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K29

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

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

Superimposing both wave equations:

[itex]\tau(x,t)=a*cos[(k+\delta k)x-(\omega +\delta \omega)t]+ a*cos [kx-\omega t][/itex]

[itex]=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][/itex]

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

The distance between two successive wavegroups is [itex]\Delta x[/itex]

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

(1)

[itex]\frac{1}{2}\delta k \Delta x = \pi[/itex]

Thus

[itex]\Delta x = \frac{2\pi}{\delta k}[/itex]

They go on with the same thing for [itex]\Delta t[/itex] and then get the Group Velocity.

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

Any assistance would be appreciated.