# Traveling and Standing Wave Equations

1. Feb 21, 2017

### Taniaz

1. The problem statement, all variables and given/known data
I am really confused with the equations.
For travelling waves, in some places they write y = A cos (kx - wt) for waves travelling in the positive x-direction. Then some write y = A sin (wt - kx) for waves travelling in the positive x-direction while others write
y = - A sin (wt - kx) or y = A sin (kx - wt) all for waves travelling in the positive x - direction.

Then for standing waves, in the example in our book, they used the case of an incident wave at a fixed end causing an inverted reflection so they used y = A cos (kx - wt) for positive and y = - A cos (kx + wt) for the negative and inverted reflected wave to get a standing wave equation of y = 2 A sin kx sin wt whereas in other places they used one of the million above mentioned sine functions and got an answer of y = 2 A sin kx cos wt.

So confused as to what is right! They want us to use the sine functions. Please help, thank you.

2. Relevant equations
y = A cos (kx - wt)
y = A sin (wt - kx)
y = - A sin (wt - kx) or y = A sin (kx - wt)
y = 2 A sin kx sin wt
y = 2 A sin kx cos wt

3. The attempt at a solution

As mentioned above.

2. Feb 21, 2017

### kuruman

Consider the wave described by y(x,t) = A sin(kx - ωt) at time t. At time t + T/2 (T = one period) it will be y(x, t+T/2) = - A sin(kx - ωt). It all depends on when the clock that measures time t is started. Does that help?

3. Feb 21, 2017

### Taniaz

How did you know it was -Asin(kx-wt) after t plus half a period?
And when is it A sin (wt-kx)?

4. Feb 21, 2017

### BvU

I know Kuru is helping you here, but I'd like to throw in my favourite: "Make a drawing"!

In this case a sine wave on a piece of transparent paper that you can move over an identical sine wave on another piece of paper is very helpful too.

or this animation from wiki

5. Feb 21, 2017

### kuruman

If you take a snapshot of a wave at time t = 0, it will look something like this

Is it a sine or cosine? It's neither and is called a "sinusoid" which could be either a sine or a cosine. If you were to take a second snapshot one period (or two or three) later, the picture would look exactly the same. If you were to take a snapshot half a period later, the picture would look upside down (the negative) from this.
It depends on when I choose my time t = 0 axis to be. Here is the same sinusoid chosen to be a sine

And here is the same sinusoid chosen to be a cosine

6. Feb 21, 2017

### kuruman

I was in the process of doing just that for OP when you made the suggestion.

7. Feb 21, 2017

### Taniaz

Ok that is starting to make sense but what is the difference between
y = A sin (kx -wt), y = A sin (wt-kx), y = - Asin (wt-kx) and y = -Asin (kx-wt) because people seem to be using all 4.

8. Feb 21, 2017

### kuruman

A trig identity says $\sin (-\theta) = - \sin (\theta)$ This should explain it.

9. Feb 21, 2017

### Taniaz

Yes that makes sense but between
y=Asin(kx-wt) and y=Asin(wt-kx) are both correct as they are? And obviously if you take the minus out it will become as you said.

Also how many forms can you have for the standing wave equation? The example my book took was of the superposition of an incident wave and an inverted reflected wave because of the fixed end.

I've seen other places where they've just taken the superposition of an incident wave and a non-inverted reflected pulse (free end).

Do all of them stand corrected as well?

10. Feb 21, 2017

### haruspex

Maybe you've already got this from the other responses.
Switching between sine and cosine only changes the phase.
Switching between kx-wt and wt-kx switches the sign of the amplitude if it's sine, and does nothing if it's cosine.
Switching the sign of the amplitude inverts the wave, which is equivalent to a 180 degree phase shift.
None of these changes the direction of travel. To do that, you have to make increasing time and increasing displacement do the same thing instead of opposite things, i.e. change it to kx+wt.

For standing waves, as I wrote, inverting the wave equates to a half period phase shift. If you have the two waves moving in opposite directions, sin(kx+wt)+sin(kx-wt), and move your time reference by a quarter phase, each wave moves by a quarter phase but in opposite directions, producing a net difference of a half wave shift. So if the combination is a standing wave, a half wave shift in one is still a standing wave.
The choice depends on where a node is known to be, from physical considerations.

11. Feb 22, 2017

### Taniaz

Got it, thank you!