# 2.2 Superposition of mechanical waves

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The superposition of waves gives rise to the phenomenum of beats

## Superposition

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

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

## Phase and group velocities

When we look at $y=2A{\mathrm{cos}}\left(x\frac{\Delta k}{2}-t\frac{\Delta \omega }{2}\right){\mathrm{sin}}\left(kx-\omega t\right)$ 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.

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