Question

< Pre A mass weighing 18 lb stretches a spring 6 in. The mass is attached to a viscous damper with damping constant 4lb-s/ft. The mass is pushed upward, contracting the spring a distance of 4 in, and then set into motion with a downward velocity of 5 in/s. Determine the position u of the mass at any time t. Use 32 ft/s” as the acceleration due to gravity. Pay close attention to the units. u(t) = in

Answer 1

= 18 lb weighing 18 32 kg mass = L = 6 inches 0.5 feet 18 0.5 = 36 lblft damping 8 = 4 lb.slft mass pushed up Contracting spoing a distance of 4 in downward velocity sinis Taking downwand direction as positive mu' (t) t r ult) + ku(t)=0 18 u" t) + uultet 36 ultro 32 4 (0) = - 2 ft, u'(o) = = 5 ft/s { since spring length linched normal becomes 4inch 80 6-4= 2 inches zinches) Charactentic ean 18 32 2 + un +36 = 0 1872 + 1287 + 1152 = 0 7 = - 128 + J (128)2 - 4 (18) (1152) 36 - 32 ± i 8565 7 =

C2 + S S 4. 24 9 u t) - sin ( 3 +) 4 (1) o u(0) - - 2 C C2 - - u (o) - 5. - 3 2 3 3 | 8 0 1 2 Sਹੈ । ohra Sਓ | sin (8575 t) Cos/8565 | 62 ੫੦ in feet For inches multiply by 12 19 565 sin 1850st - 2 Cos/8565€ ultl Sr (34) 2 4 ਤੋਂ G 20, inches.