Question

SYSTEMS

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fex) = sin son crox) Problem 1: (25pts) Linearization Consider a mass held by a nonlinear spring shown below. The equation of motion is given by f'(x) 3 Sin (20x) COS(20x) 20 + sinº (20) = 9, where is the stretch in the spring in meters, k is the nonlinear spring constant, m is the mass of the block, and g is the gravitational acceleration. Note that the argument of sin is in radians. This equation is to be linearized so that the equation of motion will take the form of f'cxeg) = 824 ++ ( X+) f(x) ^ f'(xey) (x - Yey) 1 + ax=b. (V1) Let m= 2 kg, k = 100 N/m, and g = 9.81 m/s2 (V2) Let m= 1 kg, k = 200 N/m, and g = 9.81 m/s?. (V3) Let m = 2 kg, k = 200 N/m, and g = 9.81 m/s?. + 824 x = b wh=824 wn28.71 radls f = 4,57 HZ T= 0.2188 sec 1.1 (10pt) The equilibrium position is at = ---- meters. (A) 0.024 (V3) C 0.031 (V1) (E) 0.027 (B) 0.019 (V2) (D) 0.012 (F) 0.045 1.2 (10pt) The constant coefficient, a, in the lin- earized equation is 5-2 (A) 1130 (V3) (B) 1320 (C) 1500 (V2) (D) 542 (E) 824 (V1) (F) 1220 1.3 (5pts) The natural period of the linearized system is seconds. (A) 0.243 (B) 0.162 (V2) (C) 0.187 (V3) (D) 0.897 (E) 0.333 (F) 0.219 (V1)

Answer 1

I Am! The equation of motion is given by ä tk sin? (20x) ag linearization of sin3 (20x) (wSins (20x) using Taylor series expansion , fex) ean be written as at x=de $ (a) = 4()+(xx) dibacat cu for linearize form fen) æ fixe) + (-mede) de la df = 3 Sin? (20x) Cos (20x) - 20 #ca f) = Si? (2ox ) + (-xe) 35 (20) . Cos(2oxe) . 20 Let, Å= 6 6 = ä = q- in fin 3 Rox) for equilibrium point i-e.xate p=0 0 = 0 = sin? (203) = mama w fin (20x) = (majes

for vi - xe t sint (2x9:81) 93 I 100 - ļox0.62 . 0.031 m Am @ for v2 - xe the sin a (1x9.8.13 20.019 m Ang 79.819 to V3 ole 200 ~ 0.024 m Aus ④ 12 for Vi- fex) = sin3 (20 80.031) + (X-0.031) 3 fin (20*0.031) 6J (20 x0.031)-20 -0.196 + (x -0.031) 16:49 E-0-315 +16.49% linearized equ becomes x + 4 h fex)=g or së + 100 (-0,315+16.49 %) = 981 or Ž +824.5X = 25.56

a value for mi [a=824-5) An® for V2 -20%= 20*0.019-0.38 f(x) = sin? (0.38) + (x-0.019) 3 Sin? (0.38). Coj (0-38).20 = 0.051 +68-0.019) #1666 9 -0,094 + 70666,2 ät kifen) og or ä t 200 (-0.094+7.666x) =9.81 or ü + 1533728 = 28.61 value of a=1533.2 Am © for V3 i 20 Xem 220 X 0 0 24 : 0 - 0 f(b).föns (0-48) + (32-01024).3 Sina0-48). Log (0,48), 20 -0.098 +(2-0-024) 11:35 = -0.174 +11.354 si te me fex) =g or ä + 200 (-0.174 +11:352) =9,81 or é + 1135 x = 27.21 a value for V3 is 1135 Am ②

1-3 linearized Natural frequency of the system (wn) = Ja Time period (t) = 20 = 2 2 2 for vi Time period (t) = Jezus T 20.219 sec two for V2 Time period (T 157 3.2 q=0.16 Sc) Ang ® for v3 for V3 Time period (T) = 2 4 5 T 0.186 sec Any ©

Answer 2

I Am! The equation of motion is given by ä tk sin? (20x) ag linearization of sin3 (20x) (wSins (20x) using Taylor series expansion , fex) ean be written as at x=de $ (a) = 4()+(xx) dibacat cu for linearize form fen) æ fixe) + (-mede) de la df = 3 Sin? (20x) Cos (20x) - 20 #ca f) = Si? (2ox ) + (-xe) 35 (20) . Cos(2oxe) . 20 Let, Å= 6 6 = ä = q- in fin 3 Rox) for equilibrium point i-e.xate p=0 0 = 0 = sin? (203) = mama w fin (20x) = (majes

for vi - xe t sint (2x9:81) 93 I 100 - ļox0.62 . 0.031 m Am @ for v2 - xe the sin a (1x9.8.13 20.019 m Ang 79.819 to V3 ole 200 ~ 0.024 m Aus ④ 12 for Vi- fex) = sin3 (20 80.031) + (X-0.031) 3 fin (20*0.031) 6J (20 x0.031)-20 -0.196 + (x -0.031) 16:49 E-0-315 +16.49% linearized equ becomes x + 4 h fex)=g or së + 100 (-0,315+16.49 %) = 981 or Ž +824.5X = 25.56

a value for mi [a=824-5) An® for V2 -20%= 20*0.019-0.38 f(x) = sin? (0.38) + (x-0.019) 3 Sin? (0.38). Coj (0-38).20 = 0.051 +68-0.019) #1666 9 -0,094 + 70666,2 ät kifen) og or ä t 200 (-0.094+7.666x) =9.81 or ü + 1533728 = 28.61 value of a=1533.2 Am © for V3 i 20 Xem 220 X 0 0 24 : 0 - 0 f(b).föns (0-48) + (32-01024).3 Sina0-48). Log (0,48), 20 -0.098 +(2-0-024) 11:35 = -0.174 +11.354 si te me fex) =g or ä + 200 (-0.174 +11:352) =9,81 or é + 1135 x = 27.21 a value for V3 is 1135 Am ②

1-3 linearized Natural frequency of the system (wn) = Ja Time period (t) = 20 = 2 2 2 for vi Time period (t) = Jezus T 20.219 sec two for V2 Time period (T 157 3.2 q=0.16 Sc) Ang ® for v3 for V3 Time period (T) = 2 4 5 T 0.186 sec Any ©