MIT 8.03SC Fall 2016 Lecture 13: Dispersive Medium, Phase Velocity, Group Velocity

University:
Massachusetts Institute of Technology
Course:
8.03SC | Physics III: Vibrations and Waves
Academic year:

2020
Views:

55

Pages:

5
Author:

afbolwo

8.03 Lecture 13 Reminder: Maxwell’s equation in vacuum ~ ·E ~ =0 ∇ ~ ·B ~ =0 ∇ ~ ~ ×E ~ = − ∂B ∇ ∂t ~ ~ ~ ×B ~ = µ0 0 ∂ E = 1 ∂ E ∇ 2 ∂t c ∂t √ Where c ≡ 1/ µ0 0 Resulting wave equations: 2~ ~ 2E ~ = 1 ∂ E ∇ c2 ∂t2 2~ ~ 2B ~ = 1 ∂ B ∇ 2 c ∂t2 We discussed plane harmonic wave solution. And you will show that in general a progressing wave solution: ~ = E0 ŷf (z − vt) E ~ field also satisfies Maxwell’s equations. and the corresponding B How do we transmit “information”? A simple harmonic wave would not be useful. We must use “pulses,” chunks of localized energy in time. For instance: We have learned: f (x − vt) or f (kx − ωt) is a traveling wave moving in the +x̂ direction and its shape is kept unchanged if and only if we are working in a non dispersive medium, i.e. ω/k = v Consider an ideal string: 2 ∂2ψ 2∂ ψ = v ∂t2 ∂x2 Where ω =v= k s T ρL If we create a square pulse, the square pulse will move at constant speed v. The shape of the square pulse does not change! We call this string a non-dispersive medium and the “dispersion relation” is ω = vk. Note: the string tension is responsible for the restoring force. However, if we consider the stiffness of the string, (for example, a piano string): If we bend a piano string, even when there is no tension, the string tends to restore to its original shape. To model “stiffness”: " # 2 ∂2ψ ∂4ψ 2 ∂ ψ =v −α 4 ∂t2 ∂x2 ∂x The dispersion relation becomes (where we use A cos(kx − ωt) as a test function): ω 2 = v 2 (k 2 + αk 4 ) p ω ⇒ = v 1 + αk 2 k Not a constant versus k anymore!! Where k = 2π/λ. Large k ⇒ short λ ⇒ a lot of dispersion and a higher speed v As a consequence, components with different k will be moving at different speeds vp = ω(k)/k and we get a dispersion, or the wave loses shape: Dispersion is a variation of wave speed with wave length. Example: addition of two progressing waves: ω1 ψ1 (x, t) = A sin(k1 x − ω1 t) v1 = k1 ω2 ψ2 (x, t) = A sin(k2 x − ω2 t) v2 = k2 If we add ψ1 + ψ2 and using the trig identity 1 1 sin A + sin B = 2 sin (A + B) cos (A + B) 2 2 we get k1 + k2 ω1 + ω2 k1 + k2 ω1 + ω2 x− t cos x− t 2 2 2 2 Assuming k1 ≈ k2 ≈ k and ω1 ≈ ω2 ≈ ω we have “amplitude modulation: ψ1 + ψ2 = 2A sin Where the phase and group velocity is vp = ω k vg = (ω1 − ω2 ) dω ≈ (k1 − k2 ) dk Bounded system: ψ(x, t) = X Am sin(km x + αm ) sin(ωm t + βm ) m Where ωm = ω(km ), then evolve as a function of time! Now consider the boundary conditions of this system: ψ(0, t) = 0 & ψ(L, t) = 0 This is similar to something we have solved before, and we got: mπ km = , αm = o L Identical to the ideal string case (α = o) We learned that: 1. The boundary condition “set” the km ! Does not depend on the dispersion relation ω(k) 2. The individual normal modes are oscillating at ωm = ω(km ) as calculated by the dispersion relation: This does depend on the dispersion relation! If we plot the dispersion relation: But in general ωm is not equally spaced. Full solution: ψ(x, t) = X Am sin(km x + αm ) sin(ωm t + βm ) m = X m Example: ψ(x, t) = ψ1 + ψ2 ψm In a non-dispersive medium: the system goes back to the original shape after 2π/ω1 In a dispersive medium ω2 6= ω1 . We need to wait longer until the reaches the least common multiple of 2π/ω1 and 2π/ω2

MIT 8.03SC Fall 2016 Lecture 13: Dispersive Medium, Phase Velocity, Group Velocity

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