# Waves: Trouble with simple Group Velocity derivation

In my notes on waves (specifically water waves) there is a derivation of Group Velocity.

They consider two waveforms with the same amplitude, that differ slightly in wavelength and frequency, which are then superimposed to give wave groups.

$k$is wavenumber, $\delta k$ is how much the wavenumbers between the two waves differ. Similarly for angular frequency $\omega$. $a$ is the amplitude of both waves.

Superimposing both wave equations:
$\tau(x,t)=a*cos[(k+\delta k)x-(\omega +\delta \omega)t]+ a*cos [kx-\omega t]$

$=2a*cos[\frac{1}{2}(\delta kx-\delta \omega t)]cos[(k+\frac{1}{2}\delta k)x-(\omega+\frac{1}{2}\delta \omega )t]$

I am fine with this. This results from a trig identity. So we have a wave group with a varying amplitude given by $2a*cos[\frac{1}{2}(\delta kx-\delta \omega t)]$, wavenumber $k+\frac{1}{2}\delta k$, angular frequency$\omega +\frac{1}{2}\delta \omega$

The distance between two successive wavegroups is $\Delta x$

I am having trouble understanding where the next step comes from:

(1)
$\frac{1}{2}\delta k \Delta x = \pi$

Thus

$\Delta x = \frac{2\pi}{\delta k}$

They go on with the same thing for $\Delta t$ and then get the Group Velocity.

Where does (1) come from? I don't see it.

Any assistance would be appreciated.

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rude man
Homework Helper
Gold Member
1. What is a good way to get rid of t?
2. How does one classically find minima/maxima of a function?

Now, there may be a simpler way to do this, but that would be my approach.

As an aside: do not, unlike my Halliday & Resnick textbook ( 20-7) confuse this superposition process with modulation. No new frequencies are generated with superposition, whereas they most certainly are with modulation.

Thank you. This helped