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







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  • $begingroup$
    Related question and discussions here
    $endgroup$
    – Aaron Stevens
    29 mins ago
















3












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Related question and discussions here
    $endgroup$
    – Aaron Stevens
    29 mins ago














3












3








3


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






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asked 7 hours ago









Julia KimJulia Kim

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  • $begingroup$
    Related question and discussions here
    $endgroup$
    – Aaron Stevens
    29 mins ago


















  • $begingroup$
    Related question and discussions here
    $endgroup$
    – Aaron Stevens
    29 mins ago
















$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
29 mins ago




$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
29 mins ago










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.



enter image description here



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.
enter image description here



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.



enter image description here






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$endgroup$













  • $begingroup$
    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
    30 mins ago





















2












$begingroup$

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.






share|cite|improve this answer









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    2 Answers
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    2 Answers
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    13












    $begingroup$

    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.



    enter image description here



    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.
    enter image description here



    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.



    enter image description here






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
      30 mins ago


















    13












    $begingroup$

    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.



    enter image description here



    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.
    enter image description here



    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.



    enter image description here






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
      30 mins ago
















    13












    13








    13





    $begingroup$

    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.



    enter image description here



    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.
    enter image description here



    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.



    enter image description here






    share|cite|improve this answer











    $endgroup$



    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.



    enter image description here



    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.
    enter image description here



    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.



    enter image description here







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 6 hours ago

























    answered 6 hours ago









    FarcherFarcher

    50.9k338107




    50.9k338107












    • $begingroup$
      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
      30 mins 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
      30 mins 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
    30 mins 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
    30 mins ago













    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        RuslanRuslan

        9,57343172




        9,57343172






























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