Standing waves on a string experiment -- Relative amplitude of harmonics

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  • #1
fog37
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Hello forum,

I am wondering why the higher order (higher harmonics) standing waves developed on a string under tension generated by an oscillating mechanical vibrator (set at the same amplitude but with variable frequency) have lower amplitude when compared to the lower harmonics (the fundamental has the highest amplitude)...What are the equations explain this fact?

Also, it is easy to notice that when the string tension is low, the amplitude of all modes is larger compared to when the tension is larger. Smaller tension means smaller wave speed, etc. but I cannot find equations that justify the different amplitude with different tension ##T##...

Thanks as usual!
 

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  • #2
A.T.
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I am wondering why the higher order (higher harmonics) standing waves developed on a string under tension generated by an oscillating mechanical vibrator (set at the same amplitude but with variable frequency) have lower amplitude when compared to the lower harmonics (the fundamental has the highest amplitude)...What are the equations explain this fact?
Do you have a video of what you mean here? Higher frequencies dissipate energy faster.

Also, it is easy to notice that when the string tension is low, the amplitude of all modes is larger compared to when the tension is larger. Smaller tension means smaller wave speed, etc. but I cannot find equations that justify the different amplitude with different tension ##T##...
If you assume the same energy, higher tensions will have less amplitude.
 
  • #3
fog37
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Thanks. Don't have a video. Just noticed that, for the same tension and and driving resonator amplitude (the excitation source), the various modes have different amplitude with the fundamental mode having the largest amplitude (biggest eye pattern).
 
  • #4
A.T.
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Thanks. Don't have a video. Just noticed that, for the same tension and and driving resonator amplitude (the excitation source), the various modes have different amplitude with the fundamental mode having the largest amplitude (biggest eye pattern).
Assuming the same spanned distance and maximal length of the string, spreading it over more waves will obviously result in smaller amplitudes. This is just geometry.
 
  • #5
coquelicot
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You ask for the mathematical explanation, so here is one:
Every periodic signal can be decomposed into an infinite sum of sinusoidal signals: this is the fundamental theorem of Fourier series.
So, in a Fourier series, each multiple of the fundamental has a coefficient and if these coefficients were not approaching zero, the Fourier series would not converge. This explains why the harmonics have eventually less and less amplitude as their frequency increases. But this does not explain your claim that "a harmonic of given frequency has ALWAYS a larger amplitude than a harmonic of higher frequency". In fact, I'm not sure this is true. As a violin player, I know something about that. There are 4 strings: sol, re, la and mi. It is not unusual that if the bow is not applied correctly onto the lowest string (sol), you hear a strong harmonic and not the right tone. The question is whether the fundamental is still present in this sound (that's not obvious).
 
  • #6
fog37
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Thank you.
I see how every complex signal can be decomposed into a summation of weighted sinusoids with various different frequencies. Each sinusoidal mode participates with a different amplitude and energy in the overall signal composition.

But in my experiment (standing modes on a string), I am not exciting simultaneous modes at the same time but individual modes with a frequency generator controlling a vibrating source at one end of the string while the other string is clumped. Each standing wave mode ##n## has its own specific frequency ##f_n = n f_1## where ##f_1## is the frequency of the fundamental mode. This is all verified and matches the theory. When I separately excit the 2nd, or 3rd, etc. modes, I notice that their amplitude (size of the lobes) is much smaller than the amplitude of the fundamental mode oscillating at ##f_1## and I am wondering why it is so...A.T. suggested that it is a matter of geometry.
 
  • #7
coquelicot
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This may well depend on the way you excite the string. You should describe more precisely you experience otherwise I cannot answer. Are the modes exited by some electronic means (oscillators etc.). Please, describe the exact process and if possible give a schematic of the electronic circuit.

I also add the following: it is intuitively evident that for a same amplitude of the vibrations, a vibration of a higher frequency carries more energy than a vibration of smaller frequency; this can probably seen mathematically from the formula that gives the energy dissipated in the air by a string in function of the frequency and amplitude. So, assuming the excitation transmits the same energy for two different modes, it is clear that the amplitude will be smaller for higher frequency.
 
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  • #8
fog37
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Thank you. So you are saying that, if the fundamental mode and the higher mode have antinodes with the same size amplitude, the higher mode contains more energy than the fundamental mode. However, if the fundamental and the higher mode have the same input energy, the higher mode will have antinodes with smaller amplitude. I guess the overall energy would be the total area contained in the lobes of the modes, so, geometrically, the area comprised within the lobe of the fundamental mode is about the same as the area comprised in the lobes of the higher order modes...
 
  • #9
A.T.
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I guess the overall energy would be the total area contained in the lobes of the modes,
The exact relationship would depend on the tension/length relationship of the string.

What you can say is this: At maximum amplitude all the energy is in the tension, so with the same total energy you get the same total string length at maximum amplitude. And same total string length (& same lobe shape) implies that the smaller lobes are uniformly (bidirectionally) scaled down versions of the bigger lobes. So the amplitude is proportional to the node distance.
 
  • #10
fog37
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Thanks A.T. I am getting closer to some understanding.

But what do you mean when you say that
... At maximum amplitude all the energy is in the tension
Without any excitation, the string is kept under a certain tension ##T##. Then we introduce the excitation and the string will assume either the fundamental mode shape or other higher order mode shapes depending on the frequency of the excitation. Does the tension change (increase relative to the tension when no excitation is applied) once the string assumes a mode shape (fundamental or higher)?

In the case of a higher mode, does the string get stretched and become longer to assume the multi-lobe shape? Is that what you mean with the comment below?
... with the same total energy you get the same total string length at maximum amplitude. And same total string length (& same lobe shape) implies that the smaller lobes are uniformly (bidirectionally) scaled down versions of the bigger lobes...

...So the amplitude is proportional to the node distance.
This is a key statement for me. Indeed, that is what happens: the shorter the node distance, as it happens in higher order modes, the smaller is the amplitude at the antinodes.

Thanks for the patience.
 
  • #11
A.T.
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In the case of a higher mode, does the string get stretched and become longer to assume the multi-lobe shape?
The string must elongate to form the lobes (assuming the string was initially taut).
 
  • #12
coquelicot
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In my opinion, when you say that all the energy is in the tension, you are completely missing the point that the energy is in fact dissipated in the air. True, the potential energy of the string is in the tension, but in an ideal paradigm where there would be no friction with the air (and the support of the string), it would suffice to excite the string during a short time to transmit it this energy, and then the string would vibrate endlessly. Not so in the experiment of the OP, where the excitation is continuous. So, the energy of the excitation is entirely passing to the air, and has probably nothing (or few) to do with the inner energy of the string.
What import to be known in my opinion, is a formula that estimates the energy transmitted to the air by a string. It probably depends on several parameters like the thickness of the string, the pressure of the air etc. but it is intuitively evident that as the frequency increases, more energy is transmitted (assuming the same string, equal amplitude and same conditions), hence the (intuitive) answer to the question. Sorry, I've not the time to seek a precise formula, some time with google will probably do.
 
  • #13
A.T.
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In my opinion, when you say that all the energy is in the tension, you are completely missing the point that the energy is in fact dissipated in the air.
I mention dissipation in post #2. The string's maximal elongation, which limits the amplitude depending on the number of standing waves, is another factor.
 
  • #14
coquelicot
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Here is my derivation of the energy dissipated in the air by the string. It is not from a book or any other source, so don't take it too much seriously. I assume viscous friction with the air, that is, the force is proportional to velocity.
Let L be the length of the string. The general equation for the nth-harmonic wave on the string is
$$u_n(x,t) = A \sin(\omega_n x/c) \cos(\omega_n t + \varphi_n),$$ where ##A## is the amplitude of the wave, ##\omega_n = n\pi c /L## is the pulsation, ##c## is the celerity of the wave, ##x## is the coordinate along the string, ##t## is the time, and ##\varphi_n## is the phase.
For the sake of simplicity, we shall assume ##\varphi_n = 0##.

So, the frequency is ##\omega / (2\pi)##, the period is ##2\pi /\omega##.
Let us consider an infinitesimal element ##ds## of the string. Its projection onto the ##x## axis is ##dx##, and we shall assume that during the vibration, the transverse force of the air upon ds (the only one that perform some work on the string) is proportional to ##dx##, to the transverse speed ##v## of the string, and to a coefficient ##\alpha## that sums up all the other ignored parameters. In other words
$$dF = -\alpha v dx.$$
So, the elementary work of this force over a transverse displacement ##d\ell## of the string during the vibration is
$$ d^2W = dF\cdot d\ell = dF\cdot v dt = -\alpha v^2 dt dx. \quad (1)$$
Let us set ##r = r(x) = A \sin( wx /c)##. Then ##v = -\omega r\sin(\omega t)##.
Injecting in (1),
$$ d^2W = -\alpha r^2 \omega^2 \sin^2(\omega t) dt dx.$$
Integrating ##d^2W## over t during one period ##P##, it is found that the elementary energy dissipated in the air by ##ds## during one period is
$$ dW = -\alpha r^2 \pi \omega dx.$$
Since there are ##\omega/(2\pi)## periods in one second, we find that the energy dissipated in 1 second by ##ds## is
$$dP = \frac{\alpha r^2 \omega^2}{2} dx.$$
We now integrate over ##x## to find the total energy dissipated by the string during 1 second, and find finally, after some straightforward calculus:
$$P = \alpha A^2 L \omega^2 / 4.$$
That means that the energy dissipated, which is exactly the energy transmitted to the string by your electronic device, is proportional to the square of the frequency, and hence, answers quantitatively to your question.
 
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  • #15
A.T.
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Here is my derivation of the energy dissipated in the air by the string.
...
That means that the energy dissipated, which is exactly the energy transmitted to the string by your electronic device, is proportional to the square of the frequency, and hence, answers quantitatively to your question.
Note that doing the experiment in vacuum would also result in frequency dependent amplitudes. The string itself can dissipate energy and limits the amplitudes by its maximal elongation.
 
  • #16
coquelicot
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In the vacuum, the only possibility to pump continuously energy is by heating the string itself. So, theoretically the string would heat more and more, until it melts. I think this effect is small here, and can be neglected.
 
  • #17
coquelicot
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A. T. I would add that two things are weird for me in your answer: the first one:
What you can say is this: At maximum amplitude all the energy is in the tension.
This is true only for the elastic potential energy stored inside the string, but what about the kinetic energy of the vibrating string ?
Second,
Note that doing the experiment in vacuum would also result in frequency dependent amplitudes. The string itself can dissipate energy and limits the amplitudes by its maximal elongation.
As far as I know, there is nothing like this "maximal elongation". The string can elongate until it breaks.

For Fog37, I would like to sum up my thoughts: The energetic balance is the following:
  • The total energy of the string = potential elastic energy + kinetic energy: this energy has to be transmitted by the device only during a fraction of second, then the device needs only compensate the friction and heat losses;
  • the energy transmitted to the air by the vibration: this is the major part and has been dealt above (at least for the case where the string has reached its dynamical equilibrium).
  • the heat transmitted to the string, and then to the air: vibrations make the string heat, and this heat is transmitted to the air. Whether or not this loss can be neglected his questionable, but I think this is a reasonable approximation here.
  • the energy transmitted by electromagnetic radiation to the ambient space: this is definitely tiny and can be neglected here.
Fog37, I suggest you try to compute the first point, that is, the total energy of the string at dynamical equilibrium. This is the energy the device has to transmit to the string (inside vacuum) during a short time to make the string vibrate. This is not very difficult, if you use the equation of the wave I wrote above. There are several advantages to do so and to write your result here:
1. this is funny,
2. A.T. will be happy,
3. I will be happy too.
 
  • #18
A.T.
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A. T. said:
At maximum amplitude all the energy is in the tension.
This is true only for the elastic potential energy stored inside the string, but what about the kinetic energy of the vibrating string ?
At maximum amplitude of the standing wave the entire string is at rest. The kinetic energy is zero.

As far as I know, there is nothing like this "maximal elongation". The string can elongate until it breaks.
That is the maximal elongation. Depending on the material the tension can grow non-linearly when approaching that value, so it gets very hard to stretch it further. This can limit the amplitude.
 
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  • #19
A.T.
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In the vacuum, the only possibility to pump continuously energy is by heating the string itself. So, theoretically the string would heat more and more, until it melts.
That depends on the details of the energy input. If the string absorbs a decreasing amount of energy from the driver by getting in-sync with the driver, it could reach an equilibrium where it can give away the small energy input as heat transmitted to the attachments and radiation.
 
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  • #20
coquelicot
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I knew that you'll answer that, and that's why I can reply quickly:
If you are trying to say that the driver "feels" the amplitude (or possibly the total energy of the string), for example with a feedback loop or by another control process, and that it almost stop to deliver energy to the string whenever this later has reached its dynamical equilibrium, then you are not answering to any physical question: everything is controlled by the driver, so the amplitude or the energy of the string depends only upon how the driver has been set or built, and this is certainly not the answer that was expected by the OP (see also my post in #7 about this point). If you want to answer to a physical question, you have to start by describing how the driver interacts with the string. In my case, I have simply assumed (but this is questionable too) that the driver transmits continuously the same power, regardless of the string or the frequency. While this may not be the case (#7 again), this has the merit to answer to a physical question.
 
  • #21
coquelicot
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At maximum amplitude of the standing wave the entire string is at rest. The kinetic energy is zero.
? No, the string is not at rest. A standing wave does not mean that the wave "don't move". I have probably missed something - please, explain what you mean. Are you speaking about the extremum points of the wave in the string ? but what about the other points ?
 
  • #22
A.T.
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? No, the string is not at rest. A standing wave does not mean that the wave "don't move".
When the string is at its maximal amplitude (maximal deflection from straight line), it is momentarily at rest. It's the turnaround point at which all points of the string reverse direction.

stnd_wav1.gif
 

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  • #23
coquelicot
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When the string is at its maximal amplitude (maximal deflection from straight line), it is momentarily at rest. It the turnaround point at which all points of the string reverse direction.View attachment 237709

Nice, you're probably right about that. So, the OP need not do anything more :frown:.
 
  • #24
A.T.
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If you are trying to say that the driver "feels" the amplitude (or possibly the total energy of the string), for example with a feedback loop or by another control process
No, I wasn't talking about any control mechanism. I'm just saying that this assumption doesn't have to hold:
I have simply assumed (but this is questionable too) that the driver transmits continuously the same power,
 
  • #25
coquelicot
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No, I wasn't talking about any control mechanism. I'm just saying that this assumption doesn't have to hold:
That's exactly the meaning of what you say in the following quote:
That depends on the details of the energy input. If the string absorbs a decreasing amount of energy from the driver by getting in-sync with the driver, it could reach an equilibrium where it can give away the small energy input as heat transmitted to the attachments and radiation.
Even if the builder of the driver has not "consciously" implemented a control process, if the driver behaves as you describe, it does have a more or less natural control process.

But that's not the point I want to discuss. In your post #9, you have demonstrated the following thing: in the vacuum, if a string has a total energy E and a frequency f, then for the same energy E and an higher frequency f, the amplitude of the wave is lower (and you have even given a quantitative insight). So far so good, but who said that the string should carry the same energy at different frequencies ? if the driver is built in such a way that this occurs, let it be, but you have only explained a geometrical aspect of the question, not the "why" of the question. Note: I've added the term "vacuum", because it is clear that you've neglected the friction with the air.
 
  • #26
A.T.
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who said that the string should carry the same energy at different frequencies ?
To have equal amplitude at all frequencies, the energy stored in the string would need to have a very specific relationship with the frequency, which is unlikely to just happen. So unless you specifically design it that way, the amplitude will be frequency dependent.
 
  • #27
coquelicot
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I explain again my argument (as I don't see your answer relates to it).
In your post #9, you show that if a string has a total energy E and a frequency f, then for the same energy E and an higher frequency f', the amplitude of the wave is lower. I'm not challenging this fact: I agree with it. What I pin-pointed is that that explains nothing (or hardly anything) of the experience of the OP, unless you assume that the driver is such that it manage to set the total energy E of the string to a given value (possibly fixed by the user). I think this is very improbable though.
 
  • #28
A.T.
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What I pin-pointed is that that explains nothing (or hardly anything) of the experience of the OP, unless you assume that the driver is such that it manage to set the total energy E of the string to a given value .
The point of the vacuum scenario was to separate different effects, not to explain the whole observation of the OP.

You don't need to assume the same energy E of the string for all frequencies, in order to have the amplitude vary with frequency (what the OP observes). Depending on the material the tension can grow non-linearly, so it gets very hard to stretch it further. This can limit the amplitude too, and that limit depends on the number of nodes.
 
  • #29
coquelicot
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As I said above, there is no such thing "maximal string elongation". If the elasticity of the string is strongly not linear, then you can decide arbitrarily to invent a threshold somewhere in the non linear part, exactly like electronic engineers speak of a "diode drop", at about 0.6V. But this has not physical meaning.

Now, regarding your argument that the string reaches its "maximal elongation", understood as "it reaches the non linear part of the elasticity of the string", this simply don't hold: As a violin player, I know that I can stretch the string much, much more than the normal tone (at least an octave above, and possibly more), and it's still in the limits of the linear elasticity. It is simply impossible that the small vibrations of the string stretch the string more than that. So, it should now be clear that your explanation of the experiment is upset.
 
  • #30
A.T.
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So, it should now be clear that your explanation of the experiment is upset.
Nothing is clear about the actual experiment, because the OP didn't post any detail. In particular the OP doesn't specify it's a violin string in normal operation.

And as I said before, this wasn't supposed to be a complete explanation of the vague experiment, just one factor that might affect such experiments in general, along with others like the energy dissipation that I mentioned first.
 
  • #31
coquelicot
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Yet, your explanation has "some chance" only if the string has reached the limit of its linear elasticity, which is unlikely to be the case in general. I agree that the OP has provided insufficiently many details though (post #7 again).
 
  • #32
A.T.
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Yet, your explanation...
For the n-th time: This wasn't my entire explanation, just one of the factors I mentioned that can play a role, since we are talking generally without specific data.
 
  • #33
coquelicot
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OK. Thank you for this discussion, and especially for #22 where I've learned something.
 

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