Question

1. Consider the two degree of freedom system shown. (a) Find the natural frequencies for the system (b) Determine the modal fraction for each mode. (c) Draw the mode shapes for each mode and identify any nodes for each mode. (d) Demonstrate mode shape orthogonality. (e) If F- and the motion is initiated by giving the mass whose displacement is a velocity of 0.2 m/s when in equilibrium, determine 0) and ,0 (f) Determine the steady-state solution for both *) and *,t0) if Fa)-2sin S

Answer 1

"(a) Write the equation of motion at M = 2 kg dxý ()k(x 0)+ k3(x1-x2)=f(t) M1 dt -100(저-0)+100(저 - 2) = f(t) + dt da () -200x-100x2 f() 2 + dt ..1) f(t) 2+200x-100x = Write the equation of motion at M2 =1 kg d (t) M2 dt 0 (x-x)y+(o-x)Ey dx5 (i) +200( x-0)+100(x2-x)=0 dt (1) d5 100x+300x2 = 0 dt 2-100x300x =0 (2)

From equation (1) and (2) 2 0 0 12 f(t) -100 1 200 -100 300 0 X2 2 0 M = 0 1 -100 200 K = -100 300 Calculate the natural frequencies 200 -100 2 0 к-о?м 0 1 -100 300 202 200 -100 0 -100 300 0 200-202 -100 300-2 -100 0 |200-22 -100 =0 2 300- -100 -)(100) (100)-0 200-22(300 322.47 and 77.526 =17.95 rad/s and 8.8 rad/s

"(b) The modal fraction for the mode is A2 = A 1 (K-du0 200-22 -100 0 300 22 -100 (200-20)- 100 0 1001 200- 2o 200 22 100 Calculate the value of A for 17.95 rad/s 200 2(174.95) 200-644.94 100 100 A1-4.45 The modal fraction for 17.95 rad/s is, 1 -4.45 -17.95 rad/s Calculate the value of A for o = 8.8 rad/s 200 2(8.8) 200-155.05 100 100 20.45

The modal fraction for o 8.8 rad/s is, 1 0.45 =8.8 rad/s (c) Draw the mode shape for each mode A 0.45 m2 A2-4.45 (d) Check the mode shape Orthogonality, XX = 1

1 = -4.45 A2 -17.95 rad/s 1 X2 A2= | 0.45 a-8.8 rad/s T Calculate X X2 1 хfx, - -4.45 0.45 1 =[1 -4.45 0.45 -(1x1)+(-4.45x0.45) -1-2 хfx, --1 Check the other condition for Orthog onality, X MX2 = 0 [2 0T 1 1 х?мх, %3 -4.45 0 1 0.45 2 [1 -4.45 0.45 =(1x2)+(-4.45 x 0.45) 2-2 Both conditions are satisfied and hence verified mode shape Orthogonality