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A simple question that I was asked but could not answer.
Standing waves in a string. When you pluck the string it is not in a sinusoid, it's two straight lines to the fixed ends.
When you let go, why does it end up at the resonant standing wave frequencies? Where do the other frequencies that are reflected just the same by the fixed ends go? What physical mechanism attenuates them?
Writing that prompted the thought - do I need to consider my original triangle as repeating to infinity (a long string) due to reflections (out of phase) which Fourier analyses to harmonics so there are no "other frequencies"?
In reply to elsewhere:
Yes.
In reply to elsewhere:
I agree. Such a simple question...
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In reply to elsewhere:Is it not just destructive and constructive interference?I think by using Fourier analysis in the first place, you have already built in the assumption that your 'periodic window' repeats infinitely.
> When you let go, why does it end up at the resonant standing wave frequencies? Where do the other frequencies that are reflected just the same by the fixed ends go? What physical mechanism attenuates them?
> Writing that prompted the thought - do I need to consider my original triangle as repeating to infinity (a long string) due to reflections (out of phase) which Fourier analyses to harmonics so there are no "other frequencies"?
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In reply to elsewhere:
Interestingly if you watch a slow motion video of a string then it never looks like a sinusoid!
It's mostly down to the mass of the string and how resistant it is to bring stretched (stiffness). The string has to stretch, and so is acting like a spring. It has some mass and momentum which causes it to move past the middle once released. This momentum means it keeps missing the mid point for a while.
the speed at which this happens is the pitch you hear the string at, which depends on the stiffness and mass - more mass and there is more momentum and it takes longer, resulting in a lower pitch. if there is higher stiffness then there is more force to change the direction of the string and so it takes less time, resulting in a higher pitch.
In reply to elsewhere:
There aren't any frequencies travelling down the string.
You start with a deformed structure, the string, then remove a force and thus change it's equilibrium. The vibrations are a result of the structure trying to reach a new equilibrium.
If you want a rigorous explanation and are familiar with differential equations, there's bound to be s proof on the internet.
Regarding harmonics, they drop in amplitude by 20db per decade as you go up in frequency so each harmonic is very much smaller than the last.
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In reply to elsewhere:
My understanding is it's to do with how the plane strain energy interacts (inversely proportional to plucking amplitude) with the precession of the phase angle and the Young's Modulus of the string material. I think there is also a Wigner component causing non Newtonian frequency locking but this is very much a second order effect.
Does that help?
3
In reply to elsewhere:
Consider the string at a very small timestep after release: the element at the point where the pluck occurred is being pulled downwards by the elements on both sides. All other elements are being pulled equally upwards and downwards. Thus only the plucked element accelerates downwards. At the next timestep the adjacent elements also have a net force acting on them and so start to accelerate.
This will tend to result in a curved shape and velocity profile. If the string were perfectly elastic then it would oscillate between triangular profiles at the extremes of its travel, but damping will make this decay into shape that is closer to sinusoid.
For a full mathematical understanding you will need to look into spatial convolution.
An interesting result of the physics involved is that if a string is excited near the centre the fundamental will dominate, whereas if it is excited near an end the higher harmonics will have much greater amplitudes. This is why instruments that are bowed near an end (most Western ones) sound rich but also require resonating chambers to amplify the sound (because less sound power is generated).
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In reply to elsewhere:
> Where do the other frequencies that are reflected just the same by the fixed ends go? What physical mechanism attenuates them?
There will always be damping in the system. That is, energy lost as one part of the string affects the adjacent part. It's the same process that causes your dominant modes to die away over a few seconds. The damping is greater for non-resonant modes (that being the whole concept of resonance).
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In reply to Coel Hellier:
> The damping is greater for non-resonant modes (that being the whole concept of resonance).
The kid asked me what the selective energy loss mechanism was (not in that language) so I am trying to avoid saying "fixed ends and speed fixes resonances" with no explanation physical of mechanism selective energy loss.
I can see why air damping and bridge inelasticity might be higher for higher frequency but not why a frequency slightly lower than a resonance might be more damped than the resonance.
An experiment is always best. Has anybody got a guitar (or slack line), a bright light and a phone that does slow motion? Actually it must be on YouTube, everything else is.
In reply to elsewhere:
The damping will determine how long it takes the sound to decay but has only a little effect on the frequency (such that it is ignored completely when calculating the frequency generally). From the Wikipedia page on this:
https://en.wikipedia.org/wiki/String_vibration
The frequency is derived as relating to the tension in the string, the density and the length. It oscillates at specific frequencies due to an effect similar to a pendulum (A better analogy is a mass on a spring, such as a car suspension, the frequency it naturally moves out doesn't depend on the damping but on the mass and stiffness in the spring).
In reply to stevevans5:
> The damping will determine how long it takes the sound to decay but has only a little effect on the frequency
But the frequency does have a big effect on the damping, and thus on the timescale on which that oscillation dies away.
Post edited at 12:25
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In reply to elsewhere:
> > The damping is greater for non-resonant modes (that being the whole concept of resonance).
> I can see why air damping and bridge inelasticity might be higher for higher frequency but not why a frequency slightly lower than a resonance might be more damped than the resonance.
A structure only vibrates at a set number of frequencies with the fundamental frequency being the lowest. Above that there is a mode shape and frequency for each degree of freedom of the structure so in theory in your case there would be an infinite number of nodes however in practice, as amplitude drops by a factor of 100 every time the frequency increase by 10 the modes disappear very quickly. Further, damping, energy absorption, is often also proportional to velocity and so there is a second mechanism for killing higher modes or harmonics.
In reply to Removed User:
> A structure only vibrates at a set number of frequencies with the fundamental frequency being the lowest. Above that there is a mode shape and frequency for each degree of freedom of the structure so in theory in your case there would be an infinite number of nodes however in practice, as amplitude drops by a factor of 100 every time the frequency increase by 10 the modes disappear very quickly. Further, damping, energy absorption, is often also proportional to velocity and so there is a second mechanism for killing higher modes or harmonics.
What about a frequency slightly lower than a resonance? What physical mechanism attenuates it more than a higher but resonant frequency?
In reply to elsewhere:
If you pluck a string it doesn't resonate below the fundamental frequency.
Same as hitting a cymbal. There's a fundamental frequency it'll vibrate at plus a few harmonics superimposed but you don't get any vibrations at a lower frequency.
Another analogy is a resonant rlc circuit. If you inject a spike into it you get a fundamental frequency and maybe some harmonics but nothing else. The maths is broadly the same.
1
My thinking now...
The reflection sends an equal and opposite wave into the string travelling in the opposite direction, that is a condition of an amplitude node (a fixed end that does not move) - destructive interference of amplitude. The fixed end is a supplier of equal and opposite waves going the other way.
An equivalent supplier of equal and opposite waves travelling the other way is another string plucked in the opposite transverse direction.
Rather than reflections at each end you can think of the string as infinitely* long with an amplitude node every L (string length).
Each length L is plucked into the same stationary position but in alternating transverse direction.
When string released, waves with equal and opposite transverse displacement travelling in opposite directions maintain amplitude nodes every L along string..
If you repeat that profile infinitely (or Q times) that initial displacement profile will Fourier analyse to fundamental wavelength and harmonics only.
That means there is no physical mechanism for selective energy loss for non-resonances because all of the energy is in resonances and none of the energy is in non-resonances. If Q not infinite replace the words all/none with most/little.
*or about Q times longer where Q is resonance Q factor.
A 100Hz fundamental that lasts for 1 second has oscillated a hundred times so the wave has traveled two hundred times the length of the string. Imagining the string as much longer is reasonable as the wave travels for many wavelengths.
Post edited at 13:00
In reply to elsewhere:
I struggled to turn your question into precise Physics; here is my best take.
> Standing waves in a string. When you pluck the string it is not in a sinusoid, it's two straight lines to the fixed ends.
Correct for a small pluck.
> When you let go, why does it end up at the resonant standing wave frequencies?
On an ideal string, any shape that is plucked will split into two half-amplitude copies of the original disturbance that propagate in opposite directions, reflect off the ends and interfere half a fundamental frequency (of the string) later into a copy of the original disturbance with inverted amplitude, than another half-period later they interfere to re-create the original disturbance. This process repeats indefinitely.
The frequency with which any initial disturbance re-appears on such a string is always the fundamental frequency of the string.
A 3rd harmonic sine wave disturbed onto a string appears to resonate faster but that's due to the repetitive nature of the disturbance meaning that intermediate shapes on the string - formed at several points between the reappearance of the initial disturbance - look exactly like the initial disturbance.
In practice the shape associated with one part of the initial disturbance - if tracked - only passes the same physical point in the same direction once every fundamental frequency. If your initial 3rd harmonic had slightly different amplitudes in each peak decreasing left to right, you would see that that particular order of amplitudes only occurred at the fundamental frequency, not at the frequency of the 3rd harmonic.
I did wonder if by your question you actually meant "Why does the shape of the string end up more sinusoidal than triangular?" by which I can give the cop-out answer "Non-linearities"; your triangular disturbance contains an infinite number of sinusoidal frequencies (by Fourier theory) and the higher frequencies will be damped faster causing the triangle waveform to rapidly shed its sharp edges. I assume this is due to non-linear effects in the frictional losses of stretching the string.
This seems like a good point to chime in with my crazy letter writing campaign that the decision to discount transverse modes in the derivation of standard introductory "Wave on a string" teaching is a step too far in terms of how it's justified - the only good reason to neglect it is making the maths simple enough to not confuse everyone.
http://web.phys.ntnu.no/~stovneng/TFY4160_2010/rowland_ejp_2013.pdf
https://www.math.nyu.edu/faculty/peskin/papers/wave_momentum.pdf
Post edited at 13:14
In reply to elsewhere:
> What about a frequency slightly lower than a resonance? What physical mechanism attenuates it more than a higher but resonant frequency?
Consider pushing a child on a swing, where the swing of course has a natural frequency.
If you push in time with the natural frequency, your pushes add to the natural motion, and you get a big amplitude. But imagine pushing out of time with the natural frequency. Some of your pushes would be acting against the natural motion -- you'd be pushing the swing away while it is coming towards you, and the two would tend to cancel, and the resulting amplitude of motion would be lower.
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In reply to Coel Hellier:
> Consider pushing a child on a swing, where the swing of course has a natural frequency.
The playground equivalent of plucking is like releasing the child from stationary without periodic push resulting in pendulum frequency from gravity and length. That might be best explanation for plucking - but it's a more complicated result (multiple solutions, fundamental & harmonics) to different forces and different maths. No special energy attenuation mechanism required - just tension, mass per length & length.
Post edited at 14:10
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In reply to elsewhere:
The swing is a good analogy for why a string vibrates - Momentum from the mass being resisted by gravity in the swing and by the metal stretching in the string. It is also a good analogy for why it vibrates at a certain frequency and not at every frequency, because of this fundamental mechanism.
The maths for the string comes down to the tension, length and density of the spring. If you start with F=ma and f=kx (Hooke's law) you can derive something that looks like the wave equation, and you end up with a number of solutions to this equation that look sinusoidal (these are the different harmonics).
Practically it is probably easier to think of this as standing waves - only when the wavelength is a fraction of the string length will it resonate efficiently. The frequency (speed at which it resonates) depends on how tight the string is (which to stretch the analogy is like changing the length of your pendulum) and will also depend on how heavy the string is.
Post edited at 14:23
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In reply to elsewhere:
The ends are fixed. They cannot move up and down and so are ‘forced’ to be nodes. You can then only ever have a whole number of wavelengths on the string due to its tension being equal along the whole length.
In reply to DancingOnRock:
Multiples of half wavelenghts.
In reply to various:
The person who asked knows there are the fundamental and harmonic frequencies, but that's not the question passed on to me.
Their question was.
"I look at a string attached to the ends, a music string
I think that a pinch will introduce a packet of waves into the string. This package contains 'stationary' and non-stationary Fourier component wavelengths.
So I think we're going to have two string modes:
1) a stationary, sum of all stationary components
2) one that travels back and forth along the cord, the sum of all non-stationary components. is it right?
I have now read in my physics book that non-stationary components disappear and only the stationary ones live: is that true? If so, why and what does the mechanism look like?
Sorry for my bad English. I hope you can answer me, thank you"
I think their text book is wrong. The pluck puts energy into resonant modes only so there need for a mechanism that attenuates non-resonances selectively.
Stationary waves aren't stationary but that might be a language thing rather the physics thing - all waves, perturbations, packets, resonant or non-resonant are propagating.
An optical example is a single broad band mirror with reflectivity about 1 has no mechanism for selective absorption (as stated, a broad band mirror).
Appropriately adding a second identical mirror to create a resonant cavity does create resonances but does not change the mirror surfaces to create a new way of selectively turning non-resonant photons into heat.
Not sure if it is a very smart kid or a young student, just guessing that.
Post edited at 16:00
In reply to elsewhere:
My (slightly handwave-y) explanation for A level students is to consider the travelling waves hitting one end of the string, and reflecting (with a 180 degree phase change). Waves that are of the 'correct' length and in just the right location will interfere constructively as they reflect. Waves that don't fulfil those criteria will interfere destructively. So only waves that are of the correct lengths can propagate up and down the string to become stationary - all others will just cancel themselves out.
It doesn't stand up to much scrutiny, because it leaves you with the question of what happens to the energy in the non-stationary components. You can argue that energy could never have been put into those components in the first place but that's not wholly satisfactory. But it's enough to get past A levels!
In reply to Jamie Wakeham:
> You can argue that energy could never have been put into those components in the first place but that's not wholly satisfactory.
How dare you! That's my argument for my infinitely long string plucked in alternating directions so it is obviously very satisfactory
The kid certainly got me (& UKC) thinking.
Post edited at 15:58
In reply to cb294:
No. I meant whole Wavelengths of a harmonic, not wavelengths the length of the string. Only whole fractions of the length of the string.
1/2,1/3,1/4 etc. But not eg 1/1.1111 as you can’t have a fraction of a harmonic.
In reply to elsewhere:
I don't think it's helpful to think of stationary and static as separate. The pluck is essentially putting in energy at all frequencies. Where the wavelength corresponds to a fraction of the length, the waves in opposite directions will regularly interfere and will appear stationary and large in magnitude. These are your modes of the string. There isn't a mechanism to attenuate other frequencies, it's the other way round - there is no mechanism for the string to vibrate at these frequencies
In reply to stevevans5:
> I don't think it's helpful to think of stationary and static as separate. The pluck is essentially putting in energy at all frequencies. Where the wavelength corresponds to a fraction of the length, the waves in opposite directions will regularly interfere and will appear stationary and large in magnitude. These are your modes of the string. There isn't a mechanism to attenuate other frequencies, it's the other way round - there is no mechanism for the string to vibrate at these frequencies
Agree, although I reckon pluck can be Fourier analysed into putting energy into harmonics only.
In reply to elsewhere:
Plucking is the equivalent of a step change.
In reply to elsewhere:
Yeah so you could analyse the pluck (energy at all frequencies for an ideal pluck) and then analyse the motion of one point on the string or something similar at different times to see how the different frequencies decayed? I don't think for this kind of problem where you can work out how it would behave from the physical properties of the string you would normally get into that level of analysis though!
In reply to DancingOnRock:
> Plucking is the equivalent of a step change
That gives the content of higher harmonics.
In reply to elsewhere:
No. It injects all frequencies to the system, regardless of whether they’re supportable.
As I wrote above a whole set of wavelengths cannot and will not exist. They don’t decay, they’re just not the from the beginning as the string length can’t support them.
You need to determine the energy in the string from the tension and displacement in the initial pluck. The wavelengths will then be 1/1, 1/2, 1/3 etc and the frequencies will be determined by those wavelengths.
In reply to elsewhere:
Er, what?
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