Question

Q5 The equation of the motion of the mechanical system shown in the following figure is governed by the following differential equation d2 x dx m7+9+= -f(t) - 3kx dt2 dt where m, C and k are mass, damping coefficient and spring constant, respectively. Consider the system with m = 10 kg, c = 80 Ns/m, k = 50 N/m, and the system is at rest at time t = 0 s. f(t) is the external force acting on the mechanical system. Find the system response x(t) for f(t) = -10[e-3t +e-5t + cos(3t)] using an appropriate method except the Laplace transform. 11 m sis 2k

Answer 1

Given - 10 kg c mm K K = 50 N/m X(t) 80 N./m flt) 34 ast f(t) = -lo[e te +cos3t] XT At t =o, ult) = 0 To find ast Response x Lt) for fit) = -10 [e3+ e + cos 3t] Solution force Body diagram of Moss m. cải by Newton's second Low Kn } msi misi = – in – kn-2kn. flt) aka k f(t) msitin = -f(t)- 3km 2 inde teda at2 de - f(t)-3k2

-3t - st man dez & con dt + 3 kn = –10 [ + e + cos at 80 dn 10 dn de 2 4 150 n = - 3+ -st + e 10 [ ezt + cos 3t dt Solving By Memod of diffrential Equations finding complimentry Equation 10 dae dt² + 80 da + 150 n = dt 11 (100² + 80 D + 150 ) (+3)(D+5) = o D = – 3, D = - 5 complimitary function -3t st ce & e C.F =

finding Particular Integral. 3t P. I -10 [ + 747 + cos 37] (10 d'+ 80 D + 150 ) -st -3+ e P. I = -lo lo (-10) cos 13t) + (100+800+150) 100²+ 80D + 150 (10 D + 80 D+150) -st lote P.I = -lo test To cos 37 (20 D+8°) (200+80) -90 +150 +8OD -3t -st lot e ܘܐ cos 3t P. I = – lote - loo 180 (800+60) (60+80 ) -3t -54 lo te Cos 3t P. I - – lote + 20 (8D+6} 20 5t -37 lote P. I - – lote + (8D-6) cost (64D-36) 20 20 -5t Ite - 3+ te 18 (- 3 sin 3 t) - 6653+] P.I 2 64X(-9) -36

-st -3+) tle – e PI = + 24 sin 3t + 6 cos 3t (-612) (-612) tlustest) - 2 sinst cos 3t 2 102 So alt) = complimetary Particolar + function Integral. -st – 3t - 37 -st ce & Co t tle 234) alt) = - e -cos 3t 2 sinut SI 2 102 since system is At Rest At t=o means si Lt) to - 3t ast -st x (t) = -30, C – Sal & (e. -st - 3+) -37) + +1-5€ + 3e e 2 2.3. cost + 3sin 3 t. si 102 lut tao, =) silt) - o -30,- 50- 6 51

30, +562 = -2 At t=o, si (t)=0 o - cite 102 catch = ㅗ 102 = ₂ c = 1 12 -5 68 So system Response for given fit) is - 3t -st -st 3t xlt) = f settle – e cosat 2 sin 3t 52 12 68 2 102 Answer