Consider the two waves $$y_1=Acos(\omega_1 t+k_1 x), \tag{1}$$ $$y_2=Acos(\omega_2 t+k_2 x), \tag{2}$$

where $\omega_i=k_iv(k_i)$ for $i=1,2$ so we have a dispersive medium. Then if we take their superposition to get the wave:

$$y=2Acos(\bar {\omega}t + \bar {k}x)cos\left(\frac{\Delta \omega \ t}{2} + \frac{\Delta k \ x}{2}\right), \tag{3}$$


$$\bar {\omega} = \frac {\omega _1 + \omega _2}{2}, \ \ \ \bar {k} = \frac {k_1 + k_2}{2}, \tag{4}$$

$$\Delta \omega = \omega _1 - \omega _2, \ \ \ \Delta k = k_1 - k_2. \tag{5}$$

Then the waves in wave packets are ment to spread out. Why? In this case at any time $t=t_0$ the wavelength of the wavepacket is given by $\lambda=4 \pi /(k_1+k_2)$ which is a constant and therefore the wavepackets should stay the same length, so why do they spread out?


1 Answer 1


By definition, we say that waves are dispersive if their phase velocity if a function of wavenumber. That is, $$c\equiv \frac{\omega}{k}=c(k).$$

What you've sketched above is an argument related to the group velocity. The key property of dispersion is the functional form of the angular frequency $\omega$. The relationship $\omega$ has with $k$, called a dispersion relationship, is related to the properties of the medium. For deep water surface gravity waves $\omega=\sqrt{gk}$, hence $c$ is dispersive.

This means that waves of different wavelengths travel at different speeds, and hence if you had an initially compact wave group, it would spread out.

This is why, for example, when swell comes to shore it is the longest period waves (ie the fastest waves) that reach shore first. The swell will last a certain amount of time, with progressively shorter waves coming to shore.

For shallow water linear waves $\omega = \sqrt{gh}k$ with $h$ the depth of the water. The phase velocity of these waves is independent of $k$, hence all of these waves travel at the same speed, and a packet of waves will not disperse.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.