255538827 Theory Of Vibration With Appli

Solutions' Manual-Theory of Vibration with Applications by William T. Thomson and Marie Dillon Dahleh\nTABLE OF CONTENTS\nCHAPTER 1\t1\nCHAPTER 2\t9\nCHAPTER 3\t41\nCHAPTER 4\t76\nCHAPTER 5\t116\nCHAPTER 6\t149\nCHAPTER 7\t178\nCHAPTER 8\t201\nCHAPTER 9\t229\nCHAPTER 10\t249\nCHAPTER 11\t350\nCHAPTER 12\t366\nCHAPTER 13\t412\nCHAPTER 14\t438 1 - 1\nx = A sin ωt\nA = 0.20 cm\tγ = 0.15\nω = 2π/T = 41.89 rad/s\nẋ = \u003d Aω cos ωt\nẋmax = ωA = 8.38 cm/s\nẋmax = ω2A = 350.9 cm/s2\n\n1 - 2\n\nω = 2πf = 25 × 82 = 515.2 rad/s\nω2 = 0.265 x 106\nβ = 980.0 cm/s2\nxmax = ẋmax / ω2 = (50 x 980.7 / 0.265 x 106) = 0.184 cm\n\n1 - 3\n\nωd = 2πf = 2π × 10 = 62.83 rad/s\n\nγ = 1/ f = 0.10 s\nxmax = A = 0.57 m/s\nA = 0.07274 m\nxmax = 5.3 A = 287.1/ m/s2\n\n1 - 4\n\nx = A (sin ωt + ε sin ωqt)\n= 2A cos{[(ω1 - ω2)t]}\nA1 = A cos (ω1t)\nω1 = ω + Δω)\nω1 = 240 + Δω = 2ω\n\n1 - 5\n\nz = 4 + 3i\n\n2 + 3i\n\n4 - i\n\nSum = 6 + 2 i\n\n1 - 6\n\nz = A e^{iθ}\nA = √(x2 + z2) = 6.325\nθ = tan¯1(z2/z1)\nθ = 18°26' = 0.3217/π\nz = 6.325 e^{0.3217i}\n= 6.325 [18°26'] z = A (cos \\theta + i sin \\theta) = A e^{i\\theta}\n\\dot{z} = A (i cos \\theta - sin \\theta) = Z_1\n= A[ cos(\\theta + 90) + i sin(\\theta + 90)]\n\\text{From the above,}\nI = \\frac{\\dot{z}}{8.70}\nX(t) = \\frac{a}{\\pi}[ \\sin(\\omega t) + \\frac{1}{3} \\sin(3\\omega t) + \\frac{1}{5} \\sin(5\\omega t) + \\ldots ] x(t) = \\begin{cases} 1, & 0 \\leq \\omega t \\leq \\frac{\\pi}{2}\\ 0, & \\frac{\\pi}{2} < \\omega t < \\pi \\end{cases}\nb_m = 0 \\quad b_m = \\frac{2}{\\pi} \\int_0^{T/2} x(t) \\cos(m\\omega t) dt \\quad \\forall m = 1, 2, \\ldots\na = \\frac{F}{\\pi} \\int_0^{\\pi} x(t) \n= \\{0, & \\text{for} \; \\text{even} \; n \\quad \\frac{1}{\\pi} \\text{for} \; \\text{odd} \; m \\} x(t) = \\frac{1}{2} + \\frac{A}{\\pi} \\left( \\cos(\\omega t) + \\frac{1}{3} \\cos(3\\omega t) + \\ldots \\right) x = \\frac{\\omega}{2\\pi} \\quad 0 \\leq \\omega t \\leq 2\\pi \nC_n = \\frac{1}{2\\pi} \\int_0^{2\\pi} \\omega t e^{-jn\\omega t} d(\\omega t) = \\left[ - \\frac{1}{(2\\pi)j} \\left( -1 + (1 + i2\\pi) e^{-i2\\pi n} \\right) \\right] = \\frac{1}{2\\pi} \nX(t) = C_0 + \\frac{1}{j(2\\pi)} \\left[ e^{i\\omega t} + C_{1j} e^{i2\\pi t} + e^{i3\\pi t} + \\ldots \\right]\n= \\frac{1}{2} \\left[ \\sin(\\omega t) + \\frac{1}{2} \\sin(2\\omega t) + t + \\frac{1}{3} \\sin(3\\omega t) + \\ldots \\right]\n= \\frac{1}{2} \\left( \\sin(\\omega t) + \\frac{1}{2} \\sin(3\\omega t) + \\frac{1}{3} \\sin(5\\omega t) + \\ldots \\right) 1-14 cont.\nF.S. of saw tooth wave =\n x(t) = 1/2 [sin(ωt) + 1/3 sin(2ωt) + 1/5 sin(3ωt) + ...]\n x^2(t) = 1/4 [sin^2(ωt) + 1/3 sin^2(2ωt) + ...]\n + 1/2 [sin(ωt) + 1/3 sin(2ωt) + ...]^2\n = 1/2 [sin^2(ωt) + 1/3 sin^2(2ωt) + ...] + cross products which will integrate to zero\n 1/2 ∫(x(t))^2 dt = 1/4 [cos(ωt) + 1/4ωT cos(2ωt) + 1/4ωT cos(3ωt)]\n[0 to T]\n let ωT = 2πk where k is an integer → ∞\n = lim(k→∞) (2πk)/(2π) = 1/2 (1 + 1/3 + 1/5 + ... ) = 1/2\n 1-15\n x(t) = 1/2 (cos(ωt) + j cos(ωt) + j sin(φ) + ...)\n Fourier Spectrum = plot of coefficients. For this case b_0 = 0\n C_n = √(A_n^2 + B_n^2) = A_n\n c_0 = A_0/2\n c_n = A_n/2\n -0.5 -3 -2 -1 0 1 2 3 4 5 6\n -5.0 0.405 -0.135 0.08\n n 1-16\n k_0 = 2/3\n ω_0 = average value & k_0 = 2/3 = 5/6\n A_n = 1/(mπ) sin(mπ)\n B_n = 1/(mπ) (1 - cos(mπ)) = 1/mπ(1 - cos(2mπ))\n 2C_n = √(A_n^2 + B_n^2) = √[(m/π)^2] (1 - cos(mπ))\n φ_n = tan^{-1}(B_n/A_n)\n m C_n φ_n\n 1 0.275 60°\n 2 1.379 60°\n 3 0 0\n 4 0.661 60°\n 5 0.552 60°\n 0 0 0\n 1-17\n 2 + 4 cosφ + 8 cosφ\n S = n[ -cosφ + ... + (j sinφ)] = n\n S = n[ -cosφ + ... + (j sinφ)]\n Ratio of 2nd harmonic / 1st harmonic = 1/√2 1-18\n x_e = lim(T→∞) ∫x^2(t)dt = A^2/2 = k_e A^2 = 0.10 A^2\n rms = √(x^2) = 0.3162 A\n 1-19\n x^2 = (1 - t/π) = 0 ≤ t ≤ π\n x^2 = 1 - 2t/π + i^2/(π^2)\n 1/0 to T\n D_b = 20 log10 (x_e/x_rm) = 0.50\n log10(x_e/x_1) = 0.50 - 0.0250\n (x_e/x_1) = 10^0.0250 = 1.0593\n x_e = 1.0593 ± 1.0593 x 2.5 mm = 2.6481\n Error = 0.0593 x 2.5 mm = ± 0.148 mm\n 1-22\n D_b = 20 log10(x_e/x_1) = 32.\n log10(x_e/x_1000) = 32/20 = 1.60\n x_e/x_1000 = 10^1.60 = 39.8 1 - 23\n\nconst. \nA.R. 2.0 \nconst. upper \n\n\nconst. L\n\nconst. R\n\n\n1 - 24\nMean Value = 1*x P(0) + (-1)* P(-1) = 0\nMean square value = (1^2 * P(0) + (-1)^2 * P(-1) = 1\n\n1 - 25\nP(t) = 1/2 (f(t) - f(-t)) + 1/2 (f(t) + f(-t))\n= O(t) + F(t)\nQ(-t) = 1/2 (f(t) - f(t)) = -O(t) (odd function)\nE(-t) = 1/2 (f(t) + f(t)) = E(t) (even function)\n