Why must traveling waves have the same amplitude to form a standing wave?
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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 6 hours ago
Julia KimJulia Kim
374
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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 6 hours ago
Julia KimJulia Kim
374
$endgroup$
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 6 hours ago
Julia KimJulia Kim
374
$endgroup$
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 6 hours ago
Julia KimJulia Kim
374
asked 6 hours ago
Julia KimJulia Kim
374
asked 6 hours ago
Julia KimJulia Kim
374
asked 6 hours ago
Julia KimJulia Kim
374
asked 6 hours ago
Julia KimJulia Kim
374
374
add a comment |
add a comment |
2 Answers
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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 5 hours ago
FarcherFarcher
50.9k338107
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$begingroup$
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 1 min ago
RuslanRuslan
9,55343172
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2 Answers
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2 Answers
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$begingroup$
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 5 hours ago
FarcherFarcher
50.9k338107
$endgroup$
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 5 hours ago
FarcherFarcher
50.9k338107
$endgroup$
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 5 hours ago
FarcherFarcher
50.9k338107
$endgroup$
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 5 hours ago
FarcherFarcher
50.9k338107
edited 5 hours ago
answered 5 hours ago
FarcherFarcher
50.9k338107
answered 5 hours ago
FarcherFarcher
50.9k338107
answered 5 hours ago
FarcherFarcher
50.9k338107
50.9k338107
add a comment |
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 1 min ago
RuslanRuslan
9,55343172
$endgroup$
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 1 min ago
RuslanRuslan
9,55343172
$endgroup$
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 1 min ago
RuslanRuslan
9,55343172
$endgroup$
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 1 min ago
RuslanRuslan
9,55343172
answered 1 min ago
RuslanRuslan
9,55343172
answered 1 min ago
RuslanRuslan
9,55343172
answered 1 min ago
RuslanRuslan
9,55343172
9,55343172
add a comment |
add a comment |
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