How exactly are standing waves formed?

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I understand the equations for the nth harmonic and so on, but I really don't get how a standing wave is formed. My physics teacher said that an incident wave and an identical reflected wave interfering would cause a stationary wave, but I still don't understand how this happens. For example, what should be the phase difference between the two waves, in order for a standing wave to form?
 

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  • #2
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You can figure this out. Write down the equations for the incident wave and the reflected wave. Sum them. The sum must be independent of time. That gives you the condition for the phase difference.
 
  • #3
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You can figure this out. Write down the equations for the incident wave and the reflected wave. Sum them. The sum must be independent of time. That gives you the condition for the phase difference.

But this is beyond the scope of my syllabus
 
  • #4
But this is beyond the scope of my syllabus
Have you taken into consideration the principle of superposition?
 
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But this is beyond the scope of my syllabus
What is beyond the scope of your syllabus? Knowing the sinusoidal wave equation or being able to do simple trigonometry?
 
  • #6
AlephZero
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But this is beyond the scope of my syllabus
Why is that a problem? That doesn't mean it's too hard for you to understand!

This might help (don't be put off by the cheesy introduction to the real video).
https://www.youtube.com/watch?v=DovunOxlY1k
 
  • #7
olivermsun
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I think there was just a thread discussing this topic. :wink:
 
  • #8
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What is beyond the scope of your syllabus? Knowing the sinusoidal wave equation or being able to do simple trigonometry?
Knowing the sinusoidal wave equation and what it actually means and how to obtain the phase difference given two sinusoidal wave equations.
The math is easy, I am well trained in trigonometry, calculus and algebra. I study A level Mathematics and A level physics.
 
  • #9
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Why is that a problem? That doesn't mean it's too hard for you to understand!

This might help (don't be put off by the cheesy introduction to the real video).
https://www.youtube.com/watch?v=DovunOxlY1k
Why is a wave reflected upside down if it's a rigid end?
 
  • #10
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Why is it reflected upside down?
 
  • #12
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In simple terms, Two coherent waves (or one wave that has been reflected back upon its self), interact and under go constructive and destructive superposition. At points where the amplitudes are completely opposite (http://leiferlingsson.files.lege.net/2008/07/standing_waves_400x300.jpg) the amplitudes cancel and form a node. Where the amplitudes rare both on the same side of the x axis they reinforce each other, The standing wave is merely the resultant of two wave forms.
 
  • #13
HallsofIvy
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Why is it reflected upside down?
Here is an interesting way of seeing why:

Imagine two identical waves at opposite ends of a long string. They meet at "x= 0". If the two waves are on the same side of the string when they meet they add so that the height at x= 0 will be twice as high as the two separately. If they are on opposite sides, they will "subtract" and the point at x= 0 will not move.

Since we can think of the reflected wave as a wave coming from "beyond" the end, if the end cannot move, then the reflected wave must be upside down.
 
  • #14
sophiecentaur
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Why is it reflected upside down?
Another way to look at it:

The idea of Boundary Conditions is a useful one to get familiar with and the boundary condition is that there is no net displacement at the end. Take the example of a string with a transverse wave. The instantaneous force on the fixed end point, due to the incident wave is in one direction and with a certain value. That point is in equilibrium (is doesn't move) so there must be an equal and opposite force 'from' the fixing. This force will be generating a reflected wave with an equal and opposite force at all times. This corresponds to launching a wave that is in complete anti phase with the incident wave.
 

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