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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?

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2 Answers 2

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$\omega$ is the angular velocity:

$$ \omega = 2 \pi f $$

If we look at a typical animation of a wave moving:

Wave

(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.

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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.

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