How to determine the direction of a wave propagation?

Question

In the textbook, it said a wave in the form $y(x, t) = A\cos(\omega t + \beta x + \varphi)$ propagates along negative $x$ direction and $y(x, t) = A\cos(\omega t - \beta x + \varphi)$ propagates along positive $x$ direction. This statement looks really confusing because when it says the wave is propagating along $\pm$ x direction, to my understanding, we can drop the time term and ignore the initial phase $\varphi$ while analyzing the direction, i.e. $y(x, 0) = A\cos(\pm\beta x)$, however, because of the symmetry of the cosine function, $\cos(\beta x)\equiv \cos(-\beta x)$, so how can we determine the direction of propagation from that?

I know my reasoning must be incorrect but I don't know how to determine the direction. So if we don't go over the math, how to figure out the direction of propagation from the physical point of view? Why $-\beta x$ corresponding to the propagation on positive x direction but not the opposite?

Answer 1

For a particular section of the wave which is moving in any direction, the phase must be constant. So, if the equation says $y(x,t) = A\cos(\omega t + \beta x + \phi)$, the term inside the cosine must be constant. Hence, if time increases, $x$ must decrease to make that happen. That makes the location of the section of wave in consideration and the wave move in negative direction.

Opposite of above happens when the equation says $y(x,t) = A\cos(\omega t - \beta x + \phi)$. If t increase, $x$ must increase to make up for it. That makes a wave moving in positive direction.

The basic idea:For a moving wave, you consider a particular part of it, it moves. This means that the same $y$ would be found at other $x$ for other $t$, and if you change $t$, you need to change $x$ accordingly.

Hope that helps!

Answer 2

First the assumption/definition is that $\omega$ and $\beta$ are positive constants.
Next you are asking about the phase velocity ie the velocity of a crest, a trough, any fixed point on wave profile.
This means that $\omega t \pm \beta x + \phi$, which can be called the phase of the wave, is a constant.

If you differentiate $\omega t \pm \beta x + \phi=\rm constant $ With respect to time you get the component of the velocity of the wave in the x-direction $\dfrac{dx}{dt} = \mp \dfrac{\omega}{\beta}$.

So with the bracket $\omega t + \beta x + \phi$ the wave is travelling in the negative x direction (negative component) and with the bracket $\omega t - \beta x + \phi$ the wave is travelling in the positive x-direction (positive component).

Answer 3

$y(x,t)=A\cos(\omega t+\beta x+\phi)$ in this equation $\omega t$ and $\beta x$ symbols of the coefficient are same i.e( ++ or --) then the wave is negative direction travelling wave.

$y(x,t)=A\cos(\omega t−\beta x+\phi)$ in this equation $\omega t$ and $\beta x$ symbols of the coefficient are alternative i.e( +- or -+) then the wave is positive direction travelling wave.

Answer 4

Here is (in my opinion) an easier answer. I'm sure you know that the line y = x crosses the x-axis in the origin (0, 0). Now we shift the entire line to the right in such a way that it crosses the x-axis in the point (1, 0). The consequence of this is that the function of the line changes. It becomes y = x - 1. Here is that minus-sign again. The entire line is shifted to the right because you have subtracted $1$ from all the input values. It's the same thing for a sine function, although the input is in radians then. The -1 in y = x - 1 has the same effect as - w.t in the sine function. In wave functions we call it a phase-shift. This can be a time-dependent phase shift (+/- $\omega t$) and/or a constant phase shift ($\phi$). Easy as that ;-)