Phys 6303 Final Exam Solutions December 19, 2012

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1 Phys 633 Fial Exam s December 19, 212 You may NOT use ay book or otes other tha supplied with this test. You will have 3 hours to fiish. DO YOUR OWN WORK. Express your aswers clearly ad cocisely so that appropriate credit ca be assiged for each problem. There are 6 problems. You must do 5 for full credit. TURN IN ONLY 5 PROBLEMS - I WILL GRADE ONLY THE FIRST 5 PROBLEMS YOU SUBMIT. Full credit for each problem is 25 poits. 1) Use the techique of Lagrage multipliers to fid the ratio of the radius, r, to height, h, of a right circular cylider of fixed volume which miimizes the surface area. The Volume of the cylider is; The surface area is; V = Costat = πr 2 h A = 2πr 2 + (2πrh) Form the fuctio, J, usig the Lagrage multiplier, λ. J = [2πr 2 + 2πrh] + λ[πr 2 h V ] Set the derivatives with respect to r ad h to zero, obtaiig 2 equatios. Remove the Lagrage parameter λ J r = 4πr + 2πh + λ 2πrh = J h = 2πr + λ 2πr 2 = 2r = h 2) The equatio; 1

2 ρ t = a2 2 ρ x 2 κ ρ with κ ad a costats, represets eutro desity, ρ, i a material of thickess, L. Assume eutros are reflected at the surface so ρ = at x = ad x = L. Solve the equatio by a fiite sie expasio ad a Laplace trasform with ρ = C, a costat, whe t =. As the solutio for ρ(x, t) must vaish at x =, L it will have the form, ρ(x, t) = =1 A (t) si(πx/l) This leaves the time depedet equatio for A(t). da (t) dt Apply a Laplace trasform such that; A (p) = = a 2 (π/l) 2 A (t) κ A (t) dt A (t) e pt Multiply each term i the equatio by e pt ad itegrate from to. The derivative term must be itegrated by parts. dt da dt Trasform the boudary coditio; = A e pt + p dt A e pt A(p) [t=] = L dxc si(πx/l) = D = 2CL π D + pa (p) = [(πa/l) 2 + κ] A (p) eve A (p) = D p + (πa/l) 2 + κ The iverse trasform the gives; A (t) = 1 2πi γ+i γ i dp D e pt p + (πa/l) 2 + κ This itegral is completed i the complex plae closig the itegral aroud a loop to the left of the imagiary axis. There is oe pole; 2

3 p = [(πa/l) 2 + κ] The evaluatio of A is; A (t) = D e αt α = (πa/l) 2 + κ Fially substitute this ito the sum over above to get, ρ(x, t) 3) A log wire with vaishigly small radius, carries a charge per uit legth, λ. It is placed a distace, b, from the ceter of a log, perfectly-coductig, grouded cylider with radius a < b. Fid the electric potetial i the space for a < r < usig the Gree s fuctio method. [hit; separate variables i Laplace s equatio as Bessel fuctios are ot solutios] The Gree s fuctio is a soutio to the equatio; 2 V = δ(θ θ )δ(ρ ρ ) We seek a solutio i cylirical coordiates ad it must be idepedet of the variable, z. Thus usig separatio of variables the solutio for the homogeeous equatio has the form V (ρ, θ) = R(ρ)Θ(θ). The separated homogeeous equatio is; (1/R)[ρ d dρ (ρdr dρ )] = d2 Θ dθ 2 = ω 2 I the above ω is the separatio costat. It is selected so that Θ is cotiuous as θ 2π. Sice the solutio depeds o 2 variables, there will be oe eigevalue solutio, Θ, ad oe dual solutio for R. Build the Gree s fuctio solutio from these fuctios. The eigefuctio solutios of Θ which satisfy the boudary coditios are; si(θ) with ω =. Thus expad δ(θ θ ) i the eigefuctios i these eigefuctios to obtai; δ(θ θ ) = (1/π) si(θ) si(θ ) The homogeeous radial equatio is; ρ d dρ (ρdr dρ )] = 2 R The two solutios are; 3

4 [ Aρ R = Bρ ] Require that the Gree s fuctio solutio be fiite at ρ = ad ρ =. To match the boudary coditios; [ ] A[(ρ/a) R = ) (ρ/a) ] ρ < b Bρ ρ > b The solutio is cotiuous at ρ = ρ. However, the derivative of the solutio is discotiuous. This results i the two equatios; A [(b/a) (b/a) ] = B (ρ/a) b+ǫ b ǫ dρ ρ dρ d b+ǫ (ρdr dρ ) = dρ ρ δ(ρ ρ )/ρ b ǫ A[(b/a) (b/) ] + B(b/a) ] = 1/ A = B = (b/a) [(b/a) 2 + (b/a) + (b/a) + 1] [(b/a) (b/a) ] [(b/a) 2 + (b/a) + (b/a) + 1] The solutio for the Gree s fuctio is; G(ρ, ρ ) = [ A [(ρ/a) si(θ) si(θ ) (ρ/a) ] ρ < ρ B ρ ρ > ρ The use the Gree s fuctio to fid the potetial, V. ] 2 V = 1 4πǫ ρ dρ dθ Gλ A alterative method to obtai the Gree s fuctio is the method of images replacig the cylidrical surface with a image lie charge below the surface. This is ot developed ad was ot expected to be attempted. 4) Usig the Lagragia, fid the positio where a mass, M, slidig without frictio alog o a hemispherical surface of radius, a, leaves the surface. The mass starts from rest at the top of the hemisphere. 4

5 The Lagragia is developed from the Kietic ad Potetial eergy. Use polar coordiates. The costrait is; The Lagragia is; T = (M/2)(ṙ 2 + (r θ) 2 ) V = mgr cos(θ) f = r a = L = (M/2)(ṙ 2 + (r θ) 2 ) mgr cos(θ) Add i the Lagrage multiplier costrait, λf The Euler-Lagrage equatios become; L r d dt [ L ṙ ] + λ f r = L d θ dt [ L ] + λ f θ θ = This gives the equatios; Mr θ 2 Mg cos(θ) M r + λ = mgr si(θ) mr 2 θ 2mrṙ θ = Now r is costat, so the derivatives of r vaish. Solvig the equatios; θ = (g/a) si(θ) θ 2 /2 = ( g/a) cos(θ) + g/a λ = = mg(3 cos(θ ) 2) θ = cos 1 (2/3) 5) 5

6 Fid the Fourier compoets of a square pulse of width, d, ad height, h. The pulse has the form; x < d/2 f(x) = h d/2 < x < d/2 x > d/2 f(x) = A cos(πx/d) d/2 d/2 odd dxf(x) cos(πx/d) = (d/2)a A = h π si(π/2) However, for a truely sigle pulse use the followig Fourier trasform. A(k) = 1/2π A(k) = 2h si(kd/2) 2π k dxf(x) e ikx f(x) = h π dk eikx si(kd/2) k 6) Show that; δ(x) = (1/π 2 ) dt 1 t(t x) represets a delta fuctio by evaluatig; I = dxf(x) δ(x) Use the Priciple Value itegratio. Whe f(x) is aalytic ad has appropriate limits at ±, the; f(x ) = (1/iπ) dx f(x) (x x) 6

7 Apply this twice to evaluate each itegral (ote that usig the pricipal value of x x itroduces a egative sig for this itegral); J = dx (f(x)/π 2 ) ds 1 t(t x) 7