PART IA Mechanical Vibrations Examples Paper 1 Solutions 1. 7 1 K 4 F X Applied force f = +(x-y) Ty + y = x Where T = N/K For the step input as given, y(t) = A ++ +/7 C.F. and to satisfy you at the geren! solution is + x. . P.I. t=0 : y = x0 ( 1-e-++) 4 - - - A = - Xo 1 1 T i 2. A 0 Sum of currents at A C R U : cG +q-i =0 T? + V = iR -0 where T = RC Using the result from QI., the response to a Step input current of magnitude is is v = i. R ( 1 - e- +/7)
Port IA Mech Vib 11. 3. K f Sum of forces : f= ky + dj y Ty +9 =Â¥ where T= >K From Q1 , response to step input force of magnitude for y= f (1-e-+++) Unit impulse is the derivative of a unit step , Unit step response : y= = (1-e-+/7) + - , Unit impulse response : y = = (t/(1-e -= /s) 9 ? IT Impulse I y = ==- e-+/7 IT Unit Step response t unit impulse response t If (f) is a unit step function, then intuitively the dashport will require an infinite force when the step occurs. Substitute in equations and recall that d(step ) = 5(f) to find (+) = k/T5(+) + 17 for +>0 4. TÃ’ + O = 0; 0 4 - 35 1 T=20s from initial gradient intercept with asymptote . 20 1 20 Temperature rises at 20 ?/s : 200 + 9 = 8+ + +20 (input is a ramp) Unit ramp response is the integral of The unit step response : Q= } { (1 - e-20) de e: e = 8: ( + + 20 e = = 0 ) + const 13? 0 and 0 = 20 at t =0 - : Q= 17 + 2+ +3 ==== 0 T=20 t = Q: - 3 (1 - e- +/20)
-3- Port IA Mech Vib 5. 1 v 0 Sum of forces on the train : mi = - AU : TV+ V = 0 where To > m Either V = A + +++ At t = 0, v= 0. : V= Vo e- +/4 .: x = {vdt = - VOT e+++ + c and take x=0 at t=00 (final rest position) x = - VOT e- +/7 C=0 distance moved = VOT = or use v = o du : Ty do + 4=0 : v = S + dx = = +c x= 0 when V=0 . C = 0 x = UT and for V=U., x = NOT as above 6. 1 1 K F Ty + y = x . From Q1 where T= /k Put x = Xeinst, y= > einst : (Tiw +1) Y = x : Y = T+iWT 8 * and 0