Question

Problem 1. 128 325 feet. (1 point) A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring The ball is started in motion from the equilibrium position with a downward velocity of 8 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) Suppose that after t seconds the ball is y feet below its rost position. Find y in terms of t. (Note that this means that the positive direction for y is down.) y = Take as the gravitational acceleration 32 feet per second per second.

Answer 1

mass 31 128 32 T يح stretches a a force 128 pounds so. spring 128 325 feet d 128 325 kr 128 = K * 128 325 C from, Hooke's law, F k= 325 lb/foot. The dam damping constant is given as "=0 Differential equation is given by , my" + cylt try=o » 49"+ 4 y't 325=0 with initial condition yo)=0, Glo)= 8 Now, solve this initial value problem, its auxiliary equation is 4 m² + um +325=0 m = -1/2 #9 -t -ht So, y (4) = A e cos (at) t Ben sin (at) Yrosao =) Yeo) = A=0 =) y(t) = Besinat) y = - Bots sin (44) + 99 el Cer(at) y'(o)=8 7 yllo) = qb=8 Solution is y (t) = sin (94) =) B 9 sol _t/2 e 9

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