Question

#3. An object with mass 1 kg, as attached to a spring that satisfies Hooke's Law, with stiffness k=5 N/m. The oscillations of the object are damped, with damping coefficient 2 N sec./m. At time 1 0, the object is 2 m, to the right of the equilibrium point, and it's pushed to the left so its velocity at time =0 is -8 m/sec.(i.e. it's moving to the left at speed 8 m./sec.) () Write an initial value problem, with unknown function whose solution gives y , the displacement (in m.) from equilibrium at time 1. You may omit units here and from the rest of this problem (in) Solve the IVP from part (1) (iii) Make a rough sketch of the solution y. You don't need to write scaling on the axes. (iv) When does the mass have minimum (i.e., farthest to the left of equilibrium, aka "most negative") displacement? Give an an approximate (numerical, decimal) answer, accurate to at least 4 significant figures (digits starting with the first nonzero digit).

Answer 1

mass n = 1kg Spring Constant kasillo y co= 2 *'(o)=-8 damping b= 2 N-Seeln si mynd + sy' + ce con ( 314 + 2x! +5% =0 ; fecos 22 ; X! (0) = of ] i Asrune solutra of form . Yeert (en) "+2 (err) +5(entiso ert (04+27+5) = 0 Vers 71321 Y(+) = et (cocos (27) + C2 tin (243) y cu) = 2 =) 2=e 5 +) =) =2 Y'(3 = -et (960$(27) + C2 Siren-)) +et (-usin2+ + 2C 2. Cosz vy) y (o) = - =)-8= -e ( 2 +o) te (-0 +262) 5) C2 = -3 .:: y(t) = et ( 2 COS(27) –35ín (21)) ay(t) - t

YCH)=et (200527 - 3 Sinzt) Y'(t)=-et (2c0528 -3 sin21) tet (-usinar - 6 cesar) Y'(N=et ( - scos (21-) — Sir (2+1) To get critiw point y' (t) = et (-8 cos (2 t) - Sin (2+3)=0 sin (2+) = -5 COS (24) tan (24) = -8 2x = tant (-8) = 5-tan'(s) t = 0.84757566 1+ = 0.8476/sec