Question

A moving particle is associated with a wave packet or group of waves. The group velocity is equal to:
A) Velocity of light
B) Velocity of sound
C) Velocity of particle
D) 1/Velocity of particle

Answer 1

Hint: Phase velocity is the speed at which a wave of constant phase travels as the wave propagates. Group velocity, ${v_g}$ , is the ratio of the apparent change in frequency \[\omega \] to the associated change in the phase propagation constant \[\beta \]; that is, $\dfrac{{\Delta \omega }}{{\Delta \beta {\text{ }}.}}$

Complete step by step answer:
According to the theory of wave mechanics developed by Schrodinger a material is associated with a very distinct property called wave packet. A wave packet is a form of wave function that has a well-defined position as well as momentum. Thus wave packets tend to behave classically and are easy (and fun) to visualize. Naturally, neither the momentum nor the position is precisely defined, as is governed by the uncertainty principle.

A wave packet with a very well-defined position will have a very uncertain momentum, and thus will quickly disperse as the faster components move on ahead of the slower ones. Conversely, if we construct a wave packet with a very definite momentum it will travel a long distance without dispersing, but it starts out being very broad already in position space.

The group velocity of the particle on the other hand always represents the velocity of the particle. Thus, group velocity is equal to the velocity of the particle.

Note: If the phase velocity does not depend on the wavelength of the propagating wave, then ${v_g} = {v_p}$ For example, sound waves are non-dispersive in air, i.e., all the individual components that make up the sound wave travel at same speed. Phase velocity of sound waves is independent of the wavelength when it propagates in air.