Question

Find the transfer function X (5) F(s) and X2() for the mechanical system below F(8) X (1) Ki = 4 N/m K2 = 5 N/m 0000 0000 tvi = 3 N-s/m My = 1 kg fv2 = 3 N-s/m fin 2 N-s/m M2 = 2 kg fv3 =

Can you show step by step calculation. Thank you

Answer 1

exact) + K-4 fot) My = 1kg fra Mg = 2 kg = fr = 3 funza (Time domain representation) tx XgB) K=4 000 K=5 un FS) fr = 38 M, = 1 M2=2 Lug =38 {vz =23 [ V (Laplace domain Representation ) for marss My + 38 +38 +4+5) X,6%) – (38+5) XGCB) = (39+68+8) X,(3) — 33+5)*(3) = 2. (m.ge X; 8) = 38 +5 3) ex+68+9

for mass M Q = FBS) (My g^+ 38+98 +5 ) XCS) – (38 +5) XKB) = F(S) (269+58+5) XgQ) – (38+5)*,6) putting the value of x B) an equation @ (288+58+5) X3 (8) - (35+5) estostg [(382+53 +5)$+66+9)— 987-308-as] *x) = FB) Q tog (3) EFS) ga +66+9 284 +1758 +538² +758+45-958-308-25 (5+372 May ( )= F(3) 284 +1758 +445 +A58 +20 (3+3)8 1963) = F(3) 2. Xos) 3+68+9 3 F(S) 284+178+4484+458720

from 0 xo) - 3645 1969) 37468 +9 X, 68) 38+S = х 678B+9) F68) Sthat a84+ 1753 +995 +455 +20 X, 8) = (38+5) FB) 284+175² 495 + 455+20 3sts X163) FB) 284 +178 +44587458720 S2 +6879 se? Yg 3) FB) 28471989+ 4489+458 +20