Phase of a wave - My interpretation

[Shreya]

Homework Statement

I have always wondered what the phase of a wave means. I came up with something and I would really appreciate if someone could help me verify it and possibly find out flaws in it.

Relevant Equations

My textbook says: $$y = A \sin (kx - \omega t)$$ $$ y = A \sin k (x - vt)$$ $$y=A \sin (\frac {2\pi} {\lambda} (x - vt)$$ as $\frac w k = v$

Let's begin my interpretation: (please refer the image below). There I have considered a point in the disturbance/wave (let's call it $P$),(not a particle of the medium) and I follow it as the wave progresses. The solid curve is a Pic of the wave at $t=0$ and the dotted one is its Pic at some later time. The point in the disturbance has moved by $vt$. So, the $x - vt$ in the equation can be interpreted as the intital distance of point P. On Dividing that by $\lambda$, we get 'what part of the cycle was the point at initially' or the initial state. And on multiplying it by $2\pi$, we are coverting that to 'what part of a complete revolution'. Finally taking the $\sin$ of it and multiplying with $A$ we get the $y$ value.

I would love to see comments on this & Understand any mistakes I've made. Thanks in Advance.
(Edit: Sorry for the blurry image, I have updated it)

 

[kuruman]

The definition of phase is the argument of the harmonic function, i.e. what's between the parentheses. It the angle the sine (or cosine) of which you are to consider. In a traveling wave a constant phase travels at the speed of propagation of the wave.

 

[Shreya]

The definition of phase is the argument of the harmonic function, i.e. what's between the parentheses. It the angle the sine (or cosine) of which you are to consider

Yes, I understand that. But, my attempt above was to justify why it is what it is or, rather, why we put $kx-\omega t$ as the phase.

 

[kuruman]

Yes, I understand that. But, my attempt above was to justify why it is what it is or, rather, why we put $kx-\omega t$ as the phase.

Because that is how we describe a wave traveling to the right. If you look at the equivalent formulation $y=A \sin \large(\frac {2\pi} {\lambda} (x - vt)\large)$, you will clearly see why that is. As $t$ increases (which it always does) $x$ must also increase to keep the phase constant. In other words the constant phase travels at speed $v$.

Look at the square dot in your drawing. Assume that at $t=0$ its coordinates are $(x_0,y_0)$. At $t=t_1$ you want $y(x_1,t_1) = y(x_0,0)$ units. For $y$ to have the same value at the two different times and positions, the phase must be the same, $$\frac {2\pi} {\lambda} (x_1 - vt_1)=\frac {2\pi} {\lambda} (x_0 - 0) \implies x_1-x_0=vt_1$$This says that the square dot is at distance $vt_1$ away from where it started. In other words, the constant phase travels at speed $v$.

You can consolidate your understanding and use similar reasoning to convince yourself that $y=A \sin \large(\frac {2\pi} {\lambda} (x + vt)\large)$ describes a wave traveling to the left.

On edit: Deleted extraneous $=0$ in the equation.

 

[Shreya]

Because that is how we describe a wave traveling to the right. If you look at the equivalent formulation $y=A \sin \large(\frac {2\pi} {\lambda} (x - vt)\large)$, you will clearly see why that is. As $t$ increases (which it always does) $x$ must also increase to keep the phase constant. In other words the constant phase travels at speed $v$.

Look at the square dot in your drawing. Assume that at $t=0$ its coordinates are $(x_0,y_0)$. At $t=t_1$ you want $y(x_1,t_1) = y(x_0,0)$ units. For $y$ to have the same value at the two different times and positions, the phase must be the same, $$\frac {2\pi} {\lambda} (x_1 - vt_1)=\frac {2\pi} {\lambda} (x_0 - 0)=0 \implies x_1-x_0=vt_1$$This says that the square dot is at distance $vt_1$ away from where it started. In other words, the constant phase travels at speed $v$.

You can consolidate your understanding and use similar reasoning to convince yourself that $y=A \sin \large(\frac {2\pi} {\lambda} (x + vt)\large)$ describes a wave traveling to the left.

Thanks a million @kuruman, I had this question in the back of my mind for months. Now I can see why the phase is so, and I can also understand what it means for phase to be constant . ☺

Ps: Sorry for the late reply. I'm from a different time zone.