Traveling wave solution notation

In summary, the conversation discusses the difference between writing ##f(-kx + \omega*t)## and ##f(kx - \omega*t)## for right traveling waves. It is concluded that there is no difference if the function ##f## is even, but there is a difference if the function is odd. The conversation also touches on the derivation of reflectance and transmittance for a traveling wave at a boundary.
  • #1
baseballfan_ny
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TL;DR Summary
Is there a difference between writing ## f(-kx + \omega*t) ## and ## f(kx - \omega*t) ## for right traveling waves?
This is probably kind of dumb, but it's really bothering me for some reason. I originally saw traveling wave solutions to the wave equation as ##f(kx−\omega t)## for right traveling (as t gets bigger, x needs to be bigger to "match" it's previous value) and ##f(kx+\omega t)## for left-traveling waves. And that all made sense to me. Then I saw some people writing ##f(−kx+\omega t)## for right-travelling waves. I'm pretty sure it's the same thing right? Like this also says that as t gets larger x needs to be larger to match it's original value? Is it just a notation preference when choosing between the two?

My confusion sort of stemmed from seeing the derivation of the reflection and transmittance for a traveling wave at a boundary:

IMG_20210326_104726998.jpg


where ##f_i## is the incident pulse, ##f_r## is the reflected, and ##f_t## is the transmitted.

The derivation for the reflectance and transmittance is like:

Because the wave function has to be continuous at the boundary, x = 0:
## f_i(\omega t) + f_r(\omega t) = f_t(\omega t) ##

Because it's space derivative has to be continuous at x = 0 so it doesn't have infinite acceleration:
## -k_1 f'_i(\omega t) + k_1 f'_r(\omega t) = -k_2 f'_t(\omega t) ##

And then integrating both sides of the second equation with respect to t
## \frac {-k_1} {\omega} f_i(\omega t) + \frac {k_1} {\omega} f_i(\omega t) = \frac {-k_2} {\omega} f_t(\omega t) ##

And then by the dispersion relation ## \omega = v*k ## so ## \frac {k} {\omega} = \frac {1} {v} ## ...
## \frac {-1} {v_1} f_i(\omega t) + \frac {1} {v_1} f_r(\omega t) = \frac {-1} {v_2} f_t(\omega t) ##

which gives ## -v_2 (f_i(\omega t) - f_r(\omega t)) = -v_1 (f_t(\omega t)) ##

and then we can solve this equation and the continuity of f and get

## f_r(\omega t) = \frac {v_2 - v_1} {v_1 + v_2} f_i(\omega t) = R f_i(\omega t) ##
## f_t(\omega t) = \frac {2v_2} {v_1 + v_2} f_i(\omega t) = T f_i(\omega t) ##

The problem is is that if I do this whole thing by saying the incident, right traveling wave is ##f_i(k_1x - \omega t)##, the reflected left traveling wave is ##f_r(k_1 + \omega t) ##, and the transmitted wave is ##f_t(k_2x - \omega t)##

I get these two equations to solve (using the same methods as above)

##f_i(\omega t) + f_r(\omega t) = f_t(\omega t)##
##v_2(f_i(\omega t) + f_r(\omega t)) = v_1 f_t(\omega t)##

which gives me ##f_r = \frac {v_2 - v_1} {v_1 - v_2} f_i = -f_i## and ##f_t = 0##.

So why is it necessary to write the right traveling waves as ##f_i(-kx + \omega t) ## and why does ##f_i(kx - \omega t) ## not work instead?
 

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  • #2
baseballfan_ny said:
Summary:: Is there a difference between writing ## f(-kx + \omega*t) ## and ## f(kx - \omega*t) ## for right traveling waves?
It depends if ##f## is even or not. If ##f## is even then there is no difference. Otherwise there is a difference.
 
  • #3
Dale said:
It depends if ##f## is even or not. If ##f## is even then there is no difference. Otherwise there is a difference.

Ok. I'm not sure I 100% understand the difference. I drew some odd waves and they still seem to shift right with both forms.

The wave I was dealing with in post 1 is even, so I based on what you wrote I would expect both forms to give the same R and T. Turns out I made a stupid algebra mistake in this line of post 1:

baseballfan_ny said:
fi(ωt)+fr(ωt)=ft(ωt)

The ##f_i## and ##f_t## should have a negative argument, and then I follow through with the rest of the algebra and get the same answer -- as they're both even.
 
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  • #4
baseballfan_ny said:
I drew some odd waves and they still seem to shift right with both forms.
Yes, they will still shift right, but they will have a different shape with the two forms.
 
  • #5
baseballfan_ny said:
Summary:: Is there a difference between writing ## f(-kx + \omega*t) ## and ## f(kx - \omega*t) ## for right traveling waves?
They both represent a wave traveling to the right. Obviously, for a given wave, you'd use different ##f##s in the two cases.
 
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Related to Traveling wave solution notation

1. What is the meaning of the notation "u(x-ct)" in a traveling wave solution?

The notation "u(x-ct)" represents a traveling wave solution where "u" is the wave amplitude, "x" is the position variable, and "ct" is the product of the wave speed "c" and time "t". This notation indicates that the wave is moving in the positive direction with a constant speed of "c".

2. How is the direction of the wave determined by the notation "u(x-ct)"?

The direction of the wave is determined by the sign of the coefficient of "x" in the notation "x-ct". If the coefficient is positive, the wave is traveling in the positive direction, and if it is negative, the wave is traveling in the negative direction.

3. What does the "u" represent in the notation "u(x-ct)"?

The "u" represents the wave amplitude, which is the maximum displacement of the wave from its equilibrium position. This value remains constant as the wave travels through space and time.

4. How does the notation "u(x-ct)" relate to the wave equation?

The notation "u(x-ct)" is a solution to the wave equation, which describes the behavior of waves in a given medium. It represents a traveling wave solution where the wave moves with a constant speed of "c" and maintains a constant amplitude "u".

5. Can the notation "u(x-ct)" be used to represent any type of wave?

No, the notation "u(x-ct)" is specific to traveling wave solutions and cannot be used to represent other types of waves such as standing waves or damped waves. These types of waves have different mathematical representations and cannot be described by the same notation.

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