Standing waves - Wave Equation

  • Thread starter elemis
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Main Question or Discussion Point

I don't completely understand how equation 4.4.4 was derived and determined. I understand the derivation behind the basic wave equation 4.3.4 but not what happened in 4.4.4. Why is there a need for all the negative signs ? Would a simple phase change suffice ?

Please do be a bit detailed in your explanation.... Pretend I'm an idiot. Thank you !




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Answers and Replies

  • #2
Redbelly98
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Since the reflected wave travels in the opposite direction as the incident wave, you need to flip the sign on either the kx or the ωt term***. The author chose to reverse the sign of kx to make it -kx. There was already a - sign with -ωt. For the phase term, the choice of sign is arbitrary.

Just changing the phase would express a wave traveling in the same direction as the incident wave, but shifted in phase.

Hope that helps.

EDIT added:
*** This is because a rightward-traveling wave has the form f(kx-ωt) or f(-kx+ωt). A leftward-traveling wave has the form f(kx+ωt) or f(-kx-ωt).
 
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So I could equally have written kx+wt ? Why is there a need for the phase change ∅ ?
 
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Redbelly98
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So I could equally have written kx+wt ?
Yes; see the edit added to my earlier post.
Why is there a need for the phase change ∅ ?
As the book says, the type of boundary will determine ∅. Usually the boundary is either fixed or has a maximum amplitude. Also, the location of the boundary plays a role in what ∅ is.

Eg., for a fixed end located at x=x0:

[tex]\cos(kx_0 + \phi /2) = 0[/tex]

You'd solve that for ∅, given k and x0. Your book is taking the fixed end to be at x=0, so that simplifies things somewhat.
 

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