[SOLVED] Regarding the Superposition of Two Plane Waves

[Slimy0233]

My professor was teaching me about the superposition of two waves and after this derivation, he marked ##2Acos(\frac{dk}{2}x -\frac{d\omega}{2}t)## as the oscillation part and ##sin (Kx-\omega t)## as the oscillation part, I don't understand why? Any answers regarding this would be considered helpful.

My main question would be, why did he choose sin part as the oscillation and why not the cos part and more importantly, why not both? I mean, my bad intuition tells me, that I should include both.

 

[vanhees71]

The answer is in the assumption that ##\mathrm{d}k \ll k## and ##\mathrm{d} \omega \ll \omega##. First, think about ##y## as function of ##x## at a fixed time. The cos-factor has a spatial period of ##4 \pi/\mathrm{d} k## and the sin-factor one of ##2 \pi/k\ll 4 \pi/\mathrm{d} k##. So the first cos factor is much slower varying than the sin factor as a function of ##x##. So you can interpret this as something oscillating in space with a wave length ##\lambda=2 \pi/k## and a position dependent amplitude, where the dependence of this amplitude on ##x## is much slower, i.e., the corresponding wave-length of these variations is ##\lambda'=4 \pi/\Delta k \gg \lambda##.

The analogous arguments hold also for the variations of the factors with time at a fixed position in space.

 

[Slimy0233]

The answer is in the assumption that ##\mathrm{d}k \ll k## and ##\mathrm{d} \omega \ll \omega##. First, think about ##y## as function of ##x## at a fixed time. The cos-factor has a spatial period of ##4 \pi/\mathrm{d} k## and the sin-factor one of ##2 \pi/k\ll 4 \pi/\mathrm{d} k##. So the first cos factor is much slower varying than the sin factor as a function of ##x##. So you can interpret this as something oscillating in space with a wave length ##\lambda=2 \pi/k## and a position dependent amplitude, where the dependence of this amplitude on ##x## is much slower, i.e., the corresponding wave-length of these variations is ##\lambda'=4 \pi/\Delta k \gg \lambda##.

The analogous arguments hold also for the variations of the factors with time at a fixed position in space.

beautiful analogy! thank you!

 

[Slimy0233]

[SOLVED]

 

[hutchphd]

My main question would be, why did he choose sin part as the oscillation and why not the cos part and more importantly, why not both? I mean, my bad intuition tells me, that I should include both.

I am not very comfortable with the derivation. While I understand what your prof is trying to do, his/her methods seem unnecessarilly capricious. In particular exact derivations can be found for adding waves. Let's add two waves $$f(x,t)=cos(k_1x-\omega_1t)+cos(k_2x-\omega_2t) $$ then using trig identities $$ =2cos\left( \frac {(k_2-k_1)x-(\omega_2-\omega_1)t} 2 \right)cos\left( \frac {(k_2+k_1)x-(\omega_2+\omega_1)t} 2 \right)$$ $$=2f_1(x,t)f_2(x.t)$$
Typically the differences are smaller than the sums and f1 "modulates" f2. For your ear at x=0 this will give sound "beats" at the small difference frequency in the usual way.

 

[Slimy0233]

I am not very comfortable with the derivation. While I understand what your prof is trying to do, his/her methods seem unnecessarilly capricious. In particular exact derivations can be found for adding waves. Let's add two waves $$f(x,t)=cos(k_1x-\omega_1t)+cos(k_2x-\omega_2t) $$ then using trig identities $$ =2cos\left( \frac {(k_2-k_1)x-(\omega_2-\omega_1)t} 2 \right)cos\left( \frac {(k_2+k_1)x-(\omega_2+\omega_1)t} 2 \right)$$ $$=2f_1(x,t)f_2(x.t)$$
Typically the differences are smaller than the sums and f1 "modulates" f2. For your ear at x=0 this will give sound "beats" at the small difference frequency in the usual way.

thank you for this sir!