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solutions_final - Physics 604 Final Exam Fall ‘11 Dr...

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Unformatted text preview: Physics 604 Final Exam Fall ‘11 Dr. Drake 1. (40 pts ) (a) Evaluate the following integral 0° l. peg} IZIWdZ—Z—zU—COSZ) (b) The function f is defined by :18 . with a branch out between —1 and 1. Give a plausibility argument why the branch cut does not need to be extended to infinity. Evaluate flirt) where a is real and positive. 2. {40 pts) The difierential equation for the modified Bessel function is “.2 H by +zy’—(zg+u2)y:{} 4,... (a) What is the lowest order behavior of the solutions of this equation in the [$ vicinity of z = 0? Consider all possible values of real V. (b) An integral representation of the modified Bessel function of the first kind, 1,,(2) is Mg) 1. i Biz/2)(s+1/s) “35 _ 2m 0 31+“ where a cut extends from zero to negative infinity in the 3 plane and the contour C' wraps around the out, starting at negative infinity below the cut and ending at negative infinity above the cut. Evaluate the integral approximately for 2 large and positive. 3. (60 pts) Consider a two—dimensional cylindrical cavity of inner radius a and outer radius b that has an angular width $0: rI’he electric potential V is mainw tained at —V0 and V5 at (,5 :— —q§0/2 and if) : gag/2, respectively, and zero at r 2 a, b. The potential V9", 923) satisfies Poisson’s equation .18 a is2 2 A——— —— —-—— Z V V_ T3TT8TV+ T2 8¢2V 0 (a) It is convenient to change to a new variable 8 : $710") before attempting to solve this problem. What is the equation satisfied by V when expressed If in terms of 5 and Q5? (1)) Write the solution for V in terms of a separable set of eigenfunctions (Dmfit) and Rm(s) as follows: V(S: Cb) : Z Cmles)(Ilm(¢) _-~— Write equations for ‘1)le5) and R1145) and solve these equations using 2> boundary conditions appropriate for the solution V. Normalize the basis functions so they have unit norm. Hint: Use the symmetry in 95 to simplify the solution. 2 0 (c) Solve for cm by matching the form of the potential at the side boundaries. 4. (60 points) A liquid is heated in a hollow sphere of radius 1) by a pulse of heat at t : O at radius r r: n). The equation satisfied by the liquid is T . 595? M anT =- A5{t)6(r — m} where 1 8 a 2 = *w 2_ V T2 Br?” 8?‘ The temperature of the liquid is zero for t < 0 and remains zero at the boundary for all time. .—--{a) Construct a set of basis functions @5710") to describe the liquid in the cavity. :2 D W'rite the basis functions as a linearcombination of the two solutions of the Bessel equation JyU-cr) and YVUCT). Give expressions for the eigenvalues of your basis functions (in terms of the known properties of JAM) and l’AkrD and define the normalization so that the @710“) have unity norm. State why the basis functions are orthogonal {you don’t have to prove orthogonality). What is the behavior of the eigenfunctions near r = 0? Sketch the lowest three eigenfunctions. I Hint: Bessel’s equation is r2y” + ry’ + (1627"? —— 1/2)y = 0. Let qbn = gym/rU2 and-show that gn satisfies. Bessel’s equation. What is 1/? [1 g (b) Write T(7",t) as 00 T022?) 1' E Curiae) 71:1 and derive an equation for .1105). ‘W’hat is the characteristic damping rate of the eigenfunctions? [ g (c) Solve for Cn{t) and then write a solution for the complete space/time dependence of T.’ g (d) What is the lowest order non—trivial form'of the solution at late time? Ehfl/ EXflM/t g9/Zi 7Li0fi§ péyS‘TS 5536/ @ 6L) EUR/615L716 ”SEQ-$0 ~-/‘aga:j Mafia “ffizflz 7%Cmc If We Slfl§'%Z;2“{7z/ fl% 9:3 “:> m [:06 7/?2€ [9.51%GFLW [177(65r‘fc/ fiftyfiy 111nm” <3; a go {my} gp/JL /‘[email protected]—/ Zr“? 7L0 ejyomv (577([31 / A‘sfiyya’ar) {072$ l:— gaffer {z MS?) C/ *3 ..... ng’é/L - J— iii-1'9: Q 2 C‘ 5&2- u \ H -’—\ 1‘1 C/US‘E I. {57¢ m to: LIW3 “:3 {KW} [ynfmzécaficq {Ewan ggq? #[twc’f/‘6 ”L: ~—L --=} (‘3 I] 3:0 fiméé Swat” 1/29 cm 5/52;er ffi/C‘g [/27le I; {m MEL/3 22> 14?)“ [fiq%flféw {0‘4 744’967 gkmct- ("um/i “:5 {Tow/am E éf/MW? fi.‘ .-"H 1-13—Jgt: flag)!“ 1327' 8/18 (/ng I; a»: Lh‘P ‘2) f7?) Ema/035% fismfg... 96> FE?! 6 B 49/: {:57 fif/amgé 32*» g :25 E- Who-.47 ("S ‘ 61 gm-‘jff ”(fl/Q65? fib/[éf/aw: _ «5-; {tag 9197‘ Hgflé’d fifw. Edflflufiét (”UL %: («Q I é ;m# N _ E q [dz/3C #:7qu fl "€59“ /flfl/fi€ POSflliwe Q— 7LQ€ [V27lfifia/‘faflfl/fé: dgfi’mrr7/c7éfl% 5; 7716 KKWVI/fi 79%? 7:5 /?G7é 147/1 %/éci gpg 5:“: 3&3) Ln]: EEO—€33 :6 :23 3 .—_~ V 3: lg Rmés EMQCQJ 'Wavz‘f agar/[zfij Qwfi‘fivws (v: S s’mce mac? 7&0 Marz‘lécéfl % Cl £+ l 6%; 9 Rm “V git; kmgj ((73 Kim S (NWT % {QM 3'0 W‘ “S:— /MC6<I)J [146$] 6117; 5-: ””5 Maia? fiazf lea [GM aid—[WA 7—" m ”ET km: W)? '35. 1gp?" _ (”@flg) gal Aim €\ _, L "3:”; Sfi/géfigw {3 {mfikaYMméfiwd QCMag’C (€123: "'1:> EM m ‘5";«L1Qfm665 [ - mm at c2: {.2 Cage gateway/9,) Ma V9 :- figmgmca 2—,»an .652) ”1% MRS/fife] éx/ Qmfigfi 421% [m7if‘t/mmaft" 55”?” g “O zflm km 4., Li a 5‘ 14 S i J. 37‘ L M» - fi‘sf’ VJ, 1V Cp Chg? + fa L 966mm: L C‘Q/I ~«—-— V‘L .. {—EA WK ‘5 3 A ‘55} W- S; 1% :3 “(Cow ,f‘d éfic’: BC: 5 FQCQ,‘ (Em/f :0 c? 7:3 gaTg‘FJcpoF 5571; 833 E :35 QM®( 1M6 fij‘féjfcymc/ ézzfi’ang’ IWQMV" 7‘0“?) .Jmft é cams Mam 6‘2 m 3 i _.1 ‘H ...
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