Question

Equation of basic behaviour of wave packets.

Answer 1

equation ---

${ \partial^2 u \over \partial t^2 } = c^2 { \nabla^2 u}$

where c is the speed of the wave's propagation in a given medium.

Using the physics time convention, exp(−iωt), the wave equation has plane-wave solutions

$u(\bold{x},t) = e^{i{(\bold{k\cdot x}}-\omega t)},$

where

$\omega^2 =|\bold{k}|^2 c^2, and |\bold{k}|^2 = k_x^2 + k_y^2+ k_z^2.$

This relation between ω and k should be valid so that the plane wave is a solution

To simplify, consider only waves propagating in one dimension. Then the general solution is

$u(x,t)= A e^{i(kx-\omega t)} + B e^{-i(kx+\omega t)},$

as

$u(x,t) = \frac{1}{\sqrt{2\pi}} \int^{\,\infty}_{-\infty} A(k) ~ e^{i(kx-\omega(k)t)}dk.$

A(k) = $\frac{1}{\sqrt{2\pi}} \int^{\,\infty}_{-\infty} u(x,0) ~ e^{-ikx}dx.$

For instance, choosing

u(x,0) = $e^{-x^2 +ik_0x},$

we obtain

A(k) = $\frac{1}{\sqrt{2}} e^{-\frac{(k-k_0)^2}{4}},$

Answer 2

Answer:

Non-dispersive

Non-dispersiveIt is called a dispersion relation. As in the plane-wave case the wave packet travels to the right for ω(k) = kc, since u(x, t)= F(x − ct), and to the left for ω(k) = −kc, since u(x,t) = F(x + ct).

Explanation:

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