a string (heavy rope is better) with the other end tied (a node), you launch a traveling along it. Choose the right frequency of wiggling and you will get a resonance with one. two, three or more antinodes. Your hand is more or less at a node ( the movement need only be small, compared with the maximum swings at the antinodes) and you only need to be putting in a relatively small amount of power to maintain a large amplitude of resonance.

 

[mikeph]

This is incorrect. The wave reflects from the (clamped) boundary with a phase reversal. This allows the boundary condition to be satisfied for all time while conserving energy and momentum.

But from the mathematics, it's not a d'Alembert solution...?

 

[sophiecentaur]

Just looking at a couple of the examples where the strings are plucked near one end, I don't see any that turn into standing waves.

That's because they are happening slowly and, unlike a real guitar string, you don't see multiple images at once. I don't understand - are you actually disagreeing that standing waves are due to multiple traveling waves? You'll be disagreeing about Young's Slits and Lasers too, perhaps - it's all the same phenomenon.
Just 'cos you can't see something doesn't mean it's not there.

 

[thegreenlaser]

But from the mathematics, it's not a d'Alembert solution...?

I think the issue with your post was your first claim that f(x+ct)=0 and f(x-ct)=0 at the nodes. Correct me if I'm wrong, but I think the only thing you can say is f(x+ct)+f(x-ct)=0, since d'Alembert's solution is
$$u(x,t)=\frac{1}{2} \left( f(x+ct)+f(x-ct) \right)$$
assuming initial velocity is zero. So if you supply the boundary condition u(0,t)=0 for all t, you end up with f(ct)+f(-ct)=0, which does not imply that f=0. If it were true that f=0, then the solution for the displacement of the string would always be zero if no initial velocity were involved, wouldn't it?

Please correct me if I'm wrong, this is a hunch,

If you pluck the string to create an initial shape as a V shape, for example, there will only be traveling waves if you use your finger.

If you have a V-shaped mould that you push against the string, and suddenly remove it, there will be no traveling waves, even at the start. The traveling waves only arise because it takes a finite time for the signal to move from the place where the finger was to the edges. If the entire string is supported and then instantaneously let free, there are no traveling waves.

I think that's they key.

If my understanding is correct, you would still end up with traveling waves. If you have some sort of symmetry in your starting position/velocity (e.g. a triangle whose peak is exactly in the centre) I'm pretty sure you would end up with a standing wave in the ideal case. However, if you were to, say, offset the triangle, even if you do set it off in the perfect manner like you've described, it will create a traveling wave. I'm guessing that that's explained by an imbalance of tension that could probably be shown with some trig, but I do know that such a starting position will cause a horizontally moving wave, and that can be shown with the solution to the wave equation.

 

[sophiecentaur]

I really don't understand this 'obsession' with traveling waves and how it gets in the way of understanding the final condition of standing waves. If the energy distribution on the string is not uniform (if it's plucked off centre, for instance) then energy will flow, once it's been let go (no?). So the energy will be carried by waves. Before many cycles have elapsed, can it be possible to identify the full interference pattern of these waves (i.e. the modes)? Any form of resonance has a build up or decay time associated with it so I can't see that there is a paradox / clash. It takes time to define a frequency precisely and there will be a bandwidth associated with the build up of all these oscillations. You can't just discuss the Frequency domain of the oscillations unless you specify a time over which the analysis is done. Before things have settled down I think the analysis has to be done in the time domain and would involve the equations of motion of each elemental part of the string. Too damn hard I think. So you just have to think in general terms of traveling waves on the string until a suitable time has elapsed, after which the problem becomes more straightforward and the natural modes are the result.

I'm not sure that the excellent slo-mo movies actually help, either, if people have already made up their minds about what they are showing.

 

[olivermsun]

But from the mathematics, it's not a d'Alembert solution...?

It is. I think thegreenlaser explains it quite well in a post above.

I don't understand - are you actually disagreeing that standing waves are due to multiple traveling waves?

I'm disagreeing that constructive and destructive interference between traveling waves is sufficient to create a standing wave pattern. The standing wave is a very special case where the spatial dependence is completely fixed in time, that is, the shape of the string is not time dependent. For example, the maxima and nodes have to stay in exactly the same places on the string for all time. Since there is no time dependence, the standing wave pattern must exist from t = 0 onward -- there is no "set up time" for the resonance to begin.

A pair of (equal) pure sinusoidal waves traveling in opposite directions along the string will automatically create a standing wave. On the other hand, arbitrary combinations of standing wave modes will not result in standing waves. To convince yourself of this, you can try plotting the first two standing modes on the string, where the second mode oscillates at twice the frequency of the fundamental, and their sum.

Just 'cos you can't see something doesn't mean it's not there.

True, but you can also do the math and get the same result. In fact, I'm surprised it has generated this much controversy without people just breaking out the pencil and paper and trying it themselves.

 

[chingel]

Just looking at a couple of the examples where the strings are plucked near one end, I don't see any that turn into standing waves.

In the first video for example, doesn't the kink lose it's kinkiness and just a general pattern of up and down oscillation emerge? Or is that still a traveling wave, only that the kink is smoother and less kinky?

 

[sophiecentaur]

@oliversum
Any reflection will produce a standing wave but you need multiple reflections to get a resonance. Think about transmission lines. You need a matched load to avoid a standing wave, independent of the source impedance.

With no losses, the plucked string would be expected to return, every so often, to the same peaky shape as when released but it doesn't happen in practice, because the high overtones don't survive.