Question

Questions 10 points A spring is stretched 10 cm by a 20 km. The weight is released with a downward welcity of the dam constantequals 5. Tindanion for the portion of the spring any me 8.15olsin (6)=- V3138 815 3138 8/501 sin (0) - 31387 815 31387 1615. sin ()-- 31387 8V5 31387 M(O) 31387 8V5 31387

Answer 1

Data given L-10m W 20kg v=zmls r= (=5 use equation of motion mulct)+ su(t)+ Kuct = F(t) . f(t)=0 here. No external force .. muct) + 5 r(t) + kult): 0 = since this is a constant coefficient differential ean use usert vixent ulag? et ... mo?ent 5reott keato (mot 58 tk) et mo?+ 58+k=0 Put value of terms k= mg/L 2084 58+ 1962 =0 - (1) = 20x 9.8%. = 1962 kg/5² Solve eg - ci)

g.gi + X + iß in adoption g.g is written as √31387 8.55 Gede (Cicos B x + (₂ sin ßc) uct) = é 8 (c, cos (c, cos (131387 +) + į sin (131387 t 855 initial condition are - The ucolo and - ci) 2i(0) = 2 m/s - (ii) to ean apply condition ci los | 31387 855 (O) +(sin ( 13 (31387 o=e 855 (=0 + Now wt), festa C, cos gigt + ( sin g.gt] (-4 -(, sing.gt (9.9) + 9.96 (os g.gt/ e u'(o) - 9 m/sec. (+9.9(₂ - 2x 855 C 2 gog 31387 Now Put G & C in uct) we get,

sin -1/8t uct) = e 165 -J31387 CA t √31387 8 55
if negative sign indicate in option for downward position of spring.option C is correct. I hope this will help you.???