Question

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Problem 1. (1 point) A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring 178 feet. The ball is started in motion from the equilibrium position with a downward velocity of 6 feet per second. The air resistance (in pounds) of the moving ball numerically equals times its velocity (in feet per second) Suppose that after t seconds the ball is y feet below its rest 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

weight 128 feet + of the Steal ball = 128 pounds Strechul the spring The differential equation for the goren my by tky where m= uslugs System is =0 128 32 こ g b=4 , K = 128 128 17 g 17 med sine 06 -> 2 = 0 ; y(0)= 6 feels k = 17 &lf feet (t from (1) uy + 4y +7 ylo) = 0 Now solve the intral value proglem given conditions The characterstic equation 42 tus +17=0 S = -41 JO'4(4) (17) a) with the wy

_ut 16-272 8 8 the initial condition in de a SE -47 V-256 S - - 4+ 161 O acu Si= 2-stui 88 s = -0.5tai The solution to eq (2) becomes Cos (2) Age ostan (24) where a and ca are constants ubing 70)=0 => 0 = C,+0 =>G=0 y"lo) = 6 6 = G(-0.5e 05toos (2+) - 6e sinta tz (-0.5 cost an (24) + Geot boset) -0.5" yt) = Gé ou -ost

Snie C, = 0 6= C2 (2) q=3 from (2) [yos eost Sin (2t)

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