Is y = a[log(x - vt)] a Traveling Wave? Understanding the General Wave Equation

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SUMMARY

The equation y = a[log(x – vt)] does not represent a traveling wave, as it is not a harmonic function. The discussion emphasizes the importance of applying the d'Alembertian operator, represented as \Box = \nabla^2 - {1 \over c^2} \frac{\partial^2}{\partial t^2}, to verify wave characteristics. The conclusion is that while the equation can be analyzed using a non-relativistic d'Alembertian, it does not fall within the class of solutions typically associated with wave equations, which are primarily harmonic functions.

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Amith2006
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Sir,
Does the equation,
y = a[log(x – vt)]
Represent a traveling wave?
I think the answer is No, because it is not a harmonic function. Is it right?
 
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Why don't you just plug it in the d'Alembertian operator and see if both sides are equal?
[tex]\Box = \nabla^2 - { 1 \over c^2} \frac{ \partial^2} { \partial t^2}[/tex]
 
Reshma said:
Why don't you just plug it in the d'Alembertian operator and see if both sides are equal?
[tex]\Box = \nabla^2 - { 1 \over c^2} \frac{ \partial^2} { \partial t^2}[/tex]

Yes with [itex]\Box^{2}y = 0[/itex] and the c replaced by v of course so its really a non-relativistic d'Alembertian.

But more than that you should know what CLASS of solutions can be solutions of the wave equation...harmonic functions is a subset of this class.
 
Last edited:

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