Vibração - Rao - Solução - Cap 5 - 4 ed

Chapter 5 Two Degree of Freedom Systems 5.1 Equations of motion m1 x1 + (k1 + k2) x1 - k2 x2 = 0 m2 z2 + x1 x2 - k1 x1 = 0 With xi(t) = Xi cos(wt + ϕ), i = 1, 2. Eqn. (E1) give the frequency equation | -wsq m1 + k1 + k2 -k2 | = 0 | -k2 -wsq m2 + k2 | or wsq (k1 + k2 / m1 ) wsq m2 / m1 = 0 Roots of Eq. (E2) are wsq1, wsq2 = k1 + k2 k2 + √ k1 + k2 k2 k1 k2 k2 ----- ---- ---- ---- 2 m1 2 m2 m1 m2 m1 m2 If x(1) = | x1 | and x(2) = | x1 | | x2 | | 0 | Equations of motion (E1) becomes r1 = x1/x(1) = -m2 wsq + k1 + k2/k2 = k2 / m2 wsq + k2 r2 = x2/x(2) = -m1 wsq + k1 + k2/k2 = k2 / m1 wsq + k2 General solution of Eq. (1) is e1(t) = x1(1) cos(w1t + ϕ) + x1 (2) cos(w2t + ϕ) x2(t) = r1 x1 (1) cos(w1t + ϕ) + r1 x1 (2) cos(w2t + ϕ) Where X1, X2, φ 1 and φ 2 can be found using Eqns. (5.18) For m1 = m, m2 = 2m, k1 = k and k2 = 2k. Equation (E1) gives wsq1 = (2 - √5) k / m, wsq2 = (2 + √5) k / m When k = 1000 N/m and m = 20 kg wsq 1 = 3.6603 rad/sec and wsq2 = 13.6603 rad/sec r1 = -m2 wsq2 + k2 / k1 = - 1.36004, r2 = -m2 wsq1 + k1 = -0.36602 With x1o(0) = 1, x2(0) = 0, z(o) = 1 and z(0) = 0, Eqns (5.18) give x1(1) = -0.36602, x(1) = -1.36003, ϕ 1 = 0, ϕ 2 = 0 Response of the system is x3(t) = -0.36602 cos 3.6601t - 1.36603 cos 13.6603 t x2(t) = -0.5 cos 3.66003 t + 0.5 cos 13.6603 t 5.2 Taking moments about O and mass m1 m1L21θ 1 + = -w1(L1sinθ 1) + Q2 sinθ 2 (L1 cos θ 1) = Q cos θ 1 (L1 sin θ 1) W1L1θ 1 = Q2 (L1 − θ 1) = E3 assuming Q = W2 m2 L2 ǧ x1 + m2 L2 ǧ = -W1 (L2 sin θ 2 (L2) Using the relations θ 1 = x1 and θ 2 x2 = - x∓ 1/ L2 L2 When m1 = m2 = m, L1 = L2 = L and W1 = W2 = mg, E5(E1) and (E2) become m1l x1 + mg x1 + mg x2 = 0 mls x1 + mg x1 + mg x2 = 0 For harmonic motion x i(t) = Xi cos wti ; i = 12, Eg (E5) become wsq i m L - mg x1 + mg x1 - mg x2 = 0 - lwsq 1 x2 – mg x1 = mg x2 = 0 from which the frequency equation can be obtained as wsq mL2 - (4msq L2 wsq 2 + mg 2 g2) = 0 i.e. wsq1 = " (2 − √2 L )” _____ , wsq 2 = "2 √ 2 L )” __ Ratio of amplitudes is given by Eq. (E6) as X1 / X2 = mg (-wsq 1 m / 3mg ") In mode 1, wsq1 = 0.7654 √ T X1 X2 X1 = a = - 0.41142 No node, In mode 2, wsq2 = 1.8478  R2 = 2.4913 one node located at : 1/3 2.493 or y = 0.2930 5.3 Let R1, R2 and R3 be the restoring forces in Springs. Equations of motion of mass m in x and y directions are mx = E1 my = R1 cos α i my = R1 sin α i where Rcix (↕ x cos α i + y ∓ sin ) Eg. (E3) = 0 m equation of motion (notation is repeated, no part of the above text repeated in this image Equations of motion of system Eqs. (4) 3 5) 0 = k i (x sin  2 „ i.x2 +3k L2 i x3sinXI L3/L4 + y sin -p) x/sin  = sin =0 For αi = 45°, δ1i = 135°, δ2i = 270° and δ1, δ2, k1 = δ1 + βi = βi (i) reduce to m x + k3 = k (b) Such equations are uncoupled. For harmonic motion, x(t) = X cos [ m x/√   2i) cos x(t) = X cos mk____________ + 2k=0 m for motion in x direction cos (mk/2 _____________” + y* „ d)” formulate directions i.e. for  notation repeated) N = cant find one listed equation defining the geometry of a potential ideal energy position so undefined) equations of motion. Note, length; mode; " to left and dimension such notation means enabling or body movement(not direction fine) column from O specify simplified from column above x_1(t) = x_1^{(1)} \cos (45.2256 + \varphi_a) + x_1^{(2)} \cos (78.335 t + \varphi_a) x_2(t) = x_1^{(1)} \cos (45.2256 + \varphi_a) - x_2^{(1)} \cos (78.335 t + \varphi_a) 5.15 Eq. (5.10) gives \omega_1^2, \omega_2^2 = \frac{1}{4} \Bigg[3000(1) + 4000(1) \plusmn \left\{\frac{3}{2} \left[\frac{3000(1) + 4000(1)}{2}\right]^2 - 4\left(\frac{3000(000) - 1000}{1000}\right)^2\right\}^{1/2}\Bigg] = 1633.9746, 3366.0254 \omega_1 = 40.4225 \ \text{rad/s}, \quad \omega_2 = 58.0175 \ \text{rad/s} \bar{\kappa} = \frac{1000}{(1'3660)} = 1.3660 \eta = -\omega_1^2 + \bar{\kappa} + \bar{\kappa}_x = -2(1633.9746) + 4000 = 2 \clubsuit 3660 \bar{\eta}_x = \omega_2^2 + \bar{\kappa} + \bar{\kappa}_x = - (3666.0254) + 4000 = 20 \clubsuit \text{rad/s} \newline \text{When } \bar{\kappa} = 1, x_2(0) = x_2(0) = - \frac{366 (20)}{40 + 4225} = - 2 (0.1046 \mathit{x}^{(1)} = \frac{1}{7321} = \frac{3660 (20)}{1'7321 \times 58.0175} \text{ = -0.2719} \newline \rho = s^{-1} \tan^{-1}(\varphi_\infty) = \frac{\Pi}{2} \newline \text{Motion of the two masses is given by Eq. (5.15):} \newline x_1(t) = 0.1046 \sin 40 + 4225 t + 0.2719 \sin 58.0175 t \newline x_1(t) = (1.1660)(0.1046)\sin 40+0.4225 t - (-0.3660)(0.2719)\sin 58.0175 t \newline = 0.1429 \sin 40+0.4225 t - 0.0959 \text {sin 58.0175t} \newline\newline\n5.16 (a)\omega_1^2 = 971.2, \omega_2^2 = 13029.2, \Pi_2 = 11805, \Pi_2 = 0.8471 \newline x_2(0) = 01, \text{x}_2(t) = \bar{x}(0) \bar{x}_1(0) = 0 Eq. (5.18) gives \newline\nx_k^{(1)}= \frac{(-0.8471)(0.21)}{-0.1471-0.1805} = 0.08356 \newline \nx_1^{(2)} = \frac{1}{(-0.8471-1.1805)} \left[\frac{(-1:1805)(0.21)}{1}\right] = 0.11644 \newline \varphi_a = \frac{a^{-1}}{\Pi_2} \tan^{-1}(\varphi_\infty) = 0 \newline x_k(t) = 0.09356 \cos 30.285 t + 0.11644 \cos 114 + 39.01t \newline \text {x}_1(t) = (1:1805)(0.08356) \cos 30+0.285t + (-0.8471)(0.11644)\cos 114.3901t \newline \newline (b) \quad \quad \text {x}_1(0) = 0.2, \text{x}_2(0) = 0, \bar{x}_2(0) = 0, \Pi_2 = 5.0 Eq. (5.19) gives \newline \mathit{x}^{(1)} = -2.0276 \left[ \frac{(- 0.9471 (0 = 2))^2 + \frac{(0)^3}{917.2}\right]^{1/2} = - 0.1167 \newline\newline \text{297}