This might seem like a trivial question but it is not for me. So, I was reading on group and phase velocities from A.P. French where he calculates the phase and group velocities for a superposition of sine waves of different speed and wavelength. I will write down a brief analysis:
$$y(x,t)=A\sin(k_1x-\omega_1t) + A\sin(k_2x-\omega_2t) $$ which simplifies to
$$y(x,t)=2A\sin\left(\frac{k_1+k_2}{2}x-\frac{\omega_1+\omega_2}{2}t\right)\cos\left(\frac{k_1-k_2}{2}x-\frac{\omega_1-\omega_2}{2}t\right)$$
Now, what I usually find in literature is that for a general wave,the velocity of a wave is defined as $v=\omega/k$ and here we see that by simplifying the superposition, we get a slow moving and a fast moving term and
(1) For the slow moving wave which represents the group envelope, we call the velocity as group velocity $v_g=\Delta \omega/\Delta k=\partial\omega/\partial k$ (for waves with small differences in $\omega$ and $k$).
(2) For the fast moving wave which represents the ripples, we call the velocity as phase velocity $v_g=\bar \omega/\bar k=\omega/ k$ (if $\omega$ is given as a function of $k$).
What I don't understand in this analysis is
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Why this definition of velocity? Why did we just divide the factors of x and t and call that as velocity? For a single sine wave, I understand how we can find the displacement of a maxima or minima and see how much it moves in some time t and define that as velocity (just like mentioned here). But is there any similar treatment possible for this?
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How did we identify which one was the group and which one was the phase velocity? Also, it is not very intuitive at first site for a person who didn't know this before that there are actually 2 velocities embodied in such a solution?
I would be very grateful if someone could possibly have an answer to these questions.
Best Answer
The top two graphs from the MakeaGIF.com website are for waves of one frequency/wavelength travelling at different phase speed as shown by the motion of the red and blue dot sitting on top of a crest.
The term phase is used because you are observing the particle which make up the medium at their maximum upward excursion from the equilibrium position and the speed of that crest is measured as the distance moved by a crest divided by the time taken to move that distance.
You could have equally well chosen to follow a trough or when the particles had zero displacement or the phase $kx-\omega t = \text{constant}$.
Differentiating this expression gives the phase speed as $\left (\dfrac{dx}{dt}\right)_{\rm phase} = \dfrac \omega k$
The bottom graph is the addition of the top two graphs and you will note that a modulating envelope the peak of which as shown by the red dot travels at the group velocity where group refers to the motion of a number (group) of waves added together and $\left (\dfrac{dx}{dt}\right)_{\rm group} = \dfrac {\Delta\omega} {\Delta k}$.
This is derived from French's cosine term where you want the term in the bracket to be a maximum with $\omega = \omega_1-\omega_2$ and $\Delta k = k_1- k_2$ and follow the movement of that maximum.
Hopefully the gif animations below from the Institute of sound and Vibration Research (isvr) will help you to differentiate between group velocity and phase velocity.