homework 6

Stephen Randall PHYS 411 – Homework #6

9.1 By explicit differentiation, check that the functions f1, f2, and f3 in the text satisfy the waveequation. Show that f4 and f5 do not.

Since this is just straightforward computation, we do it in Mathematica:

where the first three “True”s show that f1, f2, and f3 satisfy the wave equation, and the last two“False”s show that f4 and f5 do not.

9.2 Show that the standing wave f (z, t) = A sin(kz) cos(kvt) satisfies the wave equation, andexpress it as the sum of a wave traveling to the left and a wave traveling to the right.

∂ f∂z

= Ak cos(kz) cos(kvt) ,∂2 f∂z2 = − Ak2 sin(kz) cos(kvt) ,

∂ f∂t

= − Akv sin(kz) sin(kvt) , and∂2 f∂t2 = − Ak2v2 sin(kz) cos(kvt) ,

so yes, this function satisfies the wave equation. Using the trigonometric identity

sin u cos b =12[sin(u + v) + sin(u− v)] ,

we have (with u = kt and v = kvt)

f (z, t) =12[sin(k(z + vt))− sin(k(z− vt))]


9.3 Use Equation (9.19) to determine A3 and δ3 in terms of A1, A2, δ1, and δ2.

Taking Equation (9.19), multiplying each side by its complex conjugate, using the definition ofcosine in terms of complex exponentials, and taking a square-root gives

A3 = A21 + A2

2 + 2A1A2 cos(δ1 − δ2)

To get δ3, we expand Equation (9.19) into sinusoids:

A3(cos δ3 + i sin δ3) = (A1 cos δ1 + A2 cos δ2) + i(A1 sin δ1 + A2 sin δ2) .

Then

tan δ3 =A1 sin δ1 + A2 sin δ2

A1 cos δ1 + A2 cos δ2⇒ δ3 = arctan

(A1 sin δ1 + A2 sin δ2

A1 cos δ1 + A2 cos δ2

)

9.8 Equation (9.36) describes the most general linearly polarized wave on a string. Linear polarizationresults from the combination of of horizontally and vertically polarized waves on the same phase. If thetwo components are of equal amplitude, but out of phase by 90◦, the result is a circularly polarized wave.In that case:

a) At a fixed point z, show that the string moves in a circle about the z-axis. Does it go clockwise orcounterclockwise as you look down the axis towards the origin? How would you construct a wave goingthe other way?

At a fixed z, the horizontally and vertically polarized parts, fh and fv, of the wave are A cos(kz−ωt + π/2)y and A cos(kz − ωt)x, respectively. The 90◦ phase shift allows us to change thehorizontal part to a sine wave (instead of cosine), giving that A2 = f 2

h + f 2v . Thus the total wave

fh + fv lives on a circle of radius A. X

Plugging in a few values for t, we see that it circles counterclockwise. To go the other way, we’dneed to reverse the phase difference; 90◦ → −90◦.

b) Sketch the string at time t = 0.

c) How would you shake the string in order to produce a circularly polarized wave?

Our hand should move in a circle of radius A.


9.23 a) Shallow water is nondispersive; waves travel at a speed that is proportional to the squareroot of the depth. In deep water, however, the waves can’t “feel” all the way down to the bottom — theybehave as though the depth were proportional to λ. Show that the wave velocity of deep water waves istwice the group velocity.

Since v = a√λ for some constant a, we have that ω = a

√2πk. The group velocity is given by

vg =dω

dk=

a2(2πk)−1/2(2π) =

a2

√2πk

=a2

√λ =

12

v X

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

Ψ(x, t) = Aei(px−Et)/h ,

where p is the momentum and E = p2/2m is the kinetic energy. Calculate the group velocity and thewave velocity. Which one corresponds to the classical speed of the particle?

From the expression above, we can read off the wave number and angular velocity:

k =ph

and ω =Eh

.

Then

v =ω

k=

Ep

=p

2m

and

vg =dω

dk= h

ddp

[p2

2mh

]=

pm

Classically, p = mv, so the classical velocity is given by vg .

9.31 Work out the theory of transverse magnetic (TM) modes for a rectangular wave guide. Inparticular, find the longitudinal electric field, the cutoff frequencies, and the wave and group velocities.Find the ratio of the lowest TM cutoff frequency to the lowest TE cutoff frequency, for a given wave guide.

Simply proceeding exactly as in Section 9.5.2 — although now with Bz = 0 and Ez 6= 0 — wehave, from analogy with Equation (9.186) that the longitudinal electric field is

Ez = E0 cos(mπx

a

)cos

(nπyb

)for integers m and n. The cutoff frequencies are exactly as in Equation (9.188);

ωmn = cπ

√(ma

)2+(n

b

)2


The wave and group velocities are just as in Equations (9.191) and (9.192);

v =c√

1− (ωmn/ω)2and vg = c

√1− (ωmn/ω)2

The boundary conditions place a different lower bound on the TM modes, so for a given waveguide, the lowest TM cutoff frequency divided by the lowest TE cutoff frequency is

lowest TMlowest TE

=ω11

ω10=

√1 + (a/b)2

9.37 A microwave antenna radiating at 10 GHz is to be protected from the environment by a plasticshield of dielectric constant 2.5. What is the minimum thickness of this shielding that will allow perfecttransmission (assuming normal incidence)?

From Equation (9.199), we know that T = 1 when sin(n2ωd/c) = 1. This occurs at a shortestdistance d = cπ/n2ω. Since n2 =

√εµ/ε0µ0, assuming that µ ∼= µ0 allows us to say that

n2∼=√

εr =√

2.5. This gives

d =(3× 108)π√

2.5(2π)(10× 109)≈ 9.5 mm

9.38 Light from an aquarium goes from water (n = 4/3) through a plane of glass (n = 3/2) into air(n = 1). Assuming it’s a monochromatic plane wave and that it strikes the glass at normal incidence,find the minimum and maximum transmission coefficients. You can see the fish clearly; how well can itsee you?

Directly from Equation (9.199), we have the transmission coefficient

T =48

49 + (85/36) sin2(3ωd/2c).

Letting the sine piece vanish gives the maximum transmission coefficient, and letting it be 1gives the minimum:

Tmax =4849≈ 0.980 and Tmin =

17281849

≈ 0.935

Noting that Equation is invariant under switching n1 and n3, the fish and I can see each otherequally well.

Bonus In solving for the transverse electric waves in a rectangular wave guide, the text informs usin Equation (9.182) that using

1X

d2Xdx2 = − k2

x and1Y

d2Ydy2 = − k2

y


allows the derivation of a solution. Describe what happens if we assume instead that

1X

d2Xdx2 = − k2

x + ∆ and1Y

d2Ydy2 = − k2

y − ∆

where ∆ is some constant. How does this change the solutions? Do solutions exist for non-vanishing ∆?

This changes the solutions by a modification of the argument to the sinusoids:

kx 7→ (k2x − ∆)1/2 and ky 7→ (k2

y + ∆)1/2 .

If ∆ is such that either kx or ky maps to an imaginary quantity, we lose the sinusoidal behavior.So yes, solutions exist for non-vanishing ∆, but ∆ must satisfy the conditions

k2x − ∆ ≥ 0 and k2

y + ∆ ≥ 0 .

]

homework 6

Documents