Solutions to Problems in Chapter 2: Waves

Transcription

1 Solutions to Problems in Chapter 2: Waves [Figure and equation numbers in the answers are preceded by ii so as to avoid confusion with figure and equation numbers in the text.] 2.1 A chain of masses m are situated on the x-axis at equally-spaced points x = na. They are connected by springs having springconstant K and are restrained to move only along the x-axis. Show that longitudinal waves can propagate along the chain, and that their dispersion relation is ω =2(K/m) 1 2 sin 1 2 ka. (ii.1) Explain in physical terms why the dispersion relation is periodic. Calculate the phase and group velocities when ω 0andω = 2(K/m) 1 2. The dispersion relation is periodic because the masses are discrete, and the displacement is only sampled at the lattice points, separated by a. Any continuous wave which has the same set of sampled values represents the same set of displacements and therefore has the same frequency. The waves corresponding to wave-numbers k and k + 2πp/a, (p integer) when sampled at x = na are exp[ikna] and exp[i(k +2πp/a)na] whichare clearly identical because both p and n are integers. When ω 0, ω/k = ω/ k (Ka 2 /m) 1/2. When ω =2(K/m) 1 2, k = π/a and ω has its maximum value. Since the wavelength is 2a, neighbouring masses move in antiphase, and the nodes, half-way between them, are stationary. Thus the group velocity is zero. 2.2 Flexural waves on a bar have a wave-equation of the form 2 y t 2 = B2 4 y x 4, (ii.2) where B is a constant. Find the dispersion relation. Under what conditions are the waves evanescent? Using the substitutions of 2.3.1, we find that ω = ±Bk 2 for which there are four solutions: k = ± Bω, k = ±i Bω. (ii.3) These represent two waves, one of which is evanescent in x, travelling in each direction. Any complete solution of a wave problem on a bar must therefore involve a superposition of all four waves. It must therefore have four boundary conditions, e.g. values of y(0),y(a), dy/dx(0) and dy/dx(a) for a bar clamped at both ends. 1

2 2 Waves 2.3 X-rays in a medium have refractive index n =(1 Ω 2 /ω 2 ) 1/2 ( ) Show that the product of group and phase velocities is c 2. We have n =(1 Ω 2 /ω 2 ) 1/2 v = ω/k = c/n = c(1 Ω 2 /ω 2 ) 1/2 ; 1 = dk dω = 1 ( n + ω dn ) c dω v g (ii.4) (ii.5) (ii.6) (ii.7) Carrying out the differentiations in the last equation for (ii.4) we find v g = c(1 Ω 2 /ω 2 ) 1/2, (ii.8) so that vv g = c Waves on the surface of water, with depth h and surface tension σ, have a dispersion relation ω 2 =(gk + σk 3 /ρ) tanh(kh), (ii.9) where g is the gravitational acceleration and ρ is the density. Find the group velocity as a function of k and show that it has a turning point at a certain point related to the depth h. What is a typical value for this group velocity? It is easiest to differentiate the function numerically. The value of the surface tension for water is 72 d/cm and ρ = 1 g/cm 3 giving the results in the figure for the wave and group velocities for depth h=1 cm. Then the minimum group velocity is 20 cm/s at k =2cm 1 (λ =0.3cm.) 2.5 Two media, with wave velocities v 1 and v 2 are separated by a plane boundary. A source point A andanobserverb are on opposite sides of the boundary, and not on the same normal. Derive Snell s law from Fermat s principle by finding the minimum time for a wave to propagate from A to B. The plane interface I separates regions with refractive indices n 1 = c/v 1 and n 2 = c/v 2 (Fig. ii.2). We consider an optical path AB from point A in the first medium to B in the second, consisting of straight lines in each of the media. The angles of incidence and refraction at the interface, measured from the normal to I, areî and ˆr respectively. The time from A to B is AB/c. Now we proceed to minimize AB by varying the angles. The projected distance A 1 B 1 on the interface is given by A 1B 1 = h 1 tan î + h 2 tan ˆr, (ii.10)

3 Waves 3 Fig. ii.1. Wave (black) and group (red) velocities for water with depth 1 cm. Fig. ii.2. Snell s law proved by Fermat s principle. and the optical path from A to B is AB = h1n1 cos î + h2n2 cos ˆr. We now differentiate (ii.10) and (ii.11). Since A 1 B 1 differential is zero: (ii.11) is a constant, its 0=h 1 sec 2 î dî + h 2 sec 2 ˆr dˆr ; (ii.12) dˆr dî = h1 sec2 î h 2 sec 2 ˆr (ii.13)

4 4 Waves and dab dî = h1n1 sin î cos 2 î + h2n2 sin ˆr cos 2 ˆr dˆr dî. (ii.14) The extremum value of AB is obtained when (ii.14) is equated to zero, which gives: 0= h1n1 sin î cos 2 î h2n2 sin ˆr cos 2 ˆr h 1 sec 2 î h 2 sec 2 ˆr. (ii.15) This simplifies to Snell s law of refraction: n 1 sin î = n 2 sin ˆr. (ii.16) 2.6 Use Fermat s principle and the properties of conic sections to prove that a point source at one focus of an ellipsoidal or hyperbolic mirror is imaged at the other focus (the most common example is the paraboloid, which has one focus at infinity). The Cassegrain and Gregorian reflecting telescopes use both concave and convex mirrors to obtain high magnification (like a telephoto lens). What profiles of mirror should ideally be used? Note a basic difference between the two foci of a lens and the two foci of a conic section: The latter are images of each other, the former are not. A common way to draw an ellipse is to anchor at F 1 and F 2 astring with length >F 1 F 2, and to take up the slack with a pencil. Moving the pencil along the string draws an ellipse (Fig. ii.3(a)) This means that the (optical) path from F 1 to F 2 is a constant, and therefore F 1 is the image of F 2. Analytically, one uses the formula for an ellipse x 2 a 2 + y2 b 2 = 1 (ii.17) which has foci at (±(a 2 b 2 ) 1/2, 0) and calculates the sum of the distances from the foci to an arbitrary point on the ellipse. A hypothetical construction with string similar to the above one gives a hyperbola (Fig. ii.3b), where a + b c d = a dconst. Therefore a bundle of rays converging onto F 2 is reflected to converge on F 1. The Cassegrain telescope should ideally be constructed from one paraboloidal and one hyperboloidal mirror, and is the mirror equivalent of a telephoto lens. The two mirrors have a common focus, as in Fig. ii.4(a). The Gregorian telescope uses a paraboloidal and an ellipsoidal mirror, as shown in Fig. ii.4(b).

5 Waves 5 Fig. ii.3. Construction of (a) ellipse and (b) hyperbola with the aid of a string. The hyperbola is constructed as follows: A string of length a + b + c + d is attached to F 1 and F 2.AknotK is made so that a + b = c + d+constant. A small ring R slides on the string so that b = c. ThenF 1 R F 2 R = a d =constant, i.e. the locus of the ring defines a hyperbola. 2.7 Show that the multiple images caused by a gravitational lensing mass of finite size a correspond to both maximum and minimum optical paths. Use a simplified model consisting of a region of thickness a normal to the propagation direction having refractive index n(r) =1+2MG/c 2 (a 2 + r 2 ), where r is measured from the observation axis in the plane normal to it. Show that the optical path from source to observer has both maximum and minimum values, which correspond to the images observed. (The different optical paths result in light variations appearing with different delays, which are indeed observed. The optical path from source A to observer B (at z = u and v respectively) is, using binomial expansion up to the fourth order, AB = u(1 + r2 2u r4 r2 +...)+v( u4 2v r4 2 8v +...) 4 + a[1 + 2MG r2 (1 c 2 a 2a + 3r4 2 8a )+...] 4 v u + MGar2 2c 2 a 3 (ii.18) 3MGar4 4c 2 a (ii.19) when u, v. This function has maximum value at r = 0 and minimum at r 2 = a 2 /3. These two values correspond to the axial image and the Einstein ring in Fig The path difference is MG/12c 2, leading to a time delay MG/12c 3. This allows the lensing mass M to be weighed by correlating fluctuations in the received waves.

6 6 Waves Fig. ii.4. Cassegrain and Gregorian telescopes (see Fig in the book). 2.8 A non-planar wavefront propagates in the z direction according to Huygens principle. The wavefront has radii of curvature R x in the x, z plane and R y in the y, z plane. The intensity in plane z is I(z). By using conservation of energy for a rectangular region of the wavefront, relate the change in intensity after a further short distance δz R 1, R 2 to the radii of curvature. Show that this can be expressed as an radiative transport equation di dz = Iκ = I 2 w, (ii.20) where w(x, y) is an algebraic form for the wavefront, and κ R1 1 + R2 1 is its Gaussian curvature. The intensity transport equation can be obtained as follows. Consider a small rectangular region on the wavefront with dimension X and Y. The x dimension lies on a circle of radius R x and so after propagation for asmalldistanceδz it has increased by factor δz/r x. Likewise for the y dimension. The area has therefore increased by factor δz(1/r x +1/R y ). Since the wave energy propagates normal to the wavefront, the energy passing through the rectangle is conserved. Thus δi = Iδz(1/R x +1/R y). (ii.21) Using a quadratic approximation for the curvatures, 1/R x = 2 w x 2 have the intensity transport equation: etc., we

7 Waves 7 Fig. ii.5. Form of the guitar string reconstructed att =0and three later times from its Fourier series, using N =9. δi δz di dz = I 2 w. (ii.22). 2.9 Confirm the demonstration in the box on p. 30 using Fourier analysis from Chap. 4. Analyze the triangular wave into Fourier components analytically, and then let each one evolve, as described in 3.7. You should get the same result as in the box! We define the triangular wave in the region x ( π, π) asy = π/2 x. We calculate the Fourier components from (4.11), using the fact that the the wave is symmetrical and therefore the components are real: F n = π π (π/2 x )cos(nx)dx =2 π 0 (π/2 x)cos(nx)dx (ii.23) = 4 n 2 for odd n. (ii.24) We can confirm that N 1 F n cos(nx) indeed represents the triangular wave (Fig. ii.5 (blue curve) Now we let each component wave propagate in both the x and x directions for a time t, i.e. we replace the term cos(nx) by 1 2 {cos[n(x vt)] + cos[n(x + vt)]} and sum the series again. The result is shown in Fig. ii.5 for three values of t, which reproduce the experimental result.

8 8 Waves Fig. ii.6. Passage of a particle near to a lensing mass According to Newton, light particles of mass m passed near to a heavy body of mass M and were deflected due to its gravitational attraction GMm/r 2. Show (for small angles) that the result (2.73) is obtained, independent of m, which can then be taken as zero for a photon! A particle of mass m moves at constant velocity v along an essentially straight path whose closest approach A to the fixed mass M is at distance b (Fig. ii.6). The force at the point represented by θ is GMm cos 2 θ/b 2 and its component normal to the path is F = GMm cos 3 θ/b 2.Theposition on the path, relative to A is b tan θ, where its time is t = b tan θ/v, for which dt = bdθ/v cos 2 θ. The total transverse momentum acquired as the particle passes M, whenθ goes from π/2 toπ/2 isthen p = π/2 π/2 π/2 GMm cos 3 θ b F dt = π/2 b 2 v cos 2 θ dθ (ii.25) = 2GMm. (ii.26) vb The small angle of deflection is α = p /mv =2GM/bv 2, independent of the mass of the deflected particle. Putting v = c gives the result 2GM/bc 2 (2.73).