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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.» — “WmW3 [(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
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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: stitor(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
"113 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 .
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‘ H ‘ \
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_] 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 frequenty 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 _ 31” 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'ui — 310 "ifs — (1. III mfs +0150 mfs) — 9" v= ‘.
_:smm;. — (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 Kband 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 1091;] = 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 uImilivs: 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 frnqmm'y L‘ ﬁn,“ =2—K’Fﬁﬁllz 9:01;! Hz. Finally. tlu tl’Nnmv lntwmn :uljzulut 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 cntrgy 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 vmrgy 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).
 Spring '08
 THORNE

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