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Unformatted text preview: «seemsmwlsl, .. .szwwwmm 6f— a ‘100:
INTRODUCTION 7o
DIFF ER EIY‘H AL 5004 7mm I SECTION 3.8]
w_ Problem. Determine we, R, and 6 so as to write the expression u = — cos (t)+
\/§ sin (t) in the form u 2 R008 (wot — 6). Solution. Suppose one is able to write u = Rcos (wot — 6). Then, by using Section 3.8, Problem 2 the cosine angle subtraction formula ( cos (a — b) = cos (1;) cos (b) + sin (a) sin (b) ),
one could rewrite that expression for u as u = Rcos (wot) cos (6) + sin (wot) sin (5) =
R cos (6) cos (wot) + Rsin (6) sin (wot). Notice the similarity in form, now: The problem gives one the expression
u = — cos (t) + ViiSin (t)
and one hopes to put the expression in the form
u = Boos (6) cos (wot) + Rsin (6) sin (wot) In order for these two expressions for u to equal one another (for all time t),
the corresponding parts must be equal. (This is worth thinking about, to convince
oneself, but it does not take a deep proof — there’s no other way the two expressions
could be equal for all t.) So, wt = 1t, or else the components inside the sin and cos functions Would be
different. One knows then that we = 1. To make thecoefﬁcients in front of the sin and cos equal, one must have that
Rcos (6) = —1 and Rsin (6) = (/5, By dividing one expression by the other, one
concludes that 535% = tan (6) = ~31§ = —\/5, and so 6‘ = 43—“. To ﬁnd R, one may use the facts R2 cos (6)2 + 1122s,in(6)2 = R2(cos (6)2 +
sin (6)2) = R2, and R2 cos (5)2 + R2 sin (6)2 = (—1)2 + \/§2 = 1 + 3 = 4, so then
R2 = 4 and R = 2. So another expression for the given u is 2cos (t — g). ! 3L8p.1’ (a) m= .003 mags"? was I L: 5cm 1 LensW' QM,
mo): 0 K 4:6“  sec , w
u (0) ‘ l0 Cum/Sec, X‘ 0 I (no damp/Is, {I} no cu} NSCstana.) From ézbucdzon ('1): mu,[e)+ xu'(t)+.kult)= F00
No 44,”me _—> yao .J no “WA; gun a: Fc+.3=o.
‘30 ‘44“? Saver/aw?) eauaq4m 175: mu," + Jzu. =0. The mass m I‘s Swen, Smce mS=JaL’ 42.: :3 =7 JZ= (Iooﬂqglo) 3m/saL> .
5 cm 3° ul€)= Esth. /‘
114(0): HB=10=27 3: Li" %. W5: %$m Ht  (cm) Now, ale) ts {rouse earn. when Me = Tr “5 4416’ 4mm 08: (duck We mass Qwst W5 +° VG" 
aﬁuﬁhbndm pom't (u=o\,> ' ('1) my 325.
L ‘ 3m ‘ 1% 6t. 74(0)‘ "Luk. =_ v—l‘ibt Mo) = z 6* /sec X ‘ 0 (no ads Wasﬁance. ‘Vﬁo Mmﬁ) bekrmme 14W}; dekrrnme, ‘HSe £7 eﬁumcs‘ ampuludg Percod) g‘ phase
0g “he Moicm. ‘ ' ‘ .__....3 (“A m 3 I
m = . k _ % _.. g “L
32 @t/sec") J " "L = . '1... ‘2' ”a Solve uSmré utk)= e't  r” * 12% o a r= 1:21 :23 J =ti3ri’ 1'7 we)? A cosémt  3 sm (“at .
 Use we when! candﬁtwns +0 dekrmme Am 13; , 7110) = A: 112, u) 111(k): 'T‘i‘CoslgfBt + Bsuﬁgmt‘. 2 Mo): (8E)B= 1‘7 3‘ '27:: 1E
5; ult) = ’7'; (:05 (Nat + j}; sentemt. [Emma]
Z I 29(‘1 (4) mzzoz L ’ 5—0“ mass  stde 7 s #00 dsne‘sec/cm, drspvarzment — (gm, #9:“  dw home ' sec 11(0)= Z cm U'IO) ‘' 0 (mac: rs "released” qu u. It); Dekrmme chat!“ ““5“”?! and twist peeved) Behrmm'e
“at (duke ob We 9‘44“: 'penod 1b the paced agﬂae undaw mutton. W E0. '7‘. mu? *1“: o Jzu = F'Cé).
Sm no Miami Quay (Fm) Us appléd, 4w sonarnuns Issuedton vs; II I
mm + Nu vJzu=o. M= 20 3 vs 8””) b"— 400 damsealem vs gown. TO dekrmme J2 ’. "15:421. =‘7 J" T3 = (25351903: 3920
SO II I (Z0371, +(qu)u 4 (3420):; =0
9) u” + (20);; « (Ha)u=o. Solve Mama Solem who ﬁrm ul£3=ew 3 ﬁne» 1  .
r «ear mm :0 =—’> r‘~ ZOtJtm"q(l%3
" ——————* ﬂail76 = 401:5, IH'é‘. So 'ulb) =5 8,” [/4 0:5 (W251? +3 sin (QW\t] . (I: CowHMedl] 3' a 3» a n. u.M¢m.zua«ahm.4 : an V Wow. use we ma: conddzms 4° damn Am: 3, ’ Iot ulo): A’=z ? ale): e [Zooswrmtyasmlqma The w/wspmdu'vs waned when would be decrcbea b3 ‘lhe
Solu‘Han in w» ejuaﬁmzv u,”* (Hem=0 . awn Kc“ we +13 am 1% So we = undanytd vagumcg = I“; and T’ undanpld paced ’zﬁts: Mn TL 35.3.7.» 1.,
‘ 'Tzm ‘7’ Z"? @ A certao". whatM5 33:". mm. u"*ru'+u=o Fad 45¢ valu g . e o
W damixnj cwepfcwu X 5°" WV“, 3mm: Pound «pike dam/led mm
Is 5070 armlerﬂw‘lhe Feud :6 we corrrsiamdma wvdampd mowcan W“. all? {u 111‘0 '1 u: ere
rt... Kr " .
* o 57 r‘= ”1+ 1"
L’ ‘1 f 5' am. ‘2
e [Acole‘ljérBsulJl—Z‘Jt]. The  I I,"
$uﬂSc 9N3 $5 /‘= 1" . ‘Ih T _._..__
" ‘3 9 , e rmi Feud rs d: 211' 1? f
UM“ Mm u
u fuse .. u:— e” I.
r+.° 27 rs ti ,9, g ~ ~
g ult) A @sf + B amt
O ‘H‘? Undapw
##3va wt)":— 'OndWhound
I , Md paced T 3:" 311 “M V we want 40 ’
. d ’
/ 6'” X so wrt Ta‘ “.m pr
=(I'9W I z .
x? a ~ ’ ’1‘ ‘ ,
r: _ (W ‘ q a “kW—217‘“)
=> 11‘: ‘4‘"? 4* _ [MUST4 94
u.s)»"““"':"““"'s‘=i£
(LS) an" (1.5? 4.0.9”
Balm? :1  @ m3= 8th ”£115
washi  ‘ lb
dltplatemﬂs“ ' — 24: bcmc " sec L‘ LSL‘n. ‘' th. . Y = dampmg Mcheu‘l DeJerMme X Sowt W Sgsiem Cs ad‘h‘mws damped. Caowmmb Fined1m: Mu“ h" + Jzu =0.
m = .3.” _ ' L 15"“! 32 wm‘ ' q 3* '
Ja= 1%: Eﬁh = (,q 95 ’— ‘i 6e ‘8 Wm; “5ch (a e ‘11 ‘é‘ [I u ”ma zseu=o =27 r‘uIxrvrzsoao .5? r'= “95»; W: 'ZHquhzrb. Section 3.8, Problem 21 Problem. a. For the damped oscillation described by Equation 26, to Re“??? cos (pt —— 6), show that the time between successive maxima is T, = 277'. Solution. Maxima of u(t) occur at alternating roots of its ﬁrst derivative, ll u’(t) —— that is, the values of t for which u’ (t) = 0 will represent a maximum, then a
minimum, then a new maximum, then a new minimum, et cetera. The derivative 11/ (t) = —§:—n(R'ye'§1rtﬁ cos (pt — 6)) '— Rue57% sin (,ut — 6). Its
zeroes, then, occur when ‘67}: cos (pt — 6) — Resin (,ut — 6) = 0 (as €577: cannot equal zero for ﬁnite values of t). From this, one concludes that sin (Mt — 6) = —$n1; cos (,ut — 6) and so solutions
occur whenever tan (pt — 6) = ~ﬁﬁ. Here, some cleverness allows one to solve the remainder of the problem: one
knows that tangent is a periodic function, with period 7r: if tan (,uT —— 6) = —2—;(7,
then one also knows tan (,uT — 6 + 7r) = ~73”? So the difference between one root
T1 and the next T2 is f. (One knows this because ,u,T2 ~ 6 = uTl — 6 + 7r, and so
uTg—pT1=7r.andT2—T1=%.) , _ , However, as noted above, successive roots of the derivative represent a maxi
mum, then a minimum — the object models oscillates from top to bottom. It takes
two intervals, two roots, to get from one maximum to the next. So if T is one
maximum, then the next maximum will occur at T + g + $ = T + 2779 The interval 
between roots is 3E. b. Show that the ratio of the displacements at two successive maxima is
given by 6%}. Observe that this ratio does not depend on which pair of maxima is
chosen. The natural logarithm of this ratio is called the logarithmic decrement and
is denoted by A. Solution. Suppose that T is a maximum. Then u(T) = Re‘gg cos (uT — 6).
What is u(T + Td), the next maximum? By substituting t = T + Td into u(t), one
ﬁnds it should be equal to ReJngrmlﬂ cos (MT + Td) — 6). The ratio of u(T+Td) to
u(T) does not seem particularly simple, until one makes an observation — cos (,u.(T + Td) — 6) = cos (,u(T + 2715) — 6) = cos (,uT + 27r — 6) — which is of course equal to cos (/rT — 6). Therefore the expressions simplify: . lcommueﬂl
l. '2 Q n. :1 Cl u(T) Rea—9% cos (MT  6) = «2‘23» '= e;%_(_va;ran)) = 6%"? “(T + To!) = Re‘ﬂginlﬂ cos (,u(T — 6) e‘ﬂngﬂ c. Show that A = 777% Since m, u, and A are quantities that can be mea—
sured easily for a mechanical system, this result provides a convenient and practical
method for determining the damping constant of the system, which is more difﬁcult
to measure directly. In particular, for the motion of a vibrating mass in a viscous
ﬂuid the damping constant depends on the viscosity of the ﬂuid; for simple geomet
ric shapes the form of this dependence is known, and the preceding relation allows
the determination of the viscosity experimentally. This is one of the most accurate ways of determining the viscosity of a gas at high pressure. Solution. A = log (6%?) = 35%. As Td = 27“, then A = g  2; = a]. Salve 75370:" f/éCI;O M/0519» val/0) :V
'6. _ [.66 q: 6 . (“D/056216! Mfa #6: DE‘C9WQ& %e C“) AamcknB/TC e8, 05,7804 u/033'Q= C1
(4V0)! V Ca‘
Cine/1 Cuban” block a? scale ﬂ
ma 5
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dfsplaced kiwi Lee ‘lhe Moot be gubmrsga “b a. depth L ed:
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 Spring '04
 Yoon
 Differential Equations, Equations

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