hwsols6 - Physics 214: Homework Solutions #6 Ffiday, March...

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Unformatted text preview: Physics 214: Homework Solutions #6 Ffiday, March 7, 2008 Prepared by: Kendra Letchworth Weaver Physics 214 : Homework Solutions #6 Problem 1 Problem 1 00 mm) = 2—“ + 2 [as cos(kna:) + b" Sin(knfl';)] 11:1 fats) = 5'23 + Z on, cos(kna: + 9%) n=1 foes) = Z on exp(iknx) Rift” (a) The easiest way to proceed with this calculation is to write the cosine in f B (x) in terms of complex exponentials using WM cos(kn3: + 95”) I 2 This gives 133(3) = 229 + 11:1 2 co m exp(iqbn) exp(’ikn:c) exp(ii¢n) exp(—ikna:) = 2.2; + Z cn(cos(¢)n) + isin(¢n))—2u + cn(cos(¢n) — isin(¢n))wfl exp(ikna:) + exp(—1ikn:r) : co +ic “04¢ )“H— —cnsin(¢ )W n n 2 n _ 21'. = i + {on case”) cos(kn:v) ~ an smut) Simsan Comparing the previous expression with 124(3) yields an = 6n 005w”) and b” = ——cn Simon). Solving these expressions for on and on gives cu = afi + bf; bu (15,1 = — arctan an Of course for the case n = 0 this provides co = a0 and (to 2 0. (b) In this section, begin by reexpressing the sum over all integers in fig (2:) into a sum over the positive integers. fcfi?) = Z 0n exp(iimr) n=—oo Problem 1 continued on next page... Page 2 of 18 Physics 214 1 Homework Solutions #6 In an intermediate step in part (a) of this problem f3 (cc) was expressed as: 153(56): % + :0: M expfiknm) + m1 2 2 11:1 Comparing the two expressions gives the complex amplitude 011 as C i on exan) 11— 2 ,n>0 onzaeméflmd co:§,n=o Problem 2 a!2 w A mi+b£fli+kx= : —-sinwnt Tl. 71:1,315. . . exp(—iknm) Problem 1 By the principle of superposition for differential equations, it is sufficient to solve for the response to each term 32,16) separately. magma +1)d31ch +kzr — Asinw t (it? dt " i n " damn dmn A 1-” t m dtz + + k3,; — ] daz dz A - '1"! b_” k n = _ zwnt m (it? + dt + z 118 Guess zn(t) = A(wn)ei“'“t as the complex solution, so in this case the physical solution is 32,16) =lm[zn(t)]. Plugging zn(t) into the differential equation gives — izn + iwnbz" + kzn = gem” Substituting 25,105) = A(wn)ei“’"t, k: = mg, and b = 27’” gives ,2mwn A z¢l(wfl)[w1rnw,2I + t + mug] = — r n Solving for Aha”) gives A A(wn) : “m 2% 2 -2w n10 wn+z~f Since AM”) is a. complex number, it can be written in polar form as Aw”) = meow“) Where as in the notes, A !A(wn)l 2 % (at i use + (sew ,1. Problem 2 continued on next page. . . Page 3 of 18 Physics 214 : Homework Solutions #6 Problem 2 [(a)] 22,, flea”) = — arctan [ 2 T ] “’0 “W12; Substituting Q = %‘1 into the expressions for |A(wn)l and Mid”) and writing both expressions in terms of the ratio 5% gives 2 A w” 2 tan 2 no.» — “Wm-W3 [(1 _ (5;) J + (m) J Mm”) = —— arctan “’0 —1/2 The damped oscillator’s response to each term in the Fourier series for the square wave is $n(t) : Im{2n(t)] : Im[A(w,,)eiWnt] = fA(wn)|Im[ei(“"t+¢(“’”))] and the total response of the oscillator is the sum of all these terms, or 00 .1105) = Z xn(t) n:1,3,5... (a) The homework problem asks to consider the case where wl = we, but it is informative to consider a more general case, wl = 5,9 where l is a positive integer. The case considered by the homework is the specific case where l = 1, but the case where l = 20 will also be considered. Thus, in the expressions for |A(wn)| and 45(wn), the ratio Eff can be replaced by thermal” Thus, excluding the prefactor in the amplitude 3:435, the amplitude and the phase angle can be entirely expressed as functions of n, l, and the quality factor Q. The table in Figure 1 shows the amplitude and phase angle for n = 1,3, ..., 19 for three Q values. The underdamped case has Q : 100, the critically damped case has Q = 0.5, and the overdamped case has Q = 0.01. The cases i = 1 and l = 20 are considered separately. ('0) To calculate the response to each Fourier term, plug the expressions for |A(wn)| and ¢(w,,) into the expression for 37,,(15) —1/2 (hams? W) : erg; Problem 2 continued on next page. . . Page 4 of 18 Physics 214 : Homework Solutions #6 Problem 2 (continued) Q: [0“. l=l M (n, II! .VI mmu'n ()=HKLI=20 _\(uhlfl_\flln0hfn lJKDSIm IL34llKh ([21333] tLth799 :1139333 t113n333 :1133192 (1152359 (1211877 £1537Zh7 ():1L5.I:20 .\lnhlfl.\nlnam‘n (L9975lm (1325995 (lexjjs ([1272h7 nJr424 n1m979h ILHS4U7h tLHJZhh? [[H3415 unflw 0:0." I. I:2II .\luhlfl.\flln0h‘n [[19h135 ILHJJITS n1u17994 HJHHHX HJHD4h9 n1ullh53 n1ulllx3 uluumxu tuuumuz n1uufi$4 -nluufi -n1un53 -nluuh7 -U1IBQQ -U1l|5h4 -UJIW39 JlHIIZS 41Hl7l4 41H3{m1 -n1H7l3 IHEi -KI3625 -KI3?92 -3.I3354 - 3. USSR -KI3903 -KI3923 -RI3933 -KI394I -3.I394? HJthbb njum333 HJHI297h HJHHBKU HJHHU5N {HHHHSN HJHHDUN HJHHQIH n1uun4h 0:05. |=l .\tuhVL\HIHOM"H {L5 1LH33333 HJNWh92 HJNBHS7 HJNH355 HJHNWJS {HNNHSZ HJHND95 HJNNBH} HJNNH45 Q:"."l. Izl .\lflhVL\Mln0m‘H — u . 'JI 'JI Ha Ha -L57 -14965 -17452I -2R5h2l -1913h9 -1953hs -198h4h -31xm3h -1tfl249 -31B483 .{ u r9991 4l29773 4L4899h 4Lh7335 418457I -l1ufih9 -Ll5275 -L287 -L4U899 -L5l953 “— I.— — — — Ill— — — — — 'JI 'JI —— — — — —— — —— — —— — —— — —— — —— — -L37333 -L5n572 -L5333l -LS4S73 -L55308 -L553IJ -I.Shl9l -L5h49h -l.5h753 -L5h977 HJNNH HJHHqu HJHHH33 H.33E-US iSYE-nfi 4,4E-n5 14IE46 172E46 Figure 1: Table of Amplitude and Phase Angle Values for 1:1,20 and Q=100,0.5,0.01 Problem 2 {(b)] continued on next page... Page 5 of 18 Physics 214 : Homework Solutions #6 Problem 2 Thus, the total response to the first 19 terms is 19 t: stit-or(trim‘m[arr—twat] 7513.5... (c) Figure 2 shows the square wave produced by the first ten terms of the Fourier Series with l = 1 or ml = mg. Figure 3 shows the same square wave with l = 20 or wl = g3. The period of this square wave is twenty times longer. Figure 4 shows the response to the square wave in Figure 2 with 1:1. Notice that the term with n z 1 donfinates because it is at the resonant frequency of the oscillator, thus a square wave shape is not observed for the underdamped, Critically damped, or overdamped cases. Notice however, both from the tables in Figure 1 and from Figure 4, that as the damping is increased, the higher n terms begin to hare more of an effect, causing distortion of the sinusoidal curve. Figure 5 shows the response to the square wave in Figure 3 with 1:20. These graphs appear to be much more like square waves. Note for the underdamped case, there is a lot of ringing (excess oscillation) but the square wave shape is still observed. For the critically damped case, the driving wave in Figure 3 is faithftu reproduced in the response by the oscillator. Finally, for the overdamped case, oscillation occurs, but it is no longer sinusoidal, but rather exponential. If you are familiar with electronics, the response of the oscillator looks a lot like the response of an LRC circuit to a square wave. If the resistance is too small, the inductor dominates and ringing occurs. If the resistance is equal to the critical resistance, the signal is optimal. If the resistance is too high, the capacitor dominates and exponential waveforms are observed. Page 6 of 18 Physics 214 : Homework Solutions #6 Problem 2 191 Z 7 Sim (ant) “1,317 l=1. Fourier Series Square Wave. n=1.3,..19 Figure 2: Fourier Series for Square wave 1:1 19 1 Z — 31m (URI) _ _ ":1 3n |=20 . Fourier Serles Square Wave. n=1,3,..19 J , ./" k ‘. “.5 I I ‘ (0“! 2H 4H m. m Inn llu‘ _n.5 Figure 3: Fourier Series for Square wave 1:20 Page 7 of 18 Physics 214 : Homework Solutions #6 g, \ Aka”) | 2 Im[t:xp(:'a;,1 + mm, D] |=1. Total Response. 0:100 ":73 _-l/ma§L III) —|m 1% Hm) \ _ 2 Im[cxp(icz;,r+whom] |=1. Total Response. 0:0.5 ,1'3 A/mab “.4 (00! 4!.) 4H 15' |.~1(a \ V 102 Im[exp(r‘a;1r+f0(a;,))] |=1, Total Response, Q=0.01 "11-3 A/mah onus ....I..J.I....1....:.A;JLAAAALAA(¢)0]' I 2 3 4 " 5 a itUIHS Figure 4: Oscillator Responses to Square wave 1:1 , Q:100,0.5,0.01 Problem 2 Page 8 of 18 Physics 214 : Homework Solutions #6 Problem 2 {2, |.-l((r_;1)| "EA/mag Iln[exp(r'a;1r+imam] I=20 . Total Reaponse. 0:100 I4 J J 1. . L 4 ; . J. J 4 J. mor 4n "hlll .‘ . an Inn ‘w lltll . I . w | ’1 I . I ‘ H ‘ \ ‘ ‘l , (pk ‘ ’ 1" ‘ ‘ x I _] II ‘I : v » Ev". | : II I I I H \ w W 1" .. I y -3 ' ‘ '9 |A((r_ I _ V 1”} Im[exp(icq,r+Maw] I=20 . Total Response. Q=0.5 "I; A/mcih .vx/\ PL/‘J‘ / x 1“ (L5 I ..1...1.a;|'...1...|;..l.(“or 2t! 40 m I an Inn 12H {i | Aug!) | 2 Im[exp(ia;,r + :‘o(a;,))] I=20 . Total Response. 0:001 WT; A/HHnJ “.2 / Figure 5: Oscillator Responses to Square wave 1:20, Q=100,0.5,{).{)1 W Page 9 of 18 Physics 214 : Homework Solutions #6 Problem 3 Problem 3 Recall that the frequent-y memzured by an observer fa is given h_\' e — e0 fo— f. r — r, where 1' is the \‘elm‘it_\' (If the “'m‘e. z'o is the \‘elueity of the ohserver. r. is the \‘elot'it_\.' of the souree {all measured with respeet tn the medium}. and f. is the frequeney measured by the souree. Let's take west to he the positive direetinn for \‘eloeity and define: r . . . t'b=-i.1llm,r'.~' Ls the \'(‘i()t‘lt}' of the hat measured w1th respeet to the ground em = 1.20 infa- is the \elneity of the moth llleasllred with respeet to the ground em = —l}.tit| mfs is the veloeity of the wind medium measured with tespeet tn the ground (a) We must eunsider the path of the sound wave from the halt to the mot l1. then the return path. pATH 1 But to moth: The frequeney measured by the moth is given by: r — (rm — t'n- ) f _ 3-1” mfs — (1.2llmfs + [Hill infra) — _—H2.U '.= 8278 '. r — (e5 — ru.) " 3 It) mfs — t I. It) mfs + 0.60 mfs) RH, ‘ RH! fol = PATH 2 Moth to hat: Again We ehoose West to he the positive tiiteetioll. so that the wavespeed is negatiVe [in Ieetllre we always elmse the positiw direetiun to he that of the wave. We get the same result either way.) “'e now eonsider the moth to he a souree and the hat to he the observer, The math retransmits at the same frequeney it observes. so J,“1 = fol and ,cfimfm: _1' _il'm— l'u-i — 310 "ifs — (1. III mfs +0150 mfs) — 9" v= ‘. _:s-mm;.- — (1.20mgs + [I.(sun.,x’s) “mm” 8‘ "’kH’ (b) The beat frequency is given by the different'e in the tum frequeneies 1b... = 102— L, = titiiikHz — 32 kHz = 1.551.312 Page 10 of 18 Physics 214 : Homework Solutions #6 Problem 4 Problem 4 Take v’ to be the velocity of the moving object, which we want to measure, and v : 343 m/s to be the velocity of sound in air for part (i) and v z 299, 704, 764 m/s to be the velocity of electromagnetic waves in air for parts (ii) and (iii). The Doppler effect for an observer moving in the direction of the propagating sound wave with velocity ’UO is: fo=(1w:—°)fs The Doppler effect for a source moving in the direction of the propagating sound wave with velocity us is: 1 f0 “ W}; In both of these formulae, f0 is the frequency observed by the observer and f, is the frequency emitted by the source. Initially, treat the moving object as an observer moving with velocity on =v’ [-v’] away from [toward] a stationary source emitting with frequency f“. We don’t choose a direction yet to keep the problem general. a = (1- gap, 2 (1.7%) f... Next, consider the moving object as a new source moving with velocity vs=-o’[v’] away from [toward] a stationary observer which is actually the original source of the signal. The frequency of this moving source is f32 =f01 because the frequency this new source reflects is equal to the incoming frequency it observes. 1 _ 1 w (1??) f02‘“ (1m%)f52— (1i%)f82_ f51 The shift in frequency is given by: I (14:? 1?”?! — 1i”? :F2 ' Afzf02_fai: Trig—1 f“: (—(gfik fsl: fslzlviizdfs} 'l} The velocity we wish to resolve is v’ : Av so the A f we need to observe is $2131) f viAv 31 Af=[ The smaller frequency shift would occur for the + sign in the denominator, which signifies the observer moving AWAY from the source. To verify that we can detect Av in any direction, we should choose the smaller Af. Thus, —2Av Ar= [ l o+Av For all frequencies of the source f,, H fa] _ QAU _ v+Av W Problem 4 continued on next page... Page 11 of 18 Physics 214 : Homework Solutions #6 Problem 4 (i) Since the velocity we wish to detect is Av : 1 m/s and the velocity of sound 7) = 343 m/s, A_f fol For ultrasound waves f3] =50 kHz 2(1 m/s) : — = .0 5814 343 m/s + 1 m/s 0 |A,af’|:.005814,1‘;1 : .005814(50 x 103 Hz) = 291 Hz (ii) Since the velocity we wish to detect is At: = 1 m/s and the velocity of light is v = 299, 704, 764 111/ s, Af 2(1 m/s) ,9 = ————— z .673 10 fig] 299, 704, 764 m/s + 1 m/s 6 X For K-band radio waves f51=33.4 GHZ |Af| = 6.673 x 10“9 f81 : 6.673 x 10—9(33.4 x 109 Hz) = 222.9 Hz (iii) Since electromagnetic waves travel at the speed of light, M = 6.673 x 16—9 fa] For infrared light waves f3, =332 THz IAfJ : 6.673 x 10-91;] = 6.673 x 10—57332 x 1612 Hz) = 2.215 x 106 Hz Page 12 of 18 Physics 214 : Homework Solutions #6 Problem 5 Problem 5 (a) . ._"_/**' .. .._.. _./7"_ ‘r.3.-'2. “( an gnu: tlmt lph =T— T. Thtrtfun. u _L :ph _L 7— “EL ‘ .mul (L; d ‘. r4 3 ‘. z. 3 rg.=—=— 4-“ =.— —-A-‘~'2=.—vp.. tlk ilk p 2 p 2 ('0) Recall that k = 25;“ let: (-nlrulmv: k1 ="Hfill1lllt'u1 = 153.". m‘1 #2 = 2:7,»‘1111vm = (522 m'1 and *1: —’ 3”: 137 r' P ' W 3“". .. = —'k-’ =1“ -' 2 P a “ “'9 can nuw find Ilu- plum- and gmup u-Im-ilivs: i + d-"ll ' i : — : a ("s lph A. L2H] + k2) ll.-11m‘ Am “:2 — “:1 , 0 ,. (I, — —Ak — —k2 _ kl — [LL] m“ .s . Ju' «l. . The lwm frnqm-m'y L-‘ fin,“ =2—K’Ffifil-lz 9:01;! Hz. Finally. tlu- tl’Nnm-v ln-twm-n :uljzu-l-ut maximu in tlu- spatial Imus ls tlu- alnnlun- mlut- ml 2 x Ik0“ “2 In [no ') l\tii'nwlnlw = 0. 1H In = I Ak_,-‘2 1"? m‘ vlflpr M Page 13 of 18 Physics 214 : Homework Solutions #6 Problem 6 Problem 6 (a) y [m] '04 L— S “I - [I x [m] :02 -‘{)-l r. d}' d1 [m's] .\ [m] unl l 'Znudy do: [J m] .me L -—. 1—2.] 5 Problem 6 continued on next page... Page 14 of 18 Physics 214 : Homework Solutions #6 (b) Hm] 04 .IL‘. 0 =02 -.U4 d)‘ d\ 114 —3 ‘III : -1): -.{H u II lmdg. of [J m] l‘ Problem 6 continued on next page... \ [m] \ [m] Problem 6 Page 15 of 18 Physics 214 : Homework Solutions #6 Problem 6 (c) 2 nir.f)=m.+up=%p(%) +%r(g) “'0 must intvgmtt- tlu‘sv tornv; over tht‘ lt‘ngth [If the string. Tlfl‘ nimvt- graphs allow us to do this quickly by finding the arm] nndvr tln‘ square of the plots and multiplying by tln‘ npprnprinlv factors. Tlu‘ total kinvtiv ('nvrgy is thle KP. %(.1kg,r’m )(2m( — .‘2 unis)? + 1m (‘4 Hula-)2) 20.012 J Thr tum] pml‘lltinl c-nt-rgy is PE =%lll.\'(1m( — .(I2)2+ lm(.ll‘.!}2 +1m(.01)2 + lml — 111)?) =ll.[l2[l J Till‘ tum] (‘nt'rgv is E = KE + PE = (1.032 J (61) Will: 1hr mrthnd used nhm'r. “'1‘ ran find the total vnvrgy at 1 I] s. The total kim‘tit' ('nvrgy is tln'n KE %( .lkgj'm )( lmlJ :nfs]: +2m( — Eng-".332 -+- [mi — Eml‘i‘s)2+ lml + .2111‘,-"s)2) =l].[ll(i J The total putmltinl vm-rgy is PF. =§ 1|}.\'(1mt.nn2 + '2m( — .02)? + lml 412)? + 1:14.02)”; =[Lll1IiJ Tln‘ total (‘ni‘rgy is E: KF. + PE 2 0.032 J. Sn wv find that (‘m‘rgy llfl'i hm‘n ('unst'rvmll Problem 6 continued on next page-. . Page 16 of 18 Physics 214 : Homework Solutions #6 Problem 6 (continued) Problem 7 1+3 1/2 fatfs [1_;] The velocity “Us is the relative velocity between the source and observer. It is positive when the source and observer are getting closer together and negative when the source and observer are getting further apart. (a) (He) (1* soy/2:10 (1- (“eff/2 (1—3:)? 1-”r C 144’s 1’2 fo=f5[1_3:] =fs C When 118 << (1, becomes vanishingly small, so the numerator in the formula above becomes 1. Thus the Doppler shift formula becomes: f0: 1 f3 _.2& 1 c which is the Doppler shift due to a moving source for mechanical waves. ('0) 1+E£ 1/2 z+1:[1y£] The largest redshift ever observed has redshift parameter 226.4. Since the universe is expanding, the quasar is moving away from us, thus the Sign of the velocity is switched in the equation above, giving 1Av_, 1/2 1: C 2+ l1+sl Plugging in 2:6.4 and squaring both sides of the equations, (6.4+1)2= 5476(1+%5) =1—"—: 55.76 = $53.76 Thus, 1).; = .964c away from the Earth. (0) For the quasar in part (b), vs = —.964c. The wavelength of the source is A32410.1 mu. Since f : 1/), the equation for the Doppler shift can be rewritten in terms of wavelength to be: 1 1 Hail/2 A0 AB C 1_2£ 1/2 a] Problem 7 [(c)] continued on next page... Page 17 of 18 Physics 214 : Homework Solutions #6 Problem 7 Plugging in numerical values gives: 1 + .964 1 — .964 1/2 A0 : (410.1 nm) [ ] = 7.386(410.1 11m) = 3029 run The ultraviolet light is Doppler shifted well into the infrared. (d) Since 1' = «311; where Ho = 2.5 x 10—18 s", # v m (.964)(3 x 10‘8 m/s) __ 26 r _ H0 _ 2,5 x HHS s_1 _ 1.1568 x 10 m 1 ltyr W = 10 9.46053 X 1015 m) 1.223 x 10 ltyrs Thus, the light was created 12 billion years ago. (9) The microwave radiation with wavelength 1.9 mm comes from the thermal blackbody radiation of the early universe, highly redshifted. To calculate the redshift of the UV hydrogen line at 410.1 run with the redshift parameter 221088, use A0 = Ase: + 1) = (4101 nm)(1088 + 1) = 4465989 11111 : .4455 mm Page 18 of 18 ...
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This note was uploaded on 03/31/2009 for the course PHYS 214 taught by Professor Thorne during the Spring '08 term at Cornell University (Engineering School).

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hwsols6 - Physics 214: Homework Solutions #6 Ffiday, March...

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