Is there a name for this wave effect?
So I was using this wave simulation (https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html) when this happened; using the settings: Oscillator, High tension, no damping. Any frequency or amplitude will do as long as it's divisible by 5 I think it's because of calculations.
After sometime, the wave form would collapse on itself leaving a plain rope and restarting. (It has to be with the settings since the beginning)
Is there a name for this?;
Could it happen in the real world or is it purely hypothetical since there's no system like that;
Why with low tension doesn't happen?;
Was it purely coincidental?
waves simulations interference
add a comment |
So I was using this wave simulation (https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html) when this happened; using the settings: Oscillator, High tension, no damping. Any frequency or amplitude will do as long as it's divisible by 5 I think it's because of calculations.
After sometime, the wave form would collapse on itself leaving a plain rope and restarting. (It has to be with the settings since the beginning)
Is there a name for this?;
Could it happen in the real world or is it purely hypothetical since there's no system like that;
Why with low tension doesn't happen?;
Was it purely coincidental?
waves simulations interference
1
I think that's a bug in the simulation
– John Rennie
Dec 6 at 17:45
Do you mean the points in time where the rope is straight between opposing waveforms?
– Jasper
Dec 6 at 17:46
youtube.com/watch?v=4JuDHYjfkH8&feature=youtu.be
– Alef Gabryel
Dec 6 at 18:07
@JohnRennie, I've used this PhET simulation before and I don't see the "problem" referenced by the OP as a bug. This is a perfectly ideal simulation, with no losses, and I note that the frequency and length of the string are very slightly mismatched regarding the length required to support a 4th harmonic standing wave. Thus, at a few points in time, destructive interference flattens out the wave, and the whole sequence repeats. If the frequency or tension are slightly adjusted, a permanent standing wave can be achieved.
– David White
Dec 6 at 20:19
@AlefGabryel, when I was teaching high school physics, and I first saw PhET simulations, I thought that they were more suitable for middle school. After some amount of thought, especially when I saw the simulations involving electricity and magnetism, I changed my mind. These simulations are very instructive, especially for phenomena that you can't see (e.g., electric and magnetic fields). I encourage you to "play" with them a lot, take measurements as you can, and note that your data should match the physics equations that you are studying.
– David White
Dec 6 at 20:23
add a comment |
So I was using this wave simulation (https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html) when this happened; using the settings: Oscillator, High tension, no damping. Any frequency or amplitude will do as long as it's divisible by 5 I think it's because of calculations.
After sometime, the wave form would collapse on itself leaving a plain rope and restarting. (It has to be with the settings since the beginning)
Is there a name for this?;
Could it happen in the real world or is it purely hypothetical since there's no system like that;
Why with low tension doesn't happen?;
Was it purely coincidental?
waves simulations interference
So I was using this wave simulation (https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html) when this happened; using the settings: Oscillator, High tension, no damping. Any frequency or amplitude will do as long as it's divisible by 5 I think it's because of calculations.
After sometime, the wave form would collapse on itself leaving a plain rope and restarting. (It has to be with the settings since the beginning)
Is there a name for this?;
Could it happen in the real world or is it purely hypothetical since there's no system like that;
Why with low tension doesn't happen?;
Was it purely coincidental?
waves simulations interference
waves simulations interference
edited Dec 6 at 20:00
asked Dec 6 at 17:24
Alef Gabryel
315
315
1
I think that's a bug in the simulation
– John Rennie
Dec 6 at 17:45
Do you mean the points in time where the rope is straight between opposing waveforms?
– Jasper
Dec 6 at 17:46
youtube.com/watch?v=4JuDHYjfkH8&feature=youtu.be
– Alef Gabryel
Dec 6 at 18:07
@JohnRennie, I've used this PhET simulation before and I don't see the "problem" referenced by the OP as a bug. This is a perfectly ideal simulation, with no losses, and I note that the frequency and length of the string are very slightly mismatched regarding the length required to support a 4th harmonic standing wave. Thus, at a few points in time, destructive interference flattens out the wave, and the whole sequence repeats. If the frequency or tension are slightly adjusted, a permanent standing wave can be achieved.
– David White
Dec 6 at 20:19
@AlefGabryel, when I was teaching high school physics, and I first saw PhET simulations, I thought that they were more suitable for middle school. After some amount of thought, especially when I saw the simulations involving electricity and magnetism, I changed my mind. These simulations are very instructive, especially for phenomena that you can't see (e.g., electric and magnetic fields). I encourage you to "play" with them a lot, take measurements as you can, and note that your data should match the physics equations that you are studying.
– David White
Dec 6 at 20:23
add a comment |
1
I think that's a bug in the simulation
– John Rennie
Dec 6 at 17:45
Do you mean the points in time where the rope is straight between opposing waveforms?
– Jasper
Dec 6 at 17:46
youtube.com/watch?v=4JuDHYjfkH8&feature=youtu.be
– Alef Gabryel
Dec 6 at 18:07
@JohnRennie, I've used this PhET simulation before and I don't see the "problem" referenced by the OP as a bug. This is a perfectly ideal simulation, with no losses, and I note that the frequency and length of the string are very slightly mismatched regarding the length required to support a 4th harmonic standing wave. Thus, at a few points in time, destructive interference flattens out the wave, and the whole sequence repeats. If the frequency or tension are slightly adjusted, a permanent standing wave can be achieved.
– David White
Dec 6 at 20:19
@AlefGabryel, when I was teaching high school physics, and I first saw PhET simulations, I thought that they were more suitable for middle school. After some amount of thought, especially when I saw the simulations involving electricity and magnetism, I changed my mind. These simulations are very instructive, especially for phenomena that you can't see (e.g., electric and magnetic fields). I encourage you to "play" with them a lot, take measurements as you can, and note that your data should match the physics equations that you are studying.
– David White
Dec 6 at 20:23
1
1
I think that's a bug in the simulation
– John Rennie
Dec 6 at 17:45
I think that's a bug in the simulation
– John Rennie
Dec 6 at 17:45
Do you mean the points in time where the rope is straight between opposing waveforms?
– Jasper
Dec 6 at 17:46
Do you mean the points in time where the rope is straight between opposing waveforms?
– Jasper
Dec 6 at 17:46
youtube.com/watch?v=4JuDHYjfkH8&feature=youtu.be
– Alef Gabryel
Dec 6 at 18:07
youtube.com/watch?v=4JuDHYjfkH8&feature=youtu.be
– Alef Gabryel
Dec 6 at 18:07
@JohnRennie, I've used this PhET simulation before and I don't see the "problem" referenced by the OP as a bug. This is a perfectly ideal simulation, with no losses, and I note that the frequency and length of the string are very slightly mismatched regarding the length required to support a 4th harmonic standing wave. Thus, at a few points in time, destructive interference flattens out the wave, and the whole sequence repeats. If the frequency or tension are slightly adjusted, a permanent standing wave can be achieved.
– David White
Dec 6 at 20:19
@JohnRennie, I've used this PhET simulation before and I don't see the "problem" referenced by the OP as a bug. This is a perfectly ideal simulation, with no losses, and I note that the frequency and length of the string are very slightly mismatched regarding the length required to support a 4th harmonic standing wave. Thus, at a few points in time, destructive interference flattens out the wave, and the whole sequence repeats. If the frequency or tension are slightly adjusted, a permanent standing wave can be achieved.
– David White
Dec 6 at 20:19
@AlefGabryel, when I was teaching high school physics, and I first saw PhET simulations, I thought that they were more suitable for middle school. After some amount of thought, especially when I saw the simulations involving electricity and magnetism, I changed my mind. These simulations are very instructive, especially for phenomena that you can't see (e.g., electric and magnetic fields). I encourage you to "play" with them a lot, take measurements as you can, and note that your data should match the physics equations that you are studying.
– David White
Dec 6 at 20:23
@AlefGabryel, when I was teaching high school physics, and I first saw PhET simulations, I thought that they were more suitable for middle school. After some amount of thought, especially when I saw the simulations involving electricity and magnetism, I changed my mind. These simulations are very instructive, especially for phenomena that you can't see (e.g., electric and magnetic fields). I encourage you to "play" with them a lot, take measurements as you can, and note that your data should match the physics equations that you are studying.
– David White
Dec 6 at 20:23
add a comment |
4 Answers
4
active
oldest
votes
That's a pretty neat effect! It's not a bug in the simulation, it is correctly solving the equation.
To get a feel for what's going on, recall how a standing wave is formed. If you hold one end of a string and fix the other end, and give your end a wiggle, that wiggle will propagate down the string, bounce off the fixed end, return and bounce off the end you're holding, and so on. In practice, this would damp out quickly, but for an ideal string it would go on forever.
Now, if you continuously wiggle the string, the same process happens for each wiggle. After you've wiggled the string up and down $10$ times, there are $10$ wiggles continually bouncing back and forth.
If you're driven the string at one of the standing wave frequencies, then by the time the first wiggle comes back, you will be producing another with the same phase. They reinforce each other, producing a standing wave whose amplitude grows and grows. If you don't drive at a standing wave frequency, there will be a phase difference. For example, by the time your first wiggle has come back, you might be creating another one $90^circ$ out of phase. The next will be $180^circ$ out of phase, and the next $270^circ$ out of phase, and all four of these will superpose to exactly zero, giving a stationary rope.
In general, this will happen whenever you drive at a rational multiple of the fundamental frequency. The reason the effect doesn't work when you change the tension setting is because that changes the wave speed and hence the frequencies, so you're no longer driving at a rational multiple of the fundamental. It doesn't violate conservation of energy, because for the last two you will be doing negative work on the rope. The effect may even be observable for simple multiples of the fundamental on a real string.
If you want to give this effect a name, it's simply the usual destructive interference, but with the neat twist that a wave you're putting in now is destructively interfering with a wave you put in earlier.
Thank you. Even though someone mentioned the Poincaré recurrence theorem was nine to hear a simplified answer.
– Alef Gabryel
Dec 6 at 19:43
Isn't this answer the same as mine?
– JEB
Dec 7 at 16:35
add a comment |
At first glance I was with @John Rennie: it has to be a bug, but after playing around I think this just passive cancelation (which isn't a named phenomenon, but active cancelation is).
If you add a slight amount of damping, the effect goes away, but the wave still gets pretty small. That's not what I would expect from a bug, with the caveat that predicting the behavior of bugs is tricky.
Back to no damping: if you mix in an frequency component without hitting "restart", the phenomenon disappears, but you can see it might be happening to each component at different times. This is what linear superposition should show for a real phenomenon.
Also: it's hard to land on a resonance with this simulator. With no damping, $Q=infty$, and the resonance is infinitely narrow so you will never hit one.
You might need to work out the details, but he is where I am. I put the thing in "pulse mode" and verified that you can "catch" a reflected pulse, so I believe that after enough reflections, the system "catches" itself.
You start with a pure sine wave:
$$ y_0(x, t) = sin(kx-omega t) $$
then you add a delayed reflection:
$$ y_1(x, t) = sin(-k(x-L)-omega t + phi_1) $$
where you chose $phi_1$ so that it cancels the 1st wave.
Keep doing this an at some point, does everything cancel for one "cycle", but does:
$$sum_1^{n}y_n(x, t) = 0 $$
$forall x in (0, L]$ an $ t_n < t < t_{n+1}$
where $t_i$ is the time of the $i$-th reflection?
I think the answer is "yes".
I don't get about what the answer is for. Also do you think it's possible to predict using harmonic movement? Just in case, there was 18 waves until restart and it had 7 nods, 5 in the cord. Using basic config like 1.50 Hz and 75cm amplitude.
– Alef Gabryel
Dec 6 at 19:50
Also I discover that the amplitude and frequency break the experiment because of calculations. Using multiples of 5 result in the effect tough. No matter which amplitude every 0.5 Hz adds 6 turns to reset the loops.
– Alef Gabryel
Dec 6 at 20:06
add a comment |
i see that there is a secondary wave "riding" on your wave. Its amplitudes goes higher and in a certain point two waves extinguish themselves. Since you dont have any damping, no energy loss... the simulation doesnt seem to have a bug.
add a comment |
This is a very interesting observation, and it is actually a particular case of a much more general theorem, which applies to all Halmitonian Systems or Energy Conserving Systems: The Poincaré Recurrence Theorem (PRT). This important theorem of classical mechanics states that Halmitonian Systems, will, after a sufficiently long but finite time, return to a state arbitrarily close to the initial state. You may find a prove in the book Mathematical Methods of Classical Mechanics, by Vladimir Arnold
In this particular simulation, when you switch off damping, and the tension is sufficiently large, PRT applies. But how can that be if the system motion is forced and energy does not conserve? You may imagine the forcing system as a mass-spring system with an arbitrarily large mass. In this case, the string motion will be forced by another mass in harmonic motion, but this harmonic motion will not be significantly perturbed by the string if the mass is sufficiently larger than the string mass. The system as a whole is now conservative, and PRT applies.
It is important that the frequency of the harmonic oscillation is much lower than the resonance frequency of the string. Otherwise, the string motion amplitude will rise too much and it will eventually perturb the mass-spring motion. That is why the tension of the string must be large.
Many lab real and non-trivial systems reveal this beautiful phenomenon with good approximation. I give here a link to a system of pendulums which return approximately to the initial state (link https://youtu.be/yVkdfJ9PkRQ) at video time 1:27';
1
Very interesting theorem thank you for introducing me to it.
– Alef Gabryel
Dec 6 at 19:47
2
This system doesn't have a conserved energy, since there is an external driving force. So I'm pretty sure that the Poincaré Recurrence Theorem doesn't apply, since the volume of accessible phase space isn't bounded. As an example of this, a mass on a spring that is driven at resonance, without damping, will never return to its original configuration in phase space.
– Michael Seifert
Dec 6 at 20:57
Indeed my first answer was wrong. But I refined the reasoning, and I think the PRT can still be used to explain the recurrence observed in this particular example.
– Marco Ridenti
Dec 7 at 11:11
1
Note that the driver can also remove energy from system.
– JEB
Dec 7 at 16:34
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f445625%2fis-there-a-name-for-this-wave-effect%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
That's a pretty neat effect! It's not a bug in the simulation, it is correctly solving the equation.
To get a feel for what's going on, recall how a standing wave is formed. If you hold one end of a string and fix the other end, and give your end a wiggle, that wiggle will propagate down the string, bounce off the fixed end, return and bounce off the end you're holding, and so on. In practice, this would damp out quickly, but for an ideal string it would go on forever.
Now, if you continuously wiggle the string, the same process happens for each wiggle. After you've wiggled the string up and down $10$ times, there are $10$ wiggles continually bouncing back and forth.
If you're driven the string at one of the standing wave frequencies, then by the time the first wiggle comes back, you will be producing another with the same phase. They reinforce each other, producing a standing wave whose amplitude grows and grows. If you don't drive at a standing wave frequency, there will be a phase difference. For example, by the time your first wiggle has come back, you might be creating another one $90^circ$ out of phase. The next will be $180^circ$ out of phase, and the next $270^circ$ out of phase, and all four of these will superpose to exactly zero, giving a stationary rope.
In general, this will happen whenever you drive at a rational multiple of the fundamental frequency. The reason the effect doesn't work when you change the tension setting is because that changes the wave speed and hence the frequencies, so you're no longer driving at a rational multiple of the fundamental. It doesn't violate conservation of energy, because for the last two you will be doing negative work on the rope. The effect may even be observable for simple multiples of the fundamental on a real string.
If you want to give this effect a name, it's simply the usual destructive interference, but with the neat twist that a wave you're putting in now is destructively interfering with a wave you put in earlier.
Thank you. Even though someone mentioned the Poincaré recurrence theorem was nine to hear a simplified answer.
– Alef Gabryel
Dec 6 at 19:43
Isn't this answer the same as mine?
– JEB
Dec 7 at 16:35
add a comment |
That's a pretty neat effect! It's not a bug in the simulation, it is correctly solving the equation.
To get a feel for what's going on, recall how a standing wave is formed. If you hold one end of a string and fix the other end, and give your end a wiggle, that wiggle will propagate down the string, bounce off the fixed end, return and bounce off the end you're holding, and so on. In practice, this would damp out quickly, but for an ideal string it would go on forever.
Now, if you continuously wiggle the string, the same process happens for each wiggle. After you've wiggled the string up and down $10$ times, there are $10$ wiggles continually bouncing back and forth.
If you're driven the string at one of the standing wave frequencies, then by the time the first wiggle comes back, you will be producing another with the same phase. They reinforce each other, producing a standing wave whose amplitude grows and grows. If you don't drive at a standing wave frequency, there will be a phase difference. For example, by the time your first wiggle has come back, you might be creating another one $90^circ$ out of phase. The next will be $180^circ$ out of phase, and the next $270^circ$ out of phase, and all four of these will superpose to exactly zero, giving a stationary rope.
In general, this will happen whenever you drive at a rational multiple of the fundamental frequency. The reason the effect doesn't work when you change the tension setting is because that changes the wave speed and hence the frequencies, so you're no longer driving at a rational multiple of the fundamental. It doesn't violate conservation of energy, because for the last two you will be doing negative work on the rope. The effect may even be observable for simple multiples of the fundamental on a real string.
If you want to give this effect a name, it's simply the usual destructive interference, but with the neat twist that a wave you're putting in now is destructively interfering with a wave you put in earlier.
Thank you. Even though someone mentioned the Poincaré recurrence theorem was nine to hear a simplified answer.
– Alef Gabryel
Dec 6 at 19:43
Isn't this answer the same as mine?
– JEB
Dec 7 at 16:35
add a comment |
That's a pretty neat effect! It's not a bug in the simulation, it is correctly solving the equation.
To get a feel for what's going on, recall how a standing wave is formed. If you hold one end of a string and fix the other end, and give your end a wiggle, that wiggle will propagate down the string, bounce off the fixed end, return and bounce off the end you're holding, and so on. In practice, this would damp out quickly, but for an ideal string it would go on forever.
Now, if you continuously wiggle the string, the same process happens for each wiggle. After you've wiggled the string up and down $10$ times, there are $10$ wiggles continually bouncing back and forth.
If you're driven the string at one of the standing wave frequencies, then by the time the first wiggle comes back, you will be producing another with the same phase. They reinforce each other, producing a standing wave whose amplitude grows and grows. If you don't drive at a standing wave frequency, there will be a phase difference. For example, by the time your first wiggle has come back, you might be creating another one $90^circ$ out of phase. The next will be $180^circ$ out of phase, and the next $270^circ$ out of phase, and all four of these will superpose to exactly zero, giving a stationary rope.
In general, this will happen whenever you drive at a rational multiple of the fundamental frequency. The reason the effect doesn't work when you change the tension setting is because that changes the wave speed and hence the frequencies, so you're no longer driving at a rational multiple of the fundamental. It doesn't violate conservation of energy, because for the last two you will be doing negative work on the rope. The effect may even be observable for simple multiples of the fundamental on a real string.
If you want to give this effect a name, it's simply the usual destructive interference, but with the neat twist that a wave you're putting in now is destructively interfering with a wave you put in earlier.
That's a pretty neat effect! It's not a bug in the simulation, it is correctly solving the equation.
To get a feel for what's going on, recall how a standing wave is formed. If you hold one end of a string and fix the other end, and give your end a wiggle, that wiggle will propagate down the string, bounce off the fixed end, return and bounce off the end you're holding, and so on. In practice, this would damp out quickly, but for an ideal string it would go on forever.
Now, if you continuously wiggle the string, the same process happens for each wiggle. After you've wiggled the string up and down $10$ times, there are $10$ wiggles continually bouncing back and forth.
If you're driven the string at one of the standing wave frequencies, then by the time the first wiggle comes back, you will be producing another with the same phase. They reinforce each other, producing a standing wave whose amplitude grows and grows. If you don't drive at a standing wave frequency, there will be a phase difference. For example, by the time your first wiggle has come back, you might be creating another one $90^circ$ out of phase. The next will be $180^circ$ out of phase, and the next $270^circ$ out of phase, and all four of these will superpose to exactly zero, giving a stationary rope.
In general, this will happen whenever you drive at a rational multiple of the fundamental frequency. The reason the effect doesn't work when you change the tension setting is because that changes the wave speed and hence the frequencies, so you're no longer driving at a rational multiple of the fundamental. It doesn't violate conservation of energy, because for the last two you will be doing negative work on the rope. The effect may even be observable for simple multiples of the fundamental on a real string.
If you want to give this effect a name, it's simply the usual destructive interference, but with the neat twist that a wave you're putting in now is destructively interfering with a wave you put in earlier.
answered Dec 6 at 19:29
knzhou
40.8k11114196
40.8k11114196
Thank you. Even though someone mentioned the Poincaré recurrence theorem was nine to hear a simplified answer.
– Alef Gabryel
Dec 6 at 19:43
Isn't this answer the same as mine?
– JEB
Dec 7 at 16:35
add a comment |
Thank you. Even though someone mentioned the Poincaré recurrence theorem was nine to hear a simplified answer.
– Alef Gabryel
Dec 6 at 19:43
Isn't this answer the same as mine?
– JEB
Dec 7 at 16:35
Thank you. Even though someone mentioned the Poincaré recurrence theorem was nine to hear a simplified answer.
– Alef Gabryel
Dec 6 at 19:43
Thank you. Even though someone mentioned the Poincaré recurrence theorem was nine to hear a simplified answer.
– Alef Gabryel
Dec 6 at 19:43
Isn't this answer the same as mine?
– JEB
Dec 7 at 16:35
Isn't this answer the same as mine?
– JEB
Dec 7 at 16:35
add a comment |
At first glance I was with @John Rennie: it has to be a bug, but after playing around I think this just passive cancelation (which isn't a named phenomenon, but active cancelation is).
If you add a slight amount of damping, the effect goes away, but the wave still gets pretty small. That's not what I would expect from a bug, with the caveat that predicting the behavior of bugs is tricky.
Back to no damping: if you mix in an frequency component without hitting "restart", the phenomenon disappears, but you can see it might be happening to each component at different times. This is what linear superposition should show for a real phenomenon.
Also: it's hard to land on a resonance with this simulator. With no damping, $Q=infty$, and the resonance is infinitely narrow so you will never hit one.
You might need to work out the details, but he is where I am. I put the thing in "pulse mode" and verified that you can "catch" a reflected pulse, so I believe that after enough reflections, the system "catches" itself.
You start with a pure sine wave:
$$ y_0(x, t) = sin(kx-omega t) $$
then you add a delayed reflection:
$$ y_1(x, t) = sin(-k(x-L)-omega t + phi_1) $$
where you chose $phi_1$ so that it cancels the 1st wave.
Keep doing this an at some point, does everything cancel for one "cycle", but does:
$$sum_1^{n}y_n(x, t) = 0 $$
$forall x in (0, L]$ an $ t_n < t < t_{n+1}$
where $t_i$ is the time of the $i$-th reflection?
I think the answer is "yes".
I don't get about what the answer is for. Also do you think it's possible to predict using harmonic movement? Just in case, there was 18 waves until restart and it had 7 nods, 5 in the cord. Using basic config like 1.50 Hz and 75cm amplitude.
– Alef Gabryel
Dec 6 at 19:50
Also I discover that the amplitude and frequency break the experiment because of calculations. Using multiples of 5 result in the effect tough. No matter which amplitude every 0.5 Hz adds 6 turns to reset the loops.
– Alef Gabryel
Dec 6 at 20:06
add a comment |
At first glance I was with @John Rennie: it has to be a bug, but after playing around I think this just passive cancelation (which isn't a named phenomenon, but active cancelation is).
If you add a slight amount of damping, the effect goes away, but the wave still gets pretty small. That's not what I would expect from a bug, with the caveat that predicting the behavior of bugs is tricky.
Back to no damping: if you mix in an frequency component without hitting "restart", the phenomenon disappears, but you can see it might be happening to each component at different times. This is what linear superposition should show for a real phenomenon.
Also: it's hard to land on a resonance with this simulator. With no damping, $Q=infty$, and the resonance is infinitely narrow so you will never hit one.
You might need to work out the details, but he is where I am. I put the thing in "pulse mode" and verified that you can "catch" a reflected pulse, so I believe that after enough reflections, the system "catches" itself.
You start with a pure sine wave:
$$ y_0(x, t) = sin(kx-omega t) $$
then you add a delayed reflection:
$$ y_1(x, t) = sin(-k(x-L)-omega t + phi_1) $$
where you chose $phi_1$ so that it cancels the 1st wave.
Keep doing this an at some point, does everything cancel for one "cycle", but does:
$$sum_1^{n}y_n(x, t) = 0 $$
$forall x in (0, L]$ an $ t_n < t < t_{n+1}$
where $t_i$ is the time of the $i$-th reflection?
I think the answer is "yes".
I don't get about what the answer is for. Also do you think it's possible to predict using harmonic movement? Just in case, there was 18 waves until restart and it had 7 nods, 5 in the cord. Using basic config like 1.50 Hz and 75cm amplitude.
– Alef Gabryel
Dec 6 at 19:50
Also I discover that the amplitude and frequency break the experiment because of calculations. Using multiples of 5 result in the effect tough. No matter which amplitude every 0.5 Hz adds 6 turns to reset the loops.
– Alef Gabryel
Dec 6 at 20:06
add a comment |
At first glance I was with @John Rennie: it has to be a bug, but after playing around I think this just passive cancelation (which isn't a named phenomenon, but active cancelation is).
If you add a slight amount of damping, the effect goes away, but the wave still gets pretty small. That's not what I would expect from a bug, with the caveat that predicting the behavior of bugs is tricky.
Back to no damping: if you mix in an frequency component without hitting "restart", the phenomenon disappears, but you can see it might be happening to each component at different times. This is what linear superposition should show for a real phenomenon.
Also: it's hard to land on a resonance with this simulator. With no damping, $Q=infty$, and the resonance is infinitely narrow so you will never hit one.
You might need to work out the details, but he is where I am. I put the thing in "pulse mode" and verified that you can "catch" a reflected pulse, so I believe that after enough reflections, the system "catches" itself.
You start with a pure sine wave:
$$ y_0(x, t) = sin(kx-omega t) $$
then you add a delayed reflection:
$$ y_1(x, t) = sin(-k(x-L)-omega t + phi_1) $$
where you chose $phi_1$ so that it cancels the 1st wave.
Keep doing this an at some point, does everything cancel for one "cycle", but does:
$$sum_1^{n}y_n(x, t) = 0 $$
$forall x in (0, L]$ an $ t_n < t < t_{n+1}$
where $t_i$ is the time of the $i$-th reflection?
I think the answer is "yes".
At first glance I was with @John Rennie: it has to be a bug, but after playing around I think this just passive cancelation (which isn't a named phenomenon, but active cancelation is).
If you add a slight amount of damping, the effect goes away, but the wave still gets pretty small. That's not what I would expect from a bug, with the caveat that predicting the behavior of bugs is tricky.
Back to no damping: if you mix in an frequency component without hitting "restart", the phenomenon disappears, but you can see it might be happening to each component at different times. This is what linear superposition should show for a real phenomenon.
Also: it's hard to land on a resonance with this simulator. With no damping, $Q=infty$, and the resonance is infinitely narrow so you will never hit one.
You might need to work out the details, but he is where I am. I put the thing in "pulse mode" and verified that you can "catch" a reflected pulse, so I believe that after enough reflections, the system "catches" itself.
You start with a pure sine wave:
$$ y_0(x, t) = sin(kx-omega t) $$
then you add a delayed reflection:
$$ y_1(x, t) = sin(-k(x-L)-omega t + phi_1) $$
where you chose $phi_1$ so that it cancels the 1st wave.
Keep doing this an at some point, does everything cancel for one "cycle", but does:
$$sum_1^{n}y_n(x, t) = 0 $$
$forall x in (0, L]$ an $ t_n < t < t_{n+1}$
where $t_i$ is the time of the $i$-th reflection?
I think the answer is "yes".
answered Dec 6 at 19:38
JEB
5,6431717
5,6431717
I don't get about what the answer is for. Also do you think it's possible to predict using harmonic movement? Just in case, there was 18 waves until restart and it had 7 nods, 5 in the cord. Using basic config like 1.50 Hz and 75cm amplitude.
– Alef Gabryel
Dec 6 at 19:50
Also I discover that the amplitude and frequency break the experiment because of calculations. Using multiples of 5 result in the effect tough. No matter which amplitude every 0.5 Hz adds 6 turns to reset the loops.
– Alef Gabryel
Dec 6 at 20:06
add a comment |
I don't get about what the answer is for. Also do you think it's possible to predict using harmonic movement? Just in case, there was 18 waves until restart and it had 7 nods, 5 in the cord. Using basic config like 1.50 Hz and 75cm amplitude.
– Alef Gabryel
Dec 6 at 19:50
Also I discover that the amplitude and frequency break the experiment because of calculations. Using multiples of 5 result in the effect tough. No matter which amplitude every 0.5 Hz adds 6 turns to reset the loops.
– Alef Gabryel
Dec 6 at 20:06
I don't get about what the answer is for. Also do you think it's possible to predict using harmonic movement? Just in case, there was 18 waves until restart and it had 7 nods, 5 in the cord. Using basic config like 1.50 Hz and 75cm amplitude.
– Alef Gabryel
Dec 6 at 19:50
I don't get about what the answer is for. Also do you think it's possible to predict using harmonic movement? Just in case, there was 18 waves until restart and it had 7 nods, 5 in the cord. Using basic config like 1.50 Hz and 75cm amplitude.
– Alef Gabryel
Dec 6 at 19:50
Also I discover that the amplitude and frequency break the experiment because of calculations. Using multiples of 5 result in the effect tough. No matter which amplitude every 0.5 Hz adds 6 turns to reset the loops.
– Alef Gabryel
Dec 6 at 20:06
Also I discover that the amplitude and frequency break the experiment because of calculations. Using multiples of 5 result in the effect tough. No matter which amplitude every 0.5 Hz adds 6 turns to reset the loops.
– Alef Gabryel
Dec 6 at 20:06
add a comment |
i see that there is a secondary wave "riding" on your wave. Its amplitudes goes higher and in a certain point two waves extinguish themselves. Since you dont have any damping, no energy loss... the simulation doesnt seem to have a bug.
add a comment |
i see that there is a secondary wave "riding" on your wave. Its amplitudes goes higher and in a certain point two waves extinguish themselves. Since you dont have any damping, no energy loss... the simulation doesnt seem to have a bug.
add a comment |
i see that there is a secondary wave "riding" on your wave. Its amplitudes goes higher and in a certain point two waves extinguish themselves. Since you dont have any damping, no energy loss... the simulation doesnt seem to have a bug.
i see that there is a secondary wave "riding" on your wave. Its amplitudes goes higher and in a certain point two waves extinguish themselves. Since you dont have any damping, no energy loss... the simulation doesnt seem to have a bug.
answered Dec 6 at 19:28
Andy
42
42
add a comment |
add a comment |
This is a very interesting observation, and it is actually a particular case of a much more general theorem, which applies to all Halmitonian Systems or Energy Conserving Systems: The Poincaré Recurrence Theorem (PRT). This important theorem of classical mechanics states that Halmitonian Systems, will, after a sufficiently long but finite time, return to a state arbitrarily close to the initial state. You may find a prove in the book Mathematical Methods of Classical Mechanics, by Vladimir Arnold
In this particular simulation, when you switch off damping, and the tension is sufficiently large, PRT applies. But how can that be if the system motion is forced and energy does not conserve? You may imagine the forcing system as a mass-spring system with an arbitrarily large mass. In this case, the string motion will be forced by another mass in harmonic motion, but this harmonic motion will not be significantly perturbed by the string if the mass is sufficiently larger than the string mass. The system as a whole is now conservative, and PRT applies.
It is important that the frequency of the harmonic oscillation is much lower than the resonance frequency of the string. Otherwise, the string motion amplitude will rise too much and it will eventually perturb the mass-spring motion. That is why the tension of the string must be large.
Many lab real and non-trivial systems reveal this beautiful phenomenon with good approximation. I give here a link to a system of pendulums which return approximately to the initial state (link https://youtu.be/yVkdfJ9PkRQ) at video time 1:27';
1
Very interesting theorem thank you for introducing me to it.
– Alef Gabryel
Dec 6 at 19:47
2
This system doesn't have a conserved energy, since there is an external driving force. So I'm pretty sure that the Poincaré Recurrence Theorem doesn't apply, since the volume of accessible phase space isn't bounded. As an example of this, a mass on a spring that is driven at resonance, without damping, will never return to its original configuration in phase space.
– Michael Seifert
Dec 6 at 20:57
Indeed my first answer was wrong. But I refined the reasoning, and I think the PRT can still be used to explain the recurrence observed in this particular example.
– Marco Ridenti
Dec 7 at 11:11
1
Note that the driver can also remove energy from system.
– JEB
Dec 7 at 16:34
add a comment |
This is a very interesting observation, and it is actually a particular case of a much more general theorem, which applies to all Halmitonian Systems or Energy Conserving Systems: The Poincaré Recurrence Theorem (PRT). This important theorem of classical mechanics states that Halmitonian Systems, will, after a sufficiently long but finite time, return to a state arbitrarily close to the initial state. You may find a prove in the book Mathematical Methods of Classical Mechanics, by Vladimir Arnold
In this particular simulation, when you switch off damping, and the tension is sufficiently large, PRT applies. But how can that be if the system motion is forced and energy does not conserve? You may imagine the forcing system as a mass-spring system with an arbitrarily large mass. In this case, the string motion will be forced by another mass in harmonic motion, but this harmonic motion will not be significantly perturbed by the string if the mass is sufficiently larger than the string mass. The system as a whole is now conservative, and PRT applies.
It is important that the frequency of the harmonic oscillation is much lower than the resonance frequency of the string. Otherwise, the string motion amplitude will rise too much and it will eventually perturb the mass-spring motion. That is why the tension of the string must be large.
Many lab real and non-trivial systems reveal this beautiful phenomenon with good approximation. I give here a link to a system of pendulums which return approximately to the initial state (link https://youtu.be/yVkdfJ9PkRQ) at video time 1:27';
1
Very interesting theorem thank you for introducing me to it.
– Alef Gabryel
Dec 6 at 19:47
2
This system doesn't have a conserved energy, since there is an external driving force. So I'm pretty sure that the Poincaré Recurrence Theorem doesn't apply, since the volume of accessible phase space isn't bounded. As an example of this, a mass on a spring that is driven at resonance, without damping, will never return to its original configuration in phase space.
– Michael Seifert
Dec 6 at 20:57
Indeed my first answer was wrong. But I refined the reasoning, and I think the PRT can still be used to explain the recurrence observed in this particular example.
– Marco Ridenti
Dec 7 at 11:11
1
Note that the driver can also remove energy from system.
– JEB
Dec 7 at 16:34
add a comment |
This is a very interesting observation, and it is actually a particular case of a much more general theorem, which applies to all Halmitonian Systems or Energy Conserving Systems: The Poincaré Recurrence Theorem (PRT). This important theorem of classical mechanics states that Halmitonian Systems, will, after a sufficiently long but finite time, return to a state arbitrarily close to the initial state. You may find a prove in the book Mathematical Methods of Classical Mechanics, by Vladimir Arnold
In this particular simulation, when you switch off damping, and the tension is sufficiently large, PRT applies. But how can that be if the system motion is forced and energy does not conserve? You may imagine the forcing system as a mass-spring system with an arbitrarily large mass. In this case, the string motion will be forced by another mass in harmonic motion, but this harmonic motion will not be significantly perturbed by the string if the mass is sufficiently larger than the string mass. The system as a whole is now conservative, and PRT applies.
It is important that the frequency of the harmonic oscillation is much lower than the resonance frequency of the string. Otherwise, the string motion amplitude will rise too much and it will eventually perturb the mass-spring motion. That is why the tension of the string must be large.
Many lab real and non-trivial systems reveal this beautiful phenomenon with good approximation. I give here a link to a system of pendulums which return approximately to the initial state (link https://youtu.be/yVkdfJ9PkRQ) at video time 1:27';
This is a very interesting observation, and it is actually a particular case of a much more general theorem, which applies to all Halmitonian Systems or Energy Conserving Systems: The Poincaré Recurrence Theorem (PRT). This important theorem of classical mechanics states that Halmitonian Systems, will, after a sufficiently long but finite time, return to a state arbitrarily close to the initial state. You may find a prove in the book Mathematical Methods of Classical Mechanics, by Vladimir Arnold
In this particular simulation, when you switch off damping, and the tension is sufficiently large, PRT applies. But how can that be if the system motion is forced and energy does not conserve? You may imagine the forcing system as a mass-spring system with an arbitrarily large mass. In this case, the string motion will be forced by another mass in harmonic motion, but this harmonic motion will not be significantly perturbed by the string if the mass is sufficiently larger than the string mass. The system as a whole is now conservative, and PRT applies.
It is important that the frequency of the harmonic oscillation is much lower than the resonance frequency of the string. Otherwise, the string motion amplitude will rise too much and it will eventually perturb the mass-spring motion. That is why the tension of the string must be large.
Many lab real and non-trivial systems reveal this beautiful phenomenon with good approximation. I give here a link to a system of pendulums which return approximately to the initial state (link https://youtu.be/yVkdfJ9PkRQ) at video time 1:27';
edited Dec 7 at 11:27
answered Dec 6 at 19:14
Marco Ridenti
815
815
1
Very interesting theorem thank you for introducing me to it.
– Alef Gabryel
Dec 6 at 19:47
2
This system doesn't have a conserved energy, since there is an external driving force. So I'm pretty sure that the Poincaré Recurrence Theorem doesn't apply, since the volume of accessible phase space isn't bounded. As an example of this, a mass on a spring that is driven at resonance, without damping, will never return to its original configuration in phase space.
– Michael Seifert
Dec 6 at 20:57
Indeed my first answer was wrong. But I refined the reasoning, and I think the PRT can still be used to explain the recurrence observed in this particular example.
– Marco Ridenti
Dec 7 at 11:11
1
Note that the driver can also remove energy from system.
– JEB
Dec 7 at 16:34
add a comment |
1
Very interesting theorem thank you for introducing me to it.
– Alef Gabryel
Dec 6 at 19:47
2
This system doesn't have a conserved energy, since there is an external driving force. So I'm pretty sure that the Poincaré Recurrence Theorem doesn't apply, since the volume of accessible phase space isn't bounded. As an example of this, a mass on a spring that is driven at resonance, without damping, will never return to its original configuration in phase space.
– Michael Seifert
Dec 6 at 20:57
Indeed my first answer was wrong. But I refined the reasoning, and I think the PRT can still be used to explain the recurrence observed in this particular example.
– Marco Ridenti
Dec 7 at 11:11
1
Note that the driver can also remove energy from system.
– JEB
Dec 7 at 16:34
1
1
Very interesting theorem thank you for introducing me to it.
– Alef Gabryel
Dec 6 at 19:47
Very interesting theorem thank you for introducing me to it.
– Alef Gabryel
Dec 6 at 19:47
2
2
This system doesn't have a conserved energy, since there is an external driving force. So I'm pretty sure that the Poincaré Recurrence Theorem doesn't apply, since the volume of accessible phase space isn't bounded. As an example of this, a mass on a spring that is driven at resonance, without damping, will never return to its original configuration in phase space.
– Michael Seifert
Dec 6 at 20:57
This system doesn't have a conserved energy, since there is an external driving force. So I'm pretty sure that the Poincaré Recurrence Theorem doesn't apply, since the volume of accessible phase space isn't bounded. As an example of this, a mass on a spring that is driven at resonance, without damping, will never return to its original configuration in phase space.
– Michael Seifert
Dec 6 at 20:57
Indeed my first answer was wrong. But I refined the reasoning, and I think the PRT can still be used to explain the recurrence observed in this particular example.
– Marco Ridenti
Dec 7 at 11:11
Indeed my first answer was wrong. But I refined the reasoning, and I think the PRT can still be used to explain the recurrence observed in this particular example.
– Marco Ridenti
Dec 7 at 11:11
1
1
Note that the driver can also remove energy from system.
– JEB
Dec 7 at 16:34
Note that the driver can also remove energy from system.
– JEB
Dec 7 at 16:34
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f445625%2fis-there-a-name-for-this-wave-effect%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
I think that's a bug in the simulation
– John Rennie
Dec 6 at 17:45
Do you mean the points in time where the rope is straight between opposing waveforms?
– Jasper
Dec 6 at 17:46
youtube.com/watch?v=4JuDHYjfkH8&feature=youtu.be
– Alef Gabryel
Dec 6 at 18:07
@JohnRennie, I've used this PhET simulation before and I don't see the "problem" referenced by the OP as a bug. This is a perfectly ideal simulation, with no losses, and I note that the frequency and length of the string are very slightly mismatched regarding the length required to support a 4th harmonic standing wave. Thus, at a few points in time, destructive interference flattens out the wave, and the whole sequence repeats. If the frequency or tension are slightly adjusted, a permanent standing wave can be achieved.
– David White
Dec 6 at 20:19
@AlefGabryel, when I was teaching high school physics, and I first saw PhET simulations, I thought that they were more suitable for middle school. After some amount of thought, especially when I saw the simulations involving electricity and magnetism, I changed my mind. These simulations are very instructive, especially for phenomena that you can't see (e.g., electric and magnetic fields). I encourage you to "play" with them a lot, take measurements as you can, and note that your data should match the physics equations that you are studying.
– David White
Dec 6 at 20:23