Question

Wave Help!

Answer 1

@IrishBoy123

Answer 2

I have seen in many textbooks stating that equation for a plane progressive wave is

\[y=a \sin (kx-\omega t)\] if the wave is moving in +ve direction but in some books they say it is
\[y=a \sin (\omega t -kx)\]

Are these both same or different?
@IrishBoy123 @Michele_Laino @ganeshie8

Answer 3

How can they be both same when \(\sin(x)\ne \sin(-x)\) ?

Answer 4

the first one is more orthodox as it's travelling to right from x = 0

so maybe also see if you can manipulate the second to get a phase difference. sin (π-x) = sin(-x)... rather than a backward moving which is what it looks like

Answer 5

but as a slight aside.... maybe! ....... anything in the form f(kx - wt) satisfies the wave equation as does, it would seem combinations thereof, including the negative version......

so they're all OK in that sense

Answer 6

Yes, so they both represent travelling wave equation but they are not exact if taken in same sense. I'll have to follow any one way but I can't intermix them. Right?

Answer 7

you can mix them to see the interference pattern. that's one of the cool things about it!

Answer 8

https://www.desmos.com/calculator

Answer 9

here is my reasoning:
we can write a plane wave in this form:
\[\Large a\sin \left( {kx - \omega t} \right) = \Im \left\{ {a\exp i\left( {kx - \omega t} \right)} \right\}\]
furthermore, we have:
\[\Large \begin{gathered}
\sin \left( {kx - \omega t + \pi } \right) = \sin \left( {kx - \omega t} \right) \cdot \left( { - 1} \right) = \hfill \\
\hfill \\
= - \sin \left( {kx - \omega t} \right) = \sin \left( {\omega t - kx} \right) \hfill \\
\end{gathered} \]
therefore:
\[\Large \begin{gathered}
{\psi _1} = a\exp i\left( {\omega t - kx} \right) = a\exp i\left( {kx - \omega t} \right) \cdot \exp \left( {i\pi } \right) = \hfill \\
\hfill \\
= {\psi _2} \cdot \exp \left( {i\pi } \right) \hfill \\
\end{gathered} \]
in other words the plane waves:
\[\Large {\psi _1} = a\exp i\left( {\omega t - kx} \right),\quad {\psi _2} = a\exp i\left( {kx - \omega t} \right)\]
differ each from other, by a constant phase factor, so, thanks to \(Ehrenfest\), such waves represent the same particle

Answer 10

Yes!
I get it now. Thank you @Michele_Laino @IrishBoy123 and @ganeshie8 !