hw9s - ME?“ 634;. HW 11>? So [dd-Cums PROELEM 9.3...

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Unformatted text preview: ME?“ 634;. HW 11>? So [dd-Cums PROELEM 9.3 Part a Longitudlnsl dllplacaucntt on tho rod Iatialy thc wave (quarto. 2 2 - ‘ Du-Eumdchuunu‘f-K-gi. (I) ac2 a 2 x "51919,!rlzer6(x52) - Rc[3(x)05uzl to; aghusotdal cxcltaztonl. figs!) :19 ha vffi5jen is 3(x) 9 Ctstn fix + 929q5 fig Uherc 8 §HQJB71L The two conniantl It. lound (tom ch. boundary condition- , , ‘ n 3—3- (m) - - NHL!) + Ht) ' Ch) 2 3t , 50".!) " 0- (c) Than conditions bacon. ~Hu2 Eu.) - - AK f}:- m + to (d) and 3(0) - o (a) for tinusoidll cxcttnttons. “ Now we find C2 - O and l c I ———°—~____ . 1 AE3co¢31~ .Vuzllnal, a) Hancc. 60m) - ——'&————_ mu .3“) . “swan-manna: " ° - m and T‘x't) . E g% . W Ref! .30“) AEBcosSl-Hu’ttnfil Part b A: x - l. 6(£.:) - ——L-—- net: .3‘“) ' AEScotfil-Huz ° (t) where-scotsl - wJE7§ cot (ml/07:); For "all u. COMM/(V?) ‘ l a tale --l—-- ((t) A! 2 (J) 'i ’ H“ . and 6(l.t) - p, PROBLEH 9.) (Continued) This cqulrion 11-51 uqod to describe a has: on tho Ind o! I nasal-or spring: 2 d No ;:; - - K: + {(r) (R) and '- x - Ic(§05ucl.' - H uzi - - X; + f o .' ' (I) or x - ——1—,. an m K-H u' o Coup-ring (J) and (L) we note that .. A! . ' x t and to t. (n) Our comparison is cemplctr and sinco H >> 0A! we can us. tho narrlcss aprin; nodal with $.noli No - H on the 0nd. ‘M nonim 9 .5 A rclponso tho: can be rcprclcntcd purely ll l vnvo trnvoiin; in tho negativ‘ x diroczion implies tho: thorn ho no vsvo rofloction at tho loll-hand boundary. U. nus: hove l «0.0 + —- No.0 - 0 If m or seen in Soc. 9.1.1b. This condition can to ouristiod by a viscous do-por alone: “10.0 + lv(o.:) - o ' (b) Honco, we con write a - Alp? H - o (C) K - 0. “031.84 a. 2’: This problo- lakes tho nan. point on Proba. 8.16 and 8.17. with tho ndditionnl effect 0! material notion included. Regardlesl of tho nation. with the curron: constrained an given. the nagnotic ficld intonaity is sore to the right of tho block and uniforl into rho papor (3 direction) :9 tho 10!: of the block. uhero - - 9 "-1"1 , V (a) The only contribution to an int-grarion of the otroan tonsor ovor I ourfnco enclosing the block in on ch. 11!: surlaco. Thus I 2 ' (x f d: fxx - - d1 5 no": (b) 2 d: Io . ( ) "'z"“o(r) ‘ Tho nagnetic fore. in to the right Ind indoptndon: of tho Iaznncio layuoldl number. PROBLE‘ 9.6 , I _.. . . . - tint. we can celculete the force a! umetic origin. 1". on the rod. I! we deiine 6(l.t) to he the e.c. deflection of the rod et 3 - i. then uein; Anpete'e leu end the Maxwell etreee teneor (tq. 8.5.‘1 with negnetoetrictiol ignored) we find Hokflzlz ( . (I) ‘ 2(4-5(z.¢)15 1hie result cen eleo he obtained uein; the energy nethode 9! Chap. 3 (See Appendix E. Table-3.1). Since d >> 6(l.t). we nay linearize Ix: voluzx’ quflzlz - . (x I 2‘: + T 50.0 0) The first tern represente e constant force which is balanced by e stetic deflection on the rod. If He eaeu-e that thin etatlc deflection is included in the equilibrium length t, then we need only use the last tern of I‘ to conpute the dyne-ic deflection “2.0. In the bulk of the rod we have the veve equetion: to: sinusoidal variations 6(x.t) - n.(3(x)&3“‘1 (c) we can write the conplex e-plitude‘ZH) u 3(x) - C1 ein Bx + e, cos 8: ' ' . (d) where 8-- u|%. At I - 0 we have etfixed end. eo 6(a) - 0 end 62 5 0. At I I l the boundary condition in e 38 ' I O - I: - A! 5-; (Mt). . (C) o: 2 2. , ll N3 ! o * d3 0 - J 6(x-l) - M: a: (P1) ' (I) d PROM)?! 9.6 (Continued) Substitutin; vie obtain ' ' u a;:311 ° c sin e: - c Asa cos Bl _ - t.) d! 1 1 . Our solution is 30:) - cl ein 8x and for s non-trivial solution we trust have C1 5‘ 0. So, divide (g) by C1 and obtain the resonsnce condition! u A8212 ° 3 J sin 61 - ans eos a: (h) d Substituting 3 II «mfg-sud restun'ing. 9e heve J , is? (th-g-j I tsnOnlJ-E) (1) U N'I‘l o vhich. when solved (or u. yields the eigenfrequencies. firsphicslly. the first We eigenfrequencies ere found [ton the sketch. . the slope of the straight line decreases Notice that es the current I is increased end the first eigenfrequency (denoted by ml) goes to zero and then seemingly “appears for still higher currents. Attuslly m1 now bacon-es inn-Einer and an be found from the equation 3 {gm-1124355) - my. (mind?) not! (1) Just as there are negative solutions to (i). w 1. an: .. etc., so there en now solutions 3 Hull. Thus. because «)1 is bassinet-y. the systel is unstsble, (smplitude of one solution era-ring in‘ tile). Hence when the slope at the "night line becomes less than unity. the sync. ie unstable. This condition an be stated as: J STABLE ———> E: , > 1 (k) u N‘I‘J O or ad’ - 1) UKSTABLZ ———+ 2 3 < l . Q "on X j PRGELEM 0.2“ Part I The (tn and at x - 0 input: that mm - 0 and using equations 9.1.21 through 9.1.:fi we can onuily find that veiocitv pulse. "ounce of!” x I 0 boundarv with the same sign and magnitude. the value! for v(x.t): 7or the x-t plane no can indicate Part 5 u. can make use of part (a) if we use superposition. Consider the super. position oi boundarv and initial conditions: e free end. 1(0.t) - 0 91th the initial conditions in part (a) and the ?(D.t) 1: Shaun in '13. fl.P20b with initial conditions on T and v zero. Since the syste- in linear, we can add the velocities the: result tron the tuo situations end thus have the not velocity. For the response to the second act of conditions we have 1 0/4, Add this velocity set to the net in putt (I) and we obtein: 11 "mu; 0.:3 Pair 3 mu par: 's xl-ilar to Hal». “.M ulzh tvo'unnltllcuimn V0 - 0 1rd :hc mu 1: Wu: 1: width (My) Instead of 2:1. The No "Mutt uhle yinldln; :hc natural (rc- uenctu an "3" |1a(uL «453). n ' (fl) and 20 . ~45?» - nu(u!. 0" V42 (1) yields UL W? - 9- when n - l. 2. ... and corresponds to ulucim which an "odd". or fix) - - Ef-x). 0:) can In solvul graphically and correspond: to aoluzlons vhtch an "even". or {.(x) - ((-3). I Part h The effect of taking ‘3 u to reduce the unenfrcuucnctu of tho.— "evnn" mom's. The "odd" loluttonu predicted hy (s) an: Sud-panda: of the mu :4. This» is physically. réasonahlc sine: than 13 a node at (ha WI, and “net the man doom': lovq than 1: no Mann! force. For do. “oven” uoluums prodicua by (h). u. node: that 1!." - 0 we have annually the {annual frequencies of a umbranc of length 21.. A: H - -. tho syn:- responds 11k. two “Hugs; ncuhnnu o! lcnxth L. Thu infinite nu net: 111:. c rigid boundary. . . ...
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hw9s - ME?“ 634;. HW 11&amp;gt;? So [dd-Cums PROELEM 9.3...

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