Question

When a 6 kg mass is attached to a spring whose constant is 54 N/m, it comes to rest in the equilibrium position. Starting at i = 0, a force equal to f(0) = 30e-7t cos 6t is applied to the system. In the absence of damping, (a) find the position of the mass when t= 1. (b) what is the amplitude of vibrations after a very long time?

Answer 1

Here m= 6k8, k= 54 Nm, fH)= 30 774c0967 Thus equation of motion will be given by mät Kax= fH) 2100=(0) 6 ° +542-30 ēt cos 6t ä+9x=30 e 7+ C0867 Auxilary equation correponding to homogeneous partis mig=0 OH m3=0m=t3i Thue xe = 4 colst +CSmst and 30 et coeft xp = D279 (1-7)² g = 30 277 cos67 -t casot 30e D²49-140 tg = 30 et casst = 30 et coeft D2 140+58 -36-14D+50 - 150lt - it cos 6t sit (11470) casst - 30e 22-14D. 11² 4902 t5ēt licee67 +78 (+6) Sinet 121-49 X4-63) -7t I1Cosst- 42 Sinot = 15 3 1885 377 7t e 37 (ascaret – 1265m6t) ap

Thul = actap -77 9= G COR3t+cu Sin3t+e (33cos 67 - 126 Smet enst) 317 a(o)= at 33 377 = -33 377 57+ 377 377 +7) sė = -34 Sm 3+ +3C2 col3+ + (33.00164 -126 m (83X1-C) San 67 – 120860e8CH) si(o)=0 3C2+ -ZX33 + 1 x(-126x6) = 0 377 377 11 8377+126442 77+252 B C2 = 30 3877 329 C2 3977 Thus _ 33 cel3t+ 329 Sint elt + "33coest 377 377 377 - 126 Sinot - 72 33 e x33 + 377 377 377 go After long time ent to thee de 329 -cell att - Smal 30 377 1-331 377 377 13343292 330.650d 317 372 2015)= 33 (1+277) -33 The completede_V2 je 132912 = 0.877058 meter I 87.705 cm.