Question

Question B A machine on a viscoelastic foundation (Figure 31.1), modelled as a spring mass-damper system is acted upon by a force modelled as a harmonic force: F(t) = 0.2 sin(wt) Force is given in N and time in seconds. W Figure 31.1 Nos Given numerical values: m = 10 kg C=5 M k = 1000 = 1) draw the correct Free-Body-Diagram and determine the equation of motion [2 marks) 2) determine the natural frequency and the damping ratio of the system [3 marks) 3) determine the solution of the free vibration of the system (no force acting on the system) (5 marks) 4) assume a complex force excitation: F(t)= Felt and a complex displacement response: X(t) = Xelut Determine the amplitude of the steady-state (permanent) solution X as a function of w (10 marks] 5) Determine the amplitude of the system at resonance [3 marks) 6) determine the amplitude of the system at resonance under the action of the force given in (1) if the damping coefficient changes to c = 10 [2 marks]

Answer 1

Answer Direction of motion ķx cx FREE BODY DIAGRAM D'Alembert's principle ΣFς f(t)- x- x-(x = mx më tcxt kx = f(t) Iñ+53 +1000x50:2 sin(wt) (Equation of motion) W = 1000 -lo rad /sec LX -- _5_ =0.025 - C - amwn 2x10xio Free fibration solution r(t) = A cos wnt+B sin Wnt = A cos lot +B sinlot F(t) = 0.2evet x(t) = X evet (At steady state)

i(t) = jw Xe livet ä (t) =-w²X jut 108 +5X + 1000x 02 sin(wt) -106? Xe vest sj w Xebet 1000 Xe jwt - 02 eiwe X(-100*+5jW 41000) = 0.2 x= 02 -1063+5jW+1000 Xr_0202 X (10W+1000)-5jW (-106?+1000) +5jw Glow71000)+5jW F100241000)- 5j6 x- (-262 +200)-;W (-10 to +1000) 1250* Since the forcing fuction is the imaginary term so the respons also imaginary termi ^ (-O? (0oo) 250 6 = Wn= 10 rad /sec X = 4mm c=10 X(-1062 +10jW +1000) = 0-2

0.2 ·02 1 -ow Hooo)-10jw (-106 +1000) +0j6 610@?+1000)+10j6 (100241000)-10;W (-26 +200)-2jw X = (-1062 +1000) +1006? Since the forcing fuetion is the imaginary term · So the response also imaginary term. - X = 20 (-10W2+1000)+10062 W = Wn = 10 void /sec X = 2x10 Gloxio x1000) 7100X10" X = 2mm