Why must traveling waves have the same amplitude to form a standing wave?
$begingroup$
I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?
waves
asked yesterday
Julia KimJulia Kim
826
$endgroup$
add a comment |
$begingroup$
I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?
waves
asked yesterday
Julia KimJulia Kim
826
$endgroup$
$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
19 hours ago
add a comment |
$begingroup$
I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?
waves
asked yesterday
Julia KimJulia Kim
826
$endgroup$
I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?
waves
waves
asked yesterday
Julia KimJulia Kim
826
asked yesterday
Julia KimJulia Kim
826
asked yesterday
Julia KimJulia Kim
826
asked yesterday
Julia KimJulia Kim
826
asked yesterday
Julia KimJulia Kim
826
826
$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
19 hours ago
add a comment |
$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
19 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.
In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.
If the travelling waves are of unequal amplitude then there is a net transfer of energy.
If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.
In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.
The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.
answered yesterday
FarcherFarcher
51k339107
$endgroup$
1
$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago
2
$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago
$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago
$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.
answered 19 hours ago
RuslanRuslan
9,69343173
$endgroup$
1
$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago
2
$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago
1
$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago
$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago
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%2f466359%2fwhy-must-traveling-waves-have-the-same-amplitude-to-form-a-standing-wave%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.
In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.
If the travelling waves are of unequal amplitude then there is a net transfer of energy.
If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.
In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.
The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.
answered yesterday
FarcherFarcher
51k339107
$endgroup$
1
$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago
2
$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago
$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago
$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.
In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.
If the travelling waves are of unequal amplitude then there is a net transfer of energy.
If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.
In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.
The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.
answered yesterday
FarcherFarcher
51k339107
$endgroup$
1
$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago
2
$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago
$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago
$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.
In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.
If the travelling waves are of unequal amplitude then there is a net transfer of energy.
If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.
In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.
The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.
answered yesterday
FarcherFarcher
51k339107
$endgroup$
If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.
In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.
If the travelling waves are of unequal amplitude then there is a net transfer of energy.
If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.
In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.
The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.
answered yesterday
FarcherFarcher
51k339107
edited yesterday
answered yesterday
FarcherFarcher
51k339107
answered yesterday
FarcherFarcher
51k339107
answered yesterday
FarcherFarcher
51k339107
51k339107
1
$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago
2
$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago
$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago
$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago
add a comment |
1
$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago
2
$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago
$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago
$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago
1
1
$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago
$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago
$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago
2
2
$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago
$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago
$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago
$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago
$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago
$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.
answered 19 hours ago
RuslanRuslan
9,69343173
$endgroup$
1
$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago
2
$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago
1
$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago
$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago
add a comment |
$begingroup$
You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.
answered 19 hours ago
RuslanRuslan
9,69343173
$endgroup$
1
$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago
2
$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago
1
$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago
$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago
add a comment |
$begingroup$
You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.
answered 19 hours ago
RuslanRuslan
9,69343173
$endgroup$
You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.
answered 19 hours ago
RuslanRuslan
9,69343173
answered 19 hours ago
RuslanRuslan
9,69343173
answered 19 hours ago
RuslanRuslan
9,69343173
answered 19 hours ago
RuslanRuslan
9,69343173
9,69343173
1
$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago
2
$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago
1
$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago
$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago
add a comment |
1
$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago
2
$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago
1
$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago
$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago
1
1
$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago
$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago
2
2
$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago
$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago
1
1
$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago
$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago
$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago
$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago
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.
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%2f466359%2fwhy-must-traveling-waves-have-the-same-amplitude-to-form-a-standing-wave%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
$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
19 hours ago