analytic mechanics 7E ch3

analytic mechanics 7E ch3 - 3.] 3.3 3.4 CHAPTER 3...

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Unformatted text preview: 3.] 3.3 3.4 CHAPTER 3 OSCILLATIONS x = 0.002sin[271(5125" )1] [m] m In =(0.002)(27r)(512)l:—] = 6.43 P] S S "‘1 = (0.002)(27z)2 (5 1 2)2 = 2.07 x 104 S S . m x = 0.1 (00 cos Qt[—:l S x = 0. 1 sin (out [m] . ,. In When 1= 0, x = 0 and x = 0.3 [—] = 0.1a), S 2'! (ou=53" T=4=L26S a): r 32; . ‘ x(t) = x3 cos an +—‘sm (not and we 2 27:] a). x = 0.25 cos(207r1)+ 0.00 l 59sin (20m) [m] cos(a—,B) = cosacosfl+sinasinfl x = A cos(a)ot —¢) = A cos¢ cos (081+ Asin ¢sin (out x=Acosco01+BsinwnLA=Acos¢.B=Asin¢ - 2 2 mx2 + — kxz lmil2 +ikx‘2 =1 2 2 2 [C(xl? —x§)= m(ic§ —.%,2) l 7 . ') Eli/ll“=;mxIZ+;k\l ‘.7 '1 1 2 111,2 '1 fix; —x;xI 2 A ——x +x ’ +x | l .2 .2 l k xZ—xl 3.7 For springs tied in parallel: F = —k,x— kzx = -(kI + k2 )x S l k +k, 5 (l) : m For springs tied in series: The upward force m is kw x. Therefore, the downward force on spring k2 is keg x. The upward force on the spring k2 is klx' where x’ is the displacement of P, the point at which the springs are tied. Since the spring k2 is in equilibrium, klx' = km x . Meanwhile, The upward force at P is klx’ . The downward force at P is k2 (x—x'). Therefore, klx' = k2 (x — x') , kzx x = kl + k2 And kw xzk, k2" k1 + k2 l a): k‘l = ————k‘k2 2 m (kl +k2)m 3.8 For the system (M + m), —kX =(M+ 171))? The position and acceleration of m are the same as for (M + m) : k 'm:— xm A4+m k xm=Aeos 1+5 =deos t M+m M+m The total torce on m. Fm = mxm : mg — [yr mk mkd k Fr : mg + x," = mg + cos M + m M + m M + m f For the block to just begin to leave the bottom of the box at the top ofthe vertical oscillations. F, = 0 at xm = —d: 0__ m _ mkd g M + m d _ g(M +m) " k 3.9 x: e'7'Acos(a)‘,t—¢) dx . _ :17 = —-e "Awd sm ((udl - ¢)—ye "A cos(a)l,t —¢) . dx‘ . maxnna at y = 0 = cod s1n(mdt — {25) + y cos ((udt — 45) t _ tan(a)dt —¢) = ——y— (0d . . . . . . . 272' thus the condition of relative maxunum occurs every time that I Increases by —: (0‘, 272’ ti+l : ti +— (0d For the ith maximum: xi = e”"'/1cos((odti —¢) . _ ‘Y’MA _ vyi’i'f . AM — e cos (radiiH —¢) — e .\i xi :6 ("J :6er xi+| c , k _ 3.10 (a) y=—=3s' mf=—=25s2 2m m wd2=mf— 2:165"2 c'),“!=(z)d3—}/2:75'2 a), = fisil F 48 (b) Am” = " = m = 0.2 m C(z)‘, 60.4 27a) 2 a) a) fi (0) tan¢=—,—'—2—= 7:: '=— .~.¢=41.4“ (a); #0,) 2y‘ 7/ 3 l7 7 3.11 (21) mi: + 3,6mi + —,b’“nu‘ = 0 2 3 7 17 7 =— and ()'=— ‘ 7 zfl ta 2/3 (0,3 =042 —2;/2 = 4,82 (or = 2,6 F 1 7 , 25 1 5 b A = ” (o'=(o;— '=— ‘ .'.(u =— ( ) max 2influd d y fl [I 2 fl _ 2A 15/32 4, 1 3.12 e’”=- 2 l =——ln2= 1112 7 TJ fd 1 (a) (9d = (a); —y')2 I So. (an = ((03 + 73 )2 /; =[1f+[fifla :_d[1+[1§_:)2]% j; = 100.6Hz ul— (b) (or =(wj —}/3) f, = 99.4H2 3.13 Since the amplitude diminishes by e"’T” in each complete period. e 77;," = T," 27m 1 So wn=(wj+y3)%=wl,[l+ 1 )2 3.15 Tu For large n, — z 1+ T, l 872'2n2 fl A(a))— "7 20.277 E 0(a)) 111603 z Amax}, (J1 2 a)— 3.16 (b) Q=w_d:_°y__ Zr 27 (03 —l --—R LC‘ 7 2L 3.17 F =Rsinwt=lm[Fe"‘"] EX! —: and x I is the imagina )an of the solution to: e 13' 1 mi’ + at + ICC = Rem" i.e. x(l)= Im|:Ae'(""'¢)]= Asin(wi—¢) where. as derived in the text. E A=————I : 2 7 ‘ E (k—mw ) +c‘a)’ and 7 tang»: 9‘” , (pg—(0‘ 3.18 Using the hint‘ Fm = Re(l{e/3’ ) , where ,8 = —a +10. and x(t) is the real part ofthe solution to: mi: + 61" + IO: = Fgefl' . Assuming a solution ofthc form: x = Acm'i" F , (mfl2 +Cfl+k)x 2 —° re” A 2 . 2 . I“: . . ma — 21mm» — ma) — ca +1ca) + k = 7(cos¢ + 1 sm ¢) F m(0(2 -a)2)—ca+k =7°cos¢ (u(—2ma+c) =§sin¢ ¢ ta _l (0(c—2ma) = n 7 m(a“ —a)2)—ca+k Using sin1 ¢+ cos2 ¢ =1 , 122 = [m(ax2 —a)2 )—ca + k] +602 (c— 2ma)' A2 A = F“ l 7 2 {[m(a2 — (02 ) - ca + k]2 + (02 (c — 22mg)“ and x(l) = A?“ cos (a)! — ¢) + the transient term. I 2 ‘3 3.19 (a) T:27r\/Z[l—A—] g 8 for A=£, T~27r\/le.04l 4 g (b) g = 4’2] x1084 T . I . 4/721 . Usmg T5 = 27: — glves g = T, . approxnnately 8% too small. g ,3 2 (c) B=— M, and 1:“ 32a); 6 E _A_2 A 192 for A=£, 5 =0.0032 4 A 3.20 f(t) = chei’m” n = 0,il,i2, . . . f(t) = Zen COSIzwt+Zc,,isinn(ot, n = O,i1,i2, ... I, " T and c” =% Ef‘(t)e"”“”'dt , n = O,i1,i2, . . . 1 T T c" =? Ef(z)cos(na)t)dt—? Ef(t)sin(nwz)dl The first term on c” is the same for n and —n ; the second telm changes Sign for 11 vs. —n. The same holds true for the trigonometric terms in f (I) . Therefore, when terms that cancel in the summations are discarded: f(t)=ca AZ:[(1)cos(nwt)dt]cosn(ot 7. % f(r)sin (ncat)dt)sin mot . ~ 1 Q . ? “T1 (t)d[ Now. due to the equality of terms in in : T f(l) = c; + f(! ) cos (nan) (IIJCOS 11a)! n=il.i2. and c. = ml t _/‘(t)sin (n(ot)dt]sin nwl. n=l.2.3. Equations 3.9.9 and 3.9.10 follow directly. T 3.21 f(t) = chei'm” . c” :1? Ef(t)e"”“”dt . and n = 0.i'l,i2. n 2 2/7 a) T = — so 0, w I = ; Ef(t)e’i""”dt “ w 7r (0 7’: eru =— J:(—e"“")dt+ ’“c”"di 272' a, ) 0 If : a) 1 e'imul _ 1 )"into! ‘0 27f into _n ina) 0 {U 1 +in‘ -in,‘ = [1—e "—6 ‘+1] 2m" For n even, 6"" = c""” = l and the term in brackets is zero. For n odd, 6'“ = 6"” = —l 4 c": _ . n=il.i3. 27nn 4 inru! f(I)=Z _ n=i1i3 ,, 27m: 4 1 l , _, , : ___( uwl_ lllul)‘ "=13D ,, if n 21 4 l . = ——sm(na)l). 1121.3,5. ,, 7t 11 . 4 . 1 . l . f (t) =—[sm cut+—sm 3a)! +—sm5(ut +... if 3 5 3.22 In steady state, x(t) = Z A,,ei("w'_¢") Q A” ___ ____m—I_ [((ua2 — 2120f )2 +4ylnzaf]2 Now F" n = 1.35. .. . and wt 7 9 2 Q=100z“’—° so y“ z ‘” 2y 40‘ 000 4F 1 AI = _; . I In” 2 .60: 3 (9a)2 —a)2) +4 40000 F A, z W Zmzz'a)” 4H 1 A3 = mi: 9 3 2 3 (9a)2 —9(oZ +4 'w 200 40012 A3 z ——2 27m/ra) 4F 1 A5 = _"_. Smrr 3 3 3 (9502 —25a)2 +4[—‘”] (5a))- 200 R ’15 z '—'a ZOm/Tw' i.c.. A, : A3 : A5 = 1 229.6 : 0.1 (a) + (03x = 0 y = 3% Thus y = —a)fx divide these two equations: = = —£ x cit y (b) Solving yd?) +.\'dx =0 and Integrating 60.? Lct 2C = A2 2 2 y + x = I —> an ellipse (03A2 A2 (of = 3a) .i‘:y 3.24 The equation of motion is F(.r)=x—x3 = mil. For simplicity. let m=l. Then (a) (b) = .r—x". This is equivalent to the two first order equations - - ,3 x = y and y = x — .t The equilibrium points are defined by x—x3 =x(l—.r)(1+x)= 0 Thus, the points are: (-l.0), (0,0) and (+1 ,0). We can tell whether or not the points represent stable or unstable points of equilibrium by examining the phase space plots in the neighborhood ofthe equilibrium points. We‘ll do this in part (c). . . . d ' 3 . —. 3 The energy can be found by integrating = = t r or C 7C X V [)7 = .r — x} ) cit + C or ),2 r2 ‘_4 '_.—_'__'_+C 2 a 4 ’7 4 ,- In other words IS = T + V = = C. The total energy C is 1 constant. The phase space trajectories are given by solutions to the above equation I .4 5 y = i[x2 — "\— + 2C] . 2 The upper right quadrant of the trajectories is shown in the figure below. The trajectories are symmetrically disposed about the x and y axes. They form closed paths for energies C<0 about the two points (-l.0) and (+1.0). Thus. these are points of stable equilibrium for small excursions away from these points. The trajectory passes thru the point (0.0) for C=0 and is a saddle point. Trajectories never pass thru the point (0,0) for positive energies C>O. Thus, (0,0) is a point of unstable equilibrium. 0.5 10 3.25 §+sint9=0 :> i{£—c050}=0 dt 2 92 ” Integrating: —2— = cos 6|: 0r (2’2 = 2(c030 — cos 99) O 0 T = 4 ;__fi_, 0 [2(c050 — cos 0; )13 Time for pendulum to swing from 9 = 0 t0 6 = 00 is g . (9 5m: IT Now—substitute sin ((5 = 5 so ¢ = 7 at 0 = (9° sin —“ “ H 0 and use the identity c050 = 1— 23in2 ,— T = 4! 2 “W and after some algebra ———d?—— = ‘16 I —_—__ [l—sinzé]2 [4(sinzi—sin2 2 2 (a) T=4]——d¢—l where a=sin2% 0 [I —0tsin2 ¢]3 (b) (l-asin2¢)—; ~ l+lazsinz¢+2ar2 sin4 (13+ 2 8 T-43J.d¢|:l+iasin2¢+§a2Sin‘¢+ .1 T-7/r 1+ 0 2 8 . .. . 29: a, 93 2 03 (c) a=sm —: ———+... ~— 2 2 48 4 T=27r 1+§L+ 16 d6 .26, 2 vul— - 2 8111 ll ...
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This note was uploaded on 11/21/2009 for the course CEE 432 taught by Professor Ahmed during the Spring '09 term at Rowan.

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analytic mechanics 7E ch3 - 3.] 3.3 3.4 CHAPTER 3...

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