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Unformatted text preview: 'imo the third—order Euler equﬂliun quickly yields the dCSiIcd SECTION 5.4 mechanical Vibrations in this section we discuss fourtypes of u .m.
underdajnped critically damped, and uvéﬁﬂ  V ' 3 . Inmiunn l rsl ‘11: msformed constantcoefﬁcient equation then 115 1 cases — .in which actual oscillations cc'wj age
a mm the corresponding general solution with v = In x 'i()l_;. Mimi” HUM1.11 ' ,‘m {:1 II i. Frequency: (1),, = ‘lklm = 1544?: “6 Period: P = brim“ = 23/2: "’3“. Frequency (00 =Jm =Jm=8 r'iiii
Period: P=2ﬁitﬂn =2ms= rrM see
0; rl+6r+25=0; r=3i4i . ‘ ._ .
. The spring constant is l: = ISNIMOm " gnawingmi" ~
v)+czsin(4v)] = I'3[c.c05(4lnx)+czsin(4lnx1] with 1(0): 0 and :40) = —10 15er mam11:1”Imam 
frequency is (00 — m: m =_5 rail: .1» .3236 see. {ﬁeoenierofmeeaMis g a 5111/3?
) frequency w nfa pendulum is given.
. .,_ 25d!” : ZERJL/GM. day (36400 sec) at Paris, then jls ﬁoniltﬂkﬂs 24hr2mjn405ec = ; . [e i5 [)2 = 86560”! SEQ NOW
[is and R2 its "radius" at the equalOL = 3,132 of Prob'mﬁ (“"91 Ll. = L2) yields
.pﬁsﬁc) calculation gives 7.33 ml as the thickness mm 5 (3950+x)1i100 yields x = _9795 mi E a correct clock with unknown pendulurn
.—. 35400 sec, so npl = 86400. The given “Jpnind p2 loses 10 min = 600 sec par day, so
u Milli omeblem 5 yields But the that that the bun” wiiiiti‘ i
these two results yields a a 0317'. (c) The solution of this equation with rtOj' a
when: ox. = ngR. Hence, with g = 32.2 ﬁlm
[hat the period of the particle's simple hmonio‘rifo'ﬁt'fiii'  p = 2m, = when; w 5063.10, ' 13. (a) The characteristic equation 101v1 + 9: +2 = (5‘ mm.
,— = — 2 f 5, #112. When we impose the initial oondiiio‘iﬁs ., A
general solution x(:) = o‘e'm’ +c1te'”1 we get Lhéﬁiiiﬁ .{ x“) = 50(e72115‘e—rn) (h) The derivative m) = 252"".4'20‘; _
when r = 10 mm.) as 2.23144. Henna“ ' 14. . 1: == i252"” and
91:3). 2)r3+3"+4=0 has roots r=_2 _
(I  4.
9):; ;’{O)=0 on the general solutirm 2.r 4[ ‘ _
Irmlntion :0) = 4:  29 that dcbcﬁbes {2)r2+4 =0 has roots r=1—2l‘/i gqtuti n U
101:2. x10) = D on the general 501ution E 17'
.ser]iep1rli6ﬂlar solution um = 2.10505” . sham: in the following ﬁgure. with damping The characteristic
When we impose the initial conditions .1: t hll WW I
on = (c. + c106“ we get the minim; ‘ ""“"““1rIm ‘I '
critically damped motion. I .
without damping The characteristic aquatic
we impose the initial conditions Jaw] = 5. :16
W) = A cos(4r) + B sin(4r) we get the pmicmarlm _v “(0 = 5eos(4 r) +%sin(41) a Icharactcristic equation 3r! + 30r +63 = 0 has roots r = e 3‘ _ "I.
toitialoonditiuns 1(0): 2, ITO) = 2 on the general solution
'geltltc'particular SOIUUOI'I I“) = 48—3"  28"“ that describes characteristic equation 3r2 +63 = 0 has roots r : t tJ2_1.
"conditions xﬂl) 2. x'(D) 2 on the general solution —5 ll_} wcgct the particular solution 18. With damping The characteristic aquatic ‘25:" IE!) u Ema?“ _ 02149) ‘ When we impose the initial conditions :10
I x0) = 24' (AcosdH Bsin4r) weget the I inthe ﬁgure at the top of the next page. 5 x0) = ‘ 2 8—]. sin 4: = 2!“ mﬁiﬁl  , .+169=0 hasroots r=5!2i6i. When We
=4. x’(0)=]6 on the general solution chmtnristic equation 412 +169 = 0 has roots r = 1r 13:72.
conditions 1(0) =4, {(0) = 16 on the general solution , 13:! 2)“ get the particular solution ' E] s i 233cos[£!—O.55}7 .
l3 2 21. With dampin 7
When we impose ' x(1)= 8"" (donut+1 that describes nucleidamped ml ' a
Without damping The ChandraIi];
When we impose the initial Common "(1) = A cos(2J§ t) + Bsinu‘fgilweoe; that describes underdamped motion.
Without damping The charactalistic equatiiﬁﬁ 3 [Mi], and k = 24 lblfl, the differential
o The charaCEeristic equation When we impose the initial Corlditions
.4. (Acos41J5 + Bsin4rﬁ} We get .+1921'= 414E ' ablation If!)=8 *1oos4:J§+(t/J§)sin4r‘/§]
{Wt/5):" [(Jirzxos4rJ5+(1r2)sin4iJ§]
“1215)! east4iJ§ #2116). ..‘tudeis 2N5 = 1.15 ft; the frequency is 4J5 z 6.93
' __:s :16. weget w ' JHIOD. But we are given that
'(80 cwleslmin)(21!)(l minIGO sec) = 32:13,
\elds l: 5 701311113. ;).",sec Equation (2!) in the text yields c = 372.31
 l..86l5 Finally 9 "p’ = 0.0] gives I: 247 sec. 7 35in all].
33in air] +e"’[—Aaisin mt + Braces am}. meequatiuns A = x0 and _..____ 8mk Ham—17:7—
3; If m) = Ce'F'costmu a) then x'(l) = pCe""cos(aJ.t— 0:) +00sz 1 yields tanwlf a) = p!w.. 33. If in = inl and x2 = 102) aretwo successive local so
= Cum—Phi 905mm ' all X: = Cexptptz) coswm— a) = agg HCRCB x1! X2 = erplpth '12)]. and therefore
ln(x1!x2) = p(tl '12) = 2191' 34. With n = 0.34 and r2 = 1.17 we firstusethe ‘_
Problem 32 to calculate a): = 27d(0.83) as 7.57 _ x; = 1.46, the result of Problem 33 yields p = (110.83)1n(6.73ll.4 ’ Then Equation (16) in this section gives = 2mg = 2(100 and ﬁnally Equation (21) yields mwwww ”41'0“ whenwe 'ﬁé x'(l)=0 on the general solution 311m m[(—110")r] )p,+(—l10")c, = 1
2* '5'. This gives the pmﬁcular solution wwyaxplm'”) =10"e"sinh(10'".')_ on rz+2,+(l+]o4“)=0 has roots r :—l:t10'"i. When We  "ans x(0)= o, X'(1)= 0 on the general solution g"[Aoas[ID‘"r)+Bsin(10‘"r)]
'_ Hg]: 0, —cE+10"'c1 = l with solution r:l : 0‘ c2 : [m This
" mlnlian 3:30) = lﬂ'e"sin(10"'t). I 4. (10"!) = le"lim—————smh(m_ 0 = re" and m 10'": 1 . mm” = ”shaman"n _ M,
5 m Hr": ‘ .mm = liﬂﬁiuhayﬂ = o (by L‘Hﬁpital's rule. for sgcnou 5.5 ”ONHOMOGENEOUQCl
THE METHOD OF UND " always successful — in which case the me complementary function is known. Howey". linden: well with a summing!!! large number of the nonhomugeﬁ r: if“ "'3‘!”
arise in elementary screntiﬁc applications. ' ' Ill 1 ‘31 I. In cach 0f l’mblems 1720 we give ﬁrst the form of the trial 56
the coefﬁcmnts we get when we substitute yum into the/dig?
terms. and ﬁnally the resulting particular solution y“. . L ymd = A9“; 25A : 1, yp = {”25”}:
2. J'ma, = A+Bx; —2A—B=4, ~23=3; y” = {ﬂag}! . 3, yum, = Acos3x+Bsin3x; lSA—3B=0, 3,]1'53'
ya = (cos 3::  5 sin 3x)f39 5, First we substitute sinzx = (l — ms 2x)l'2 on the right l'i’ah
equation. Then: I ‘ {J
t 3m] , A+Bc032x+Csin2Jq 14:1]2, —3B+‘2C="I ' '
yp = (13+ 3c052x— 25in2x)126 6 Jim, : A+Bx+Cx2; 7A+4B+4C =0, 73+
yp = (4  563: + 49x2)i343 7. First we substitute sinhx = (e’  9")”. 0111119;
equation. Then: I ytrial = Ae‘uBe“; I ﬁmcn'on  {war it  (”ﬁrsinhzr “J: I
amnion J" = e", +33”  Then: EB 2, _6,1=J; y, = {lrsin 3x 33:035..”6 'Ihc mmpiﬂﬂmm ﬁmction I I
itsuh: y, = (3: —2x}/s 0. 24B=l; y, = (—3x2e‘+x3e‘)/24 item. We cannot solve the characteristic equation
«My function, but we can see that it contains no
solution yakI = A leads immediately to the 17 First we note the dnptt ‘
Then: y“ = x[(A+Bx)cosx+(CqI‘J£ji .\
23+2c=o 4b=i,2A+2n.1,
In 3 (xlsinxxcosxlld 18 First we note the duplication with the comp
ye = cle” + (38' +c,e“' +c.e". Then: yimi = XML") + :03 + Car) e"; —6.4 = 1. 12'3"
,.p _ 424“! ~ 19m“ + rule‘wm ‘' 19_ First wc note the duplication with the pm ”i .+ ”a” . 1.
(which corresponds to the factor 1'2 ol'lhe characr : )‘uial : x2(/l + Bx + 062); 4At lZB =—l, 1'23
yp = (mg: — 4x] +x‘)f3 _' mu. _‘ 10. First we note that thc characteristic polynomial r5? .
to the duplicating part 9‘ of the complementary. . ytrill. : A+x(Be‘); “A=7, 33:1; ypi ln Problems 21430 we list ﬁrst the complementary function a function yi. and ﬁnally the actual trial function yII inwln
complementary function has been eliminated. ll 2]. y: e‘(c,cosx+clsinx);
yI : e‘(Acnsx+Bsinx] yp = x‘e‘ (Acosx + Bsinx) it = (A+Bx+Cx1)+(De’) .
yp = x3(A+Bx+sz)+x(De‘l‘) 22. V = (c,+c1x+cixz)+(c4e‘)+(c,e" ;, 23. y: = c.cosx+czsinx;
yi = (A+Bx)cos2x+(c+D;).".‘ y,, = x (A+Bx)cos2x+(C 33. (15 mh‘f 8, Sin 2x)
”c0052: 1 95in 2 I)
”on; + (“052’ + D 51" 2"] 34. 35. 36. complementary function ya, the trial solution y" for the
and tho corresponding general solutiOn yg : yc + yp where
outs an n, .so as to satisfy the given nonhomogeneous
11m equations obtained by imposing the given initial
' 81' solution y(x).
37. Ci :1, 3c2+215 = 0
y(I) = (lScos3x—25in3x+33in'2:c). .1": = clcosx+czsinx; y" = x.“
y“ : alcosx+czsinx+éxsinx cl = 1, 6‘2 = *1; y(x) = cosxsini
y: = e‘(clcosx+czsinx); y“ =
y“ = e'(c.cosx+clsinx)+l+xi2'
c1+1= 3, cl+c2+112 = 0 y[.\‘) = e'(4cosx—Ssinx)I2+l+xIZ_
Y: : cl+czx+cleat+cteh§ yu h 3"
yg = cl+czx+c,e‘1‘+c¢e“— ”15;
cl +cl+c4 31' 62263+2c‘=1_, ¢ 320:) = (234 +240x—9e‘2‘533e‘f x I
y: = C+Cze +0111 ’ y a? .1
ya = CI +1239" +caxle. Solution of this system gitres ﬁnal
particular solution of Problem'41? . 41 (a) cos3x+isin3x = cos’ x  3(cos x) (l — cos: x.) _ 0 ~c +c _C =0 1 and readily solve for cns’x = icosxp
’ ' I ‘ i similarly by equating imaginary pans? ' Se‘HOcosxZﬂyd
‘ J'; .= A+Bx+sz+Dx’+Ex‘+Fx5 leads to the equations A~B2C6.D+24E = 0
_BI2C6D24E+120F = 0 .
, 313123—601? = o 2. 44. We use the identity sinxsinsx = in
45—20:? _ solution yp = Acost+Bsin2x+_
I y" +y'+y = Jicoslx—écostl‘xl W
D = 2/141. The resulting general 45. We substitute sin‘x =  .'_II"I_IHn:u».h ‘ 4 Wgﬁndthﬂt A =—mo, 3:0, so when We (llsﬁcos 4x. substitutein yp I IIIEmail: 
pin 43 the differential equation can be MinBl} as $709331. Using the identities 1 {H1 0 we subsmute the trial SOlUtion the integrals and ﬁnd that  in: = cimx+czsmx' s ‘mx+(c+m)sinx]+[(E4FI)COS3X + (G + Hx)5in3x]_ i y!) = (4x 903112;: _ $11114: C=O.D=3116,E=0.F="“32'G=3"23sH=0Hence 1
byy=ye+yl+y2 where . +‘3?sinx)fl5 and y2 _— {3 sin 3x 4x cos 3x)/128_
u, = (cosSx—Scosx)l20, "I =( _ .mdepmdent solutions yI and y2 of the associated hOmogeneous #370?» h). the coefﬁcient functions ' Pa = —(U4)(c.os 1‘ “’5 x ‘ 5i“ 2.1: 5h”) + (1' = —(1I5)cos 3x (I) yltxmx) dx
Wm ‘ (1)15 and uz(x) =I WI?) 52, y. = cos 3x, y; = sinSx, "i = —(6x—sin6x)136,l u, = = "IJ’+"2J’2 aqu. (32) in the text, and finally yp itself.
_ J’P — —(.t cos 320/6 .3” = e", W = e"3r
‘ '  53. y. = cos 3):, y; = sin 3):,
u; = (2{3)tan 31:, u; = 2:3!" yI1 : (2!9)[3x sin 3:: + (cos 3:) lnlcog3 . 54. y] = cos x, yz '' sinx, H; = ucscx, 11;: cosxcsczk yp = —] — (coax) lnlcscx—cmgl _ 't 55. yl = cos 2):, 3n = 5m u; = (1!2)sin2xsin2x = ...
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 Spring '08
 Toland
 Equations

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