# Phase difference of a standing wave

Before starting the actual question: I do not want any typical answer that anybody might have thought of or criticism and downvote without even reading the question properly.

I have been googling and seeking answer to this issue for about two months, and I posted related questions to various physics forums including Stack Exchange, but still things are unclear.

1. Motivation: According to D. Halliday, in his book Fundamentals in Physics, 10th Edition, for a travelling wave described by the function $$y(x, t)=A \sin (kx - \omega t)$$, the phase is the argument $$kx - \omega t$$, and thus the phase difference is $$\Delta (kx - \omega t) = k\Delta x - \omega \Delta t = k(x_2 - x_1) + \omega (t_2 - t_1).$$

We usually consider either the position $$x$$ or the time as being constant, so either of the terms becomes zero and that usually enables us to think of the phase difference as either $$k\Delta x$$ or as $$\omega \Delta t$$. Okay. Until here everything's fine.

2. Standing Wave: The equation for the standing wave can be easily derived (using a bit of trigonometry) from combining two identical travelling waves moving in the opposite direction, and the expression is $$y(x, t) = A \cos(\omega t)\sin(kx).$$

The first confusion starts from determining which one, obviously between time and position, to take as the argument. So I would say I was unsure about whether to take $$\Delta \phi = \omega \Delta t$$ or $$\Delta \phi = k \Delta x$$. I tried either one, and I thought (also obviously) when I'm considering the phase difference of two separate points I have to refer to $$\Delta \phi = k \Delta x$$, because the instant of time at which we are investigating the phase difference, in this case, does not really matter.

2.1 The Phase Difference of a Standing Wave: I read many books and articles about this but none of them directly clarified this. Yet, I found one book called "Waves & Normal Modes" by Prof Matt Jarvis at Oxford University Department of Physics, in which he says:

All points on the string have the same phase, or are multiples of π, in terms of how the oscillations move in time. For example, all the points are at rest at the same time ,when the string is at a maximum displacement from the equilibrium position, and they all pass through the origin or equilibrium position at the same time.

I guess he is referring to the time phase difference. As the $$\cos \omega t$$ term does not depend on $$x$$ the time phase difference between two separate points is, obviously, zero. I think this is reasonable. I found a similar (and satisfying) argument from the answer by John Rennie to this question.

However, I found that it is a common convention in high school/first year undergraduate level of physics to say that

if any two arbitrary points are moving upwards at the same time or downwards at the same time, then the phase difference is called to be zero, and if they are moving in the opposite direction then the phase difference is $$\pi$$. Only two values are possible, and none of other values, even the integer multiples of $$\pi$$ are impossible.

You can see an example of a source making this argument at Isaac Physics.

3. The Problem: So, going back to the point, this argument basically says that for any two points, no matter how close or far away they are, if they are just both moving upwards or both moving downwards they are in phase and the phase difference is zero. No equation, and no reasoning. I am very unsatisfied with it but textbooks say that it's correct. Is it merely a trivial error in high school textbooks or is there something more than that?

Also sorry for the inappropriate tags. If anybody could clarify, I would send sincere thanks to him or her.

The difference arises from two different conventions for stating the amplitude of the standing wave as a function of $$x$$.

If you take the amplitude as $$A(x) = A \sin (kx)$$ then

$$y(x,t) = A(x) \cos (\omega t)$$

and any two points are in phase, since they both have the same phase angle $$\omega t$$ at any give time $$t$$.

On the other hand, if you insist that the amplitude function should always be zero or positive then you have $$A(x) = |A \sin (kx)|$$ and for half the points

$$y(x,t) = A(x) \cos (\omega t)$$

and for the other half of the points

$$y(x,t) = -A(x) \cos (\omega t) = A(x) \cos (\omega t + \pi)$$

Two given points can then either be in phase because they have the same phase angle at any given time or exactly out of phase because their phases angles always differ by $$\pi$$.

Personally I prefer the first convention (it is simpler and it generalises to complex wave functions) but it is not universal, as your examples show.

• Wow, wow, wow! Bravo! That perfectly fits and is extremely elegant. I'm just thinking like I haven't I thought of that! It's indeed very simple. Oct 2, 2020 at 17:30
• Where did you learn this? Do you have any specific source or book? Otherwise if it's your idea never mind, it's even more amazing. Oct 2, 2020 at 17:31
• @curious Sorry, I don’t have a source. I have just seen both conventions used in different places. Oct 2, 2020 at 18:05
• @gandalf61 Why instead of the factor $(kx)$, the factor $(\omega t)$ is regarded as the phase? Is there a valid justification for this? May 21 at 14:17
• @Solidification No offense, but I think you spend way too much time worrying over exactly what words are used, in cases where the words are inherently vague. "Phase" can mean over a dozen different things and is chosen by context to be whatever is most useful in any given moment. It is like how, if you buy 5 watermelons at the grocery store, that is "a lot of watermelons", but if you buy 5 bananas, it doesn't count as "a lot of bananas". Is that a logical contradiction? Maybe, but it's not worth spending time arguing about. May 23 at 17:11