Sect9.1 - Q5l%od. Sacha“ Q's — Rubia,“ (dd-Fwy...

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Unformatted text preview: Q5l%od. Sacha“ Q's — Rubia,“ (dd-Fwy Sec'fim. (“M M '3.-v -2 n .. b1‘ twkl~ «a : (1&3\y (a) ‘.1 ,br - évw( v3f4 o“ ' V : rz‘ur-z a (W-‘L‘Kr V‘: 'l)-—\ afienvdwg ~ In 2 v:2: $,-’L§z:a -> ‘g ~ (‘3 A\ ! v=‘\: 43(43120 G7 §z)= (2). i S e “j (\Mrt CW; (H Sit-Lode pun“- (wae Squaw”; Law-c oppasfifl’sfiwfl 1 Vuskm (LN Scuuzte punk an: um‘hb'e) (cl \ = (S‘WKFH 1-3 3 i-Y -; val-V1 3 : (T—lwr-fl v "t %?ficum‘u; '3? E ~ -—> - rta'" t‘ l ‘0 Z ‘ (3 % ifjcfiv-tcfw5 (M Mafia (em eMWLuu: SAM Si" “X T WASku (Qfiew Value; PesflHve) Xk Cd v24. Wp‘fifio -7 E = (H (awamqmmfi (M Vw A. ufi 0AM“; Slam 9(3)») aamp “fo’fi‘ng‘H gm C 655 w W; ) X1. X‘ ‘ (M S‘wvd ‘sofvd- (OHWWMLX- Camping). Qfijmk‘fo'fi‘catfi {HA1 ( ’YcauQ PM Mtfiq'ffae) §t Nokt (>th W *V‘SS:W3 “9-1 €301..ch are fi—“fl—‘j 1+3. . n. n mhew... g 2 ( I ' wise ram-w, {WoW£i‘u\ f‘ -‘r (-2*'i\?, :o ‘ a -5- ~ - .- at ‘ *1 1 = Y -H A‘kxw‘hwh; % 1" (-ZAU'glzo .7 Em : (2T£)V Nfi: M is +M«V5Yo?wHMW£1 fad-w Lcmm 41M fans? (b) ceng (beams: 073ng 4+: dam-L FMrMfi) Skylab. C hams: {0(1344- % ya,» ) (C) v \c’L‘ Section 9.1, Problem 17 Problem. The equation of motion of a spring-mass system with damping (see Section 3.8) is d2u du m—Cu—2+czi?+ku=0 where m, c, and k are positive. Write this second order equation as a system of first order equations for :1: = u, y = 53%. Show that x = 0,11; = 0 is a critical point, and analyze the nature and stability of the critical point as a function of the parameters m, c, and k. A similar analysis can be applied to the electrical circuit equation (see Section 3.8) d2] d1 1 ’ L— R— —I = 0. dt2 + dt + 0 Solution. The first equation in the system is easy; it will be y = %’;—‘, or in the new variables x and y, y = 5%. This is common to all problems in which one changes a single high-order differential equation into a system of first-order differential equations. To find the second equation, write the original differential equation, entirely in terms of the new variables. Instead of 93%, write 3;. Instead of 3%, write gig. This makes the second equation: dy mE+cy+kx=Q Generally, a system of differential equations is written as ‘13”; = f (as, y), it = g(ac, y), as that makes it more convenient to write the system as a matrix multi- plication. So, moving all the first derivatives to the left side of the equation, and everything else to the right, gives one the pair of equations, and so the system of first-order differential equations: d_w_ dt_y @-_£,_£ dt_ m y This can then be written in the form of a matrix equation, using the vector M rw/ x 0 1 a? = and the matrix A = k c — so that :E" = .A56. 31 ‘5 ‘R The point a; = 0, y = 0 has to be a critical point; at these values, % = 0 and 43 = 0 dt ' To analyze this system, look at the eigenvalues for this matrix — which, despite there being symbols in the second row, are calculated the same way as if there were numerals. As a result, the eigenvalues are A___c_i\/02—4mk 2m 2m It’s assumed in this problem that c is always positive, so the only thing that changes behavior is the plus or minus sign of (:2 — 4mk: If c2 -— 4mk > 0, then both eigenvalues will be distinct, real, negative numbers. This means that the equilibrium at a; = 0, y = 0 will be an asymptotically stable, proper, node. If 02 —- 4mk = 0, there is only a single, real, negative eigenvalue. This will still be an asymptotically stable node, but it will be an improper one. I If 02 ——4mk < 0, there will be a pair of complex-valued eigenvalues. The system will be an asymptocially stable (because the real component, —§%, will be less than zero) spiral. ...
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Sect9.1 - Q5l%od. Sacha“ Q's — Rubia,“ (dd-Fwy...

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