Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Shallow water is non-dispersive; waves travel at a speed that is proportional to the square root of the depth. In deep water, however, the waves can’t “feel” all the way down to the bottom—they behave as though the depth were proportional to λ. (Actually, the distinction between “shallow” and “deep” itself depends on the wavelength: If the depth is less than λ, the water is “shallow”; if it is substantially greater than λ, the water is “deep.”) Show that the wave velocity of deep water waves is twice the group velocity.

(b) In quantum mechanics, a free particle of mass m traveling in the x direction is described by the wave function

ψ(x,t)=Aei(px-Et)

wherep is the momentum, and E=p2/2mis the kinetic energy. Calculate the group velocity and the wave velocity. Which one corresponds to the classical speed of the particle? Note that the wave velocity is half the group velocity.

Short Answer

Expert verified

(a) It is proved that the wave velocity of deep water waves is twice the group velocityas vp=2vg.

(b) The group velocity and the wave velocity iskm and the group velocity corresponds to the classical speed.

Step by step solution

01

Expression forthe wave velocity for deepwater wave and phase velocity: 

Write the expression forthe wave velocity for deepwater waves.

vp=αλ …… (1)

Here, αis a constant and λis the wavelength.

Write the expression for phase velocity.

vp=ωk …… (2)

Hereω, is the angular velocity and kis the wave number.

02

Determine the relation between wave velocity and group velocity:

(a)

Equate equations (1) and (2).

…… (3)

ωk=αλω=kαλ …… (3)

Write the relation between the wavenumber in terms of wavelength.

k=2πλλ=2πk

Substituteλ=2πk in equation (3).

ω=kα2πkω=2παk

Write the equation for the group velocity.

vg=dωdk

Substitute iω=2παkn the above equation.

vg=ddk(2παk)vg=2παddk(k)vg=2πα12(k)12

On further solving, the above equation becomes,

vg=2πα21kvg=α22πkvg=α2λvg=vp2

Therefore, it is proved that the wave velocity of deep water waves is twice the group velocity.

03

 Step 3: Determine the wave velocity and group velocity:

(b)

From the given problem, the equation is given as:

ψ(x,t)=Aei(pxEt)/ …… (4)

Write the spatial representation of a wave.

ψ(x,t)=Aei(kxωt) …… (5)

Equate equations (4) and (5).

Aei(pxEt)/=Aei(kxωt)i(pxEt)=i(kxωt)k=pω=E

Write the expression for the wave velocity.

vp=ωk

Substituteω=E in the above expression.

vp=(E/)p/vp=Epvp=p2/2mpvp=k2m …… (6)

Write the expression for the group velocity.

vg=dωdk

Substituteω=E in the above expression.

vg=ddk(E)=ddk(p22m)=ddk(2k22m)

On further solving, the above equation becomes,

vg=2mddk(k2)vg=2m(2k)vg=km ……. (7)

From equations (6) and (7).

vp=k2mvp=vg2

Hence, the phase velocity is half of the group velocity.

Group velocity corresponds to the classical speed of the particle. This can be understood as follows.

Write the expression for the classical speed.

vc=pm

Substitutep=k in the above expression.

vc=kmvc=vg

Hence, the from the equation the wave velocity do not represent the classical speed of the particle.

Therefore, the group velocity and the wave velocity is kmand the group velocity corresponds to the classical speed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free