2.2 Superposition of mechanical waves

Superposition

Suppose we have two waves, with the same amplitude but different wavelengths and velocities and we add them $$y_1=A\sin \left[\frac{2\pi }{{\lambda }_1}\left(x-{\text{v}}_{1}t\right)\right]$$ $$y_2=A\sin \left[\frac{2\pi }{{\lambda }_2}\left(x-{\text{v}}_{2}t\right)\right]\text{.}$$ Then $$y_1+y_2=A\left(\sin \left[\frac{2\pi }{{\lambda }_1}\left(x-{\text{v}}_{1}t\right)\right]+\sin \left[\frac{2\pi }{{\lambda }_2}\left(x-{\text{v}}_{2}t\right)\right]\right)\text{.}$$ Lets rewrite using wave number and angular frequency $$y_1+y_2=y=A\left(\sin \left[\left(k_{1}x-{\omega }_{1}t\right)\right]+\sin \left[\left(k_{2}x-{\omega }_{2}t\right)\right]\right)\text{.}$$ Now we will use $\sin (\theta +\phi )+\sin (\theta -\phi )=2\sin \theta \cos \phi$ and set $$\theta +\phi =k_{1}x-{\omega }_{1}t$$ $$\theta -\phi =k_{2}x-{\omega }_{2}t\text{.}$$ We can rearrange to get $$2\theta =(k_1+k_2)x-({\omega }_1+{\omega }_2)t$$ $$2\phi =(k_1-k_2)x-({\omega }_1-{\omega }_2)t\text{.}$$ By substituting we can then see that $$y=2A\left(\cos \left[\frac{k_1-k_2}{2}x-\frac{{\omega }_1-{\omega }_2}{2}t\right]\times \sin \left[\frac{k_1+k_2}{2}x-\frac{{\omega }_1+{\omega }_2}{2}t\right]\right)\text{.}$$ Now set $$\Delta k=k_1-k_2$$ $$\Delta \omega ={\omega }_1-{\omega }_2$$ $$k=\frac{k_1+k_2}{2}$$ $$\omega =\frac{{\omega }_1+{\omega }_2}{2}$$ and we can rewrite the wave as $$y=2A\cos (x\frac{\Delta k}{2}-t\frac{\Delta \omega }{2})\sin (kx-\omega t)\text{.}$$

The above equation shows beats. For example you can set $t=0$ and see that you get $$y=2A\cos (x\frac{\Delta k}{2})\sin (kx)\text{.}$$ Likewise you could pick $x=0$ and get the same figure, but now the horizontal axis is time $$y=2A\cos (-t\frac{\Delta \omega }{2})\sin (-\omega t)$$ or $$y=2A\cos (t\frac{\Delta \omega }{2})\sin (-\omega t)\text{.}$$ You get a traveling wave that has an oscillating amplitude.

Phase and group velocities

When we look at $$y=2A\cos (x\frac{\Delta k}{2}-t\frac{\Delta \omega }{2})\sin (kx-\omega t)$$ we see that there are two velocities. One, referred to as the phase velocity, is the speed of the individual wavecrests: $${\text{v}}_p=\frac{\omega }{k}=\nu \lambda \text{.}$$ The group velocity is the velocity of the envelope $${\text{v}}_g=\frac{\Delta \omega }{\Delta k}\to \frac{d\omega }{dk}$$ Energy and momentum normally move with the group velocity.