Question

e.g.22 A compressor of 25 kg is mounted on isolator as shown in the figure. The isolator is idealized by a spring and a damper of damping ration { = 0.3. It is found that the compressor suffers from vibration due to unbalance of 0.001 sin50t. Also, it is found that all the maximum impressed force is transmitted to the floor. What will be the TR and the maximum Ft to the floor when the excitation force becomes 0.005 sinwt? 0.00i Sinsot compressor isolation + m 11? k & lic floor

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Answer 1

Oool Sin Sof m k 부 m = 25kg 7203 F= 0.001 sin(so) so w=50 sadra who natural pequency (Vina ) un => wna vin? wn= 52.414 rad is let Nog wa نا. so V1-0.32 82 50 = 0.954 Wn S2.414 TRE VH (220) V (182)2 + (22022 561+ (1 + X0-3 X0-954) (1-ag582)2 + (280-320958 TR=1.988 AN Now =TR Ft Fo Max transmitted forle Ft le Et = 0.005 X 1-988 Ft = 0.00994 ~ Ang