(a) Formulate an appropriate boundary condition, to replace Eq. 9.27, for the case of two strings under tension T joined by a knot of mass m.

(b) Find the amplitude and phase of the reflected and transmitted waves for the case where the knot has a mass m and the second string is massless.

Short Answer

Expert verified

(a) The boundary conditions are fz0+-fz0-=mT2ft20andfz0=fz0+

(b) The amplitude and phase of the reflected wave is AR=AIand δR=δI+tan-12β1-β2respectively, and the amplitude and phase of the transmitted wave is AT=21+β2A1andδT=δl+tan-1βrespectively.

Step by step solution

01

Expression for the derivative of f.

Consider the knot is of negligible mass, write the expression for the derivative of f.

fz0-=fz0+

As the two strings under tension T are joined by a knot of mass m, write an appropriate equation for the unbalanced forces.

Tsinθ+-Tsinθ-=m2ft20T(sinθ+-Tsinθ-)=m2ft20 …… (1)

02

Determine the boundary conditions:

(a)

Since, it is known that:

sinθ+=fz0+

sinθ+=fz0-

Substitute the values ofsinθ+andsinθ- in equation (1).

Tfz0+-fz0-=m2ft20fz0+-fz0-=m2ft20

Hence, the boundary condition will be,

fz0+-fz0-=m2ft20fz0+-fz0-

Therefore, the boundary conditions are fz0+-fz0-=m2ft20andfz0+-fz0-

03

Determine the amplitude and phase of the reflected and transmitted wave:

(b)

Write the disturbance on the string for a sinusoidal incident wave.

f-z,t=A1eiklz-ϖt+AReiklz-ϖtz<0A1eiklz-ϖtz>0

Write the expression for the outgoing amplitudes andA in terms ofA incoming one .

Al+AR=ATklAl-AR=k2AT …… (2)

From the second boundary condition,

Tik2AT-ik1Al-AR=mϖ2ATik2AT-ik1Al-AR=mϖ2ATTk1Al-AR=k2AT-mϖ2ATTk1Al-AR=k2-imϖ2TAT …… (3)

Multiply equation (2) with and add the obtained equation to equation (3).

k1Al+k1AR=k1ATk1Al+k1AR+k1Al-AR=k2-imϖ2TAT+k1AT2k1Al+k2AT-imϖ2TAT+k1ATAT=2k1k1+k2-imϖ2TAl

…… (4)

Multiply equation (2) with and add the obtained equation to equation (3).

AR=AT-AlAR=2k1k1+k2-imϖ2TAl-AlAR=k1+k2-imϖ2Tk1+k2-imϖ2TAl …… (4)

Divide the above equation by .

AR=1-k2k1-imϖ2T1+k2k1-imϖ2TAlSince, and .

Therefore, the amplitude and phase of the reflected wave is and respectively, and the amplitude and phase of the transmitted wave is and respectively.

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

By explicit differentiation, check that the functions f1, f2, and f3in the text satisfy the wave equation. Show that f4and f5do not.

Question:According to Snell's law, when light passes from an optically dense medium into a less dense one the propagation vector bends away from the normal (Fig. 9.28). In particular, if the light is incident at the critical angle

θc=sin-(n2n1)

Then , and the transmitted ray just grazes the surface. If exceeds , there is no refracted ray at all, only a reflected one (this is the phenomenon of total internal reflection, on which light pipes and fiber optics are based). But the fields are not zero in medium ; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2.26

Figure 9.28

A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with and

kT=kTsinθTx^+cosθTz^

the only change is that

sinθT=n1n2sinθI

is now greater than, and

cosθT=1-sin2θT

is imaginary. (Obviously, can no longer be interpreted as an angle!)

(a) Show that

ET(r,t)=E0Te-kzeI(kx-ωt)

Where

kωc(n1sinθ1)2-n22

This is a wave propagating in the direction (parallel to the interface!), and attenuated in the direction.

(b) Noting that (Eq. 9.108) is now imaginary, use Eq. 9.109 to calculate theirreflection coefficient for polarization parallel to the plane of incidence. [Notice that you get reflection, which is better than at a conducting surface (see, for example, Prob. 9.22).]

(c) Do the same for polarization perpendicular to the plane of incidence (use the results of Prob. 9.17).

(d) In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are

Er,t=E0e-kzcoskx-ωty^Br,t=E0ωe-kzksinkx-ωtx^+kcoskx-ωtz^

(e) Check that the fields in (d) satisfy all of Maxwell's equations (Eq. 9.67).

(f) For the fields in (d), construct the Poynting vector, and show that, on average, no energy is transmitted in the z direction.

Write down the (real) electric and magnetic fields for a monochromatic plane wave of amplitude E0, frequency ω, and phase angle zero that is (a) traveling in the negative xdirection and polarized in the direction; (b) traveling in the direction from the origin to the point(1,1,1) , with polarization parallel to thexyplane. In each case, sketch the wave, and give the explicit Cartesian components of k^andn^ .

Question: Show that the standing wave fz,t=Asinkzcoskvtsatisfies the wave equation, and express it as the sum of a wave traveling to the left and a wave traveling to the right (Eq. 9.6).

(a) Show directly that Eqs. 9.197 satisfy Maxwell’s equations (Eq. 9.177) and the boundary conditions (Eq. 9.175).

(b) Find the charge density, λ(z,t), and the current, I(z,t), on the inner conductor.

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