Why is $kx−ωt$ a constant for a travelling wave? In my physics textbook, there is a statement like this:
The motion of a fixed phase point on a progressive wave is given by
$kx−ωt$= a constant.
What does this mean? Why is it a constant?
Does fixed phase point mean that it is a particle of the medium always at the same height? Then why are we considering its motion? Are they talking about different points? Or are they talking about a point on the wave itself moving in its direction of propagation?
 A: Suppose you have a general (unity amplitude) sinusoidal wave moving along the x direction
$$\psi(x,t)=\sin(kx-\omega t-\phi_0)$$
If now for example you are interested in the locations where $\sin(\phi)=0$, then you notice that this requires the phase $\phi$ to be
$$\phi=z\pi$$
where $z$ is any integer number. Hence, if you want to investigate the movement of these points, you set
$$kx-\omega t-\phi_0=z\pi$$
or
$$kx-\omega t=\phi_0+z\pi$$
which is a constant for any arbitrarily chosen $z$, as stated. Since the wave always stays sinusoidal by definition, you could repeat this argument with any reference value other than zero. Zero value is just the easiest choice for showing the movement of the wave.
By the way, you get the velocity of these moving points by deriving
$$x(t)=\frac{1}{k}(\phi_0+z\pi+\omega t)$$
with respect to $t$
$$\frac{dx}{dt}=\frac{\omega}{k}$$
This is called the phase velocity of the wave.
A: Suppose the equation of the wave is $y=A \sin (kx - \omega t)$.
If $kx - \omega t = \pi/2$ then $y=A$ and there is a wave peak of amplitude $A$.
How might you measure the speed of a water wave?
What you might do is follow a particular peak and measure how long it took $t$ to travel a given distance $d$.
You might then say that the speed of the wave is then $d/t$.
You can also say that you are following a part of the wave where $kx - \omega t = \pi/2$ ie following a constant phase $\pi / 2$.
This is called the phase speed of the wave.
If you differentiate the expression $kx - \omega t = \pi/2$, or any other part of the wave which can be identified with a constant phase, with respect to time you get that $k\frac{dx}{dt} - \omega = 0 \Rightarrow \frac{dx}{dt} = \frac{\omega}{k} $.
$\frac {dx}{dt}$ is the speed of the wave, the rate at which a constant value of $kx - \omega t$ travels.
Note that $k=\frac{2\pi}{\lambda}$ and $\omega = 2 \pi f$ so the speed of the part of the wave which has a constant phase, eg a peak, a trough etc, the wave speed, is the familiar $f \lambda$.
A: This relation pretty much defines a travelling wave:
$$
\mathbf{k}\mathbf{x}-\omega t=const \Rightarrow \mathbf{x}=\mathbf{v}t+\mathbf{x}_0,
$$
i.e., the wave front (or the surface pf constant phase) of a wave $v(\mathbf{k}\mathbf{x}-\omega t)$ moves with a constant velocity (where $v(\phi)$ is an arbitrary shape).
