(a) A three-degree-of-freedom system is shown in Figure Q1. Figure Q1 F2 F3 x X2 X3...

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(a) A three-degree-of-freedom system is shown in Figure Q1. Figure Q1 F2 F3 x X2 X3 3k 2k 2k www www2m-www m 4m where k-48 N/m m-3kg F =F2 = 5 cos 2t N F3 = ON Knowing that the natural frequencies are: W = 8.6636 rad/s W2 = 2.3172 rad/s Ws 5.9642 rad/s and the mode shapes are: 1.0000 1.0000 1.0000 (x)= -0.2304 and (X)2- 1.2215 0.0856 and {X}3= 0.5922 0.7339 -2.6528 it can be shown that the modal masses are: M = 3.6809 M = 24.1354 M3 = 49.4337 Using modal analysis, find the steady-state response of the system. The damping ratio for each mode is = 0.05, i = 1,2,3. The mathematical solution to the equation: q; + 25wii + w q = Ficos wt, i-1,2,3 can be written as: = q(1) A cos(wt-6). i = 1,2,3 where UFMFM8-30-3 Fi 1 A wf + 3 20 and = tan 1 Wi 2 3 1 [1 Page 2 of 4 (b) What are the advantages of Theoretical Modal Analysis when it is used to find the response, to an excitation, of a MDOF system? Giving a practical example, explain the main disadvantage of the method. (a) A three-degree-of-freedom system is shown in Figure Q1. Figure Q1 F2 F3 x X2 X3 3k 2k 2k www www2m-www m 4m where k-48 N/m m-3kg F =F2 = 5 cos 2t N F3 = ON Knowing that the natural frequencies are: W = 8.6636 rad/s W2 = 2.3172 rad/s Ws 5.9642 rad/s and the mode shapes are: 1.0000 1.0000 1.0000 (x)= -0.2304 and (X)2- 1.2215 0.0856 and {X}3= 0.5922 0.7339 -2.6528 it can be shown that the modal masses are: M = 3.6809 M = 24.1354 M3 = 49.4337 Using modal analysis, find the steady-state response of the system. The damping ratio for each mode is = 0.05, i = 1,2,3. The mathematical solution to the equation: q; + 25wii + w q = Ficos wt, i-1,2,3 can be written as: = q(1) A cos(wt-6). i = 1,2,3 where UFMFM8-30-3 Fi 1 A wf + 3 20 and = tan 1 Wi 2 3 1 [1 Page 2 of 4 (b) What are the advantages of Theoretical Modal Analysis when it is used to find the response, to an excitation, of a MDOF system? Giving a practical example, explain the main disadvantage of the method.