# On the velocity of a wave

In this video doctorphys gives a proof for the general form of a wave. With that I was convinced that equation of a travelling wave can be $$\psi(x,t)=A\sin(kx-\omega t)+B\cos(kx-\omega t),$$ where $$k$$ is the angular wave number; and $$\omega$$ is the angular velocity of the wave (as per the video); and $$x$$ and $$t$$ are positions and time, respectively.

Now my textbook gives another expression for velocity of a wave. $$v=\frac{\omega}{k}$$

Now I am confused? If $$\omega$$ ins't the velocity, then what is it?

$$\omega$$ is the angular velocity:

$$\omega = 2 \pi f$$

If we look at a typical animation of a wave moving: (from the University of Southampton web site) then the velocity the wave moves right is called the phase velocity because it's the velocity a point with constant phase moves, where by the phase $$\phi$$ we mean:

$$\psi(x,t)=A\sin(\phi)+B\cos(\phi)$$

i.e.

$$\phi = kx - \omega t$$

If we choose, for example, the point with phase $$\phi= 0$$ then we get:

$$\phi = kx - \omega t = 0$$

Hence:

$$\frac xt = \frac{\omega}{k}$$

and $$x/t$$ is just the velocity. That's why the phase velocity is $$\omega/k$$.

An alternative way to see this is to note that the angular velocity is defined by $$\omega = 2\pi f$$ and the wave vector is defined by $$k = 2\pi/\lambda$$ so:

$$\frac{\omega}{k} = \frac{2\pi f}{2\pi/\lambda} = f\lambda$$

and we know $$v = f\lambda$$ is the velocity of the wave.

There is a missing step in the consequence you draw from the video.

In the video, the general solution of the wave equation was written as $$\psi(x,t)= A f(x-vt) + B g(x+vt),$$ where $$f(x-vt)$$ and $$g(x+vt)$$ represent respectively the rigid shift towards right, by a distance $$vt$$, of the function $$f(x)$$, and the rigid shift towards left, by a distance $$vt$$, of the function $$g(x)$$.

Notice that $$\psi$$ may represent many different physical quantities evolving in a wave-like, and $$f$$ and $$g$$ are generic functions of their arguments. Therefore, just by dimensional analysis, we can conclude that the dimensioned arguments $$(x+vt)$$ and $$(x-vt)$$ of the two functions $$f$$ and $$g$$ should appear in a dimensionless form $$k(x+vt)$$, where $$k$$ should be a quantity with the physical dimension length$$^{-1}$$.

In the case of sinusoidal or cosinusoidal profiles for $$f$$ and $$g$$, it is natural to use a form for $$k$$ written in a form that makes explicit the space periodicity over a distance $$\lambda$$ (the wavelength): $$k=\frac{2 \pi}{\lambda}.$$ Thus, for a sin wave propagating toward the right, we have $$f(x-vt)=\sin \left( k(x-vt) \right) = \sin \left( kx-\omega t \right),$$ by introducing $$\omega=k v$$. Similarly, for a $$\cos$$ wave or propagation towards the left. Then it is clear that there is no contradiction between the video and textbook, and the key point is that $$\omega$$ is not the velocity of the wave.