- #1

- 163

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

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- Thread starter elemis
- Start date

- #1

- 163

- 1

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

- #2

- 12,122

- 160

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*).

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(

Last edited:

- #3

- 163

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So I could equally have written kx+wt ? Why is there a need for the phase change ∅ ?

- #4

- 12,122

- 160

Yes; see the edit added to my earlier post.So I could equally have written kx+wt ?

As the book says, the type of boundary will determine ∅. Usually the boundary is either fixed or has a maximum amplitude. Also, theWhy is there a need for the phase change ∅ ?

Eg., for a fixed end located at

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

You'd solve that for ∅, given k and

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