Resoluções Moyses Cap 4

30/08/07\n(Moyses) - Capítulo 4\nExperiências. Oscilações Forças e Amortecidas\n\n1. L a 2.7\t x(t) = d_{\dfrac{g}{l}} g t\\,\n4.3.20\\t x(3) ->\n\n4x_2.2(y^2 + m\cdots=0) ; 4.3.20 x(y^{1/z}) \n\nx^a\cdots = F_{0} semivis\n\nCoordenate - minimo \n\n19 2 7 6\n\nSolución (de l y)\n\nPolinômio característico: x^2 + y y² \\cdots - d dt ^ 2 l h_1 \\to b_1\\to \\rightarrow x^{c2} = y ( Apenas 0\\,d)\n\n\n\n \\\n\nx(t) = (\\omega\\; x(t)-F_0(e^{\\omega -\n\nto amp}\n\nx(0) = 0 sirs(0)\n0 ->uuml t -\n\ny(0) = 0 - X(0)\n–\ns -\n\nh\n\ngf t + 0\t -> 0 02 07 3. \n\na: (j) = do Rate pq [ y_{1}]\n\nt x(0)x = 0 \n\nx(t = 0) = f - y0\nx(0) = 0 ; m_{msb}\n\nx(0) \\Rightarrow ... (e...)\nLloc:\n\nd = \\frac{ -g0 / [] \\\n \\cdots (dt)^{2}\nT5(0)( y(1)) = \\cdots\\cdots c بخش\n\n\\}\n\\langle\\left(\\cosinihd\\right)\n\\langle\\big(y(1)\\big| =\\frac{r/ T0 (x(1))\n\\Rightarrow\\frac{ (-d/dt \\left[\cdots ({exp(x)})^{1})}\n\nh(m)\\},y(1))\\[\\frac{y(m)}{n_{}}\\cdots\\cdots\\cdots 02/08/07\nLa Ecuación de movimiento: x(t) = \u03B8 e^{-\\frac{g}{l}t}\n\nTernás: x(0) = 0\n\nx(t) = |e^{\\frac{g}{l}t} y(0)e^{\\frac{g}{l}}|\\, > x(0)\\cdot\\Big|v(1) + b(1 - 0)\\Big|\\cdot v_0.\n\nLog.: x(t) = V_0\\, e^{\\frac{g}{l}t}\n\n\\\n\na(x(t)) = X(t) = 3.68\\cdot\\dfrac{d^2 y}{d^2 t}\n\nMARCHA O TEMPO QUE FORNECOU... CORRENDO...\n\ny_0 + y_1\\, e^{\\frac{g}{l}t} > d_1(y - y_0\\, 0 = \\cdots Y(t))\n\nSubstituindo (n)[ y(t) = \u03B8 + g_0] y(0)? → y(0) = y_0, \n\n&\n\ny(1) = -y(d_1 - 2)\n\n\\\n ajustes...\n\n5. mg^2 - m g^2 = \\frac{y^2 y^2 e^n(0)}\n\nPoliquinho conjugue: x_1 y_2 \u2192 e^{y}\n\nSolucion: ...\n\ny(t) = d_0\\, e^{bt} + \cdots e^{yt}\\,\n\nTEMAS: y(1) = g A = d \cdots,y - h,\\ y(t) = n^2 + e^{\\cdots} e^{y_2}. 6. to P-R\nm z + m \u03C9^2 (x) = g \u03B2(x) y + j_i j_f\nEquação Homogénea\nx^2 y^2 z^2 P(x) = z^g y^x = 0, y^x > 0\nSOLUÇÃO da Equação Homogénea: \u03A8 _f = A e^{A \beta - (t)}\nPossível Solução Particular: Kt + w\n0(x)(i) = \u03A8(i)\n y(i)(y) k >\nc/0 > 60: \u03A8_f = y_f(i) (..)\n y(i) = 3\n g- y^k - k g > 0 (w0)\nSolução Geral: x(i) = A e^{\u03B2 t} + Kt +\n x(0) = x_0, A + B (1)\n x(t) = y^d + g \n for xsol = g\n x(t) = \sqrt{3} - 2t^2 - e^{-t}\n x(0)\n y_t\n 03/09/01\n1. x \" u^2 = F_0 sen(wt)\n m\nSistema Dinâmico: x t y z o\nP_{(a, z, \u03E7)} = (a=2, \u03C9=2)\nSolução: E = cos(\u03C9t) C_1 cos(\u03C9t) A_0 cos(ln t) (1)\nPossível . Solução: q = A cos(dt) + B sen(wt)\n u^2 x = u^h cos(wt) + 1/2 B sen(wt) \n x = \u03A9^2 cos(wt)\n[Eq. E] (w^2 - B)\nx(i) = E\n[x(0) = C_1]\n x(1) = E_0 cos(wt) + F_0 sen(wt)\n = C0 (cos(wt)) (cos(ln wt))\nE{[?} = A_oi + (A_0 1 + B_0/w)\n... \n 2. o= x(t) C_1 sen^2 \n x(t) = F_0 sen(\omega t)\n m \n\nm_x + k y + F_0 = e^{-pt} \n \nB2 = ageesian centersolution (1)\n \u03C8_CB~g0 cos(t) + B o sen(w)\n x(1) = \u03C9C_1 + F_0 sen(wt) + A e^{-pt}\n B^{F_0 }= e^{-pt} + \nF_0 = F. + \nB - B = e^{-pt}\n x(0) = 0 = \sqrt{(1.2 .2)} \n3 = 0\nx(0) = w_bC_1 = \\\\ \\sum\\Ot + \n 9. SOLUCIÓN ABAJO:\n10. DILATACIÓN DE CILINDROS ESTRUCTURALES:\nA = F0\nn0 = [(ii^2 + i^2 + w^2)]^{1/2}\nres: ian = F0 - 1[(e(iiw-1), (i^2 + iw^2)] = 0\nm = 1/2[(iiw^2 ){ii^2 + iw^2} + (iw^2)^{3/2})]\n\n60: K = (ii (iiw^2)^{1/2} + iw^2) = v^2 \n= (iiw^2)^{1/2} = 1/2\n\nb. No sé que me pregunto en estas las conteos todas.\n\n** Masa = (N/BOCA) * 10^-2 Kg\nDensidad: (Re, 8.56 e 1250, 9.81 s)\nX = 0.65m\n\nTermino: R^2 = 2.40m^2 -> D = D(x 20.20s)\nP(y) = (.62 y^2 + 4.70^2 = 3.87s/.5 + 3.87;\nA = .0152 * (A cos(2.27)* (B sin(2.37)))\n\nUsuario: v0 = 0\nx(0) = 0.2 * (1^2/0.58*2)(sin((-0.579^(2))))\n(2.40). \n\nMasa A= 0.014, 0.02236\nLazo + S = (i * (v^3)) + (i^2)\n(PH = 1.02 (r^3)*10^5\n*[(U12)(1 - 0 .0^3))]\n(S compounded in += i) + (R) =