Question

A mass weighting 64 lbs stretches a spring 8 inches. The mass is in a medium that exerts a viscous resistance of 14 lbs when the mass has a velocity of 4 ft/sec. Suppose the object is displaced an additional 6 inches and released. Find an equation for the object's displacement, u(t), in feet after t seconds. u(t) = e *'(ä сos (8.71977t) +.0167 sin(8.719771)) x cm

Answer 1

classmate Date Page m - 64 lbs 12 k= mg AX AU - 8 in 0.667 ft -64X32.2 0.667 ebs 52 k= 3091.2 lbs/s² Fc = bv 14 = b(4) 3 b 출 = 3.5 .. By Newton's I law du du dt m + b + ku= dth džu 64 dt? + 3.5 du + 3091.2u=0 at dre 0.0 54 6875 dut 48.3 U=0 dt² Auxillary eqn is 22 +10.0546875 at 48.30 d = -0.02734 £ 6.94977

dassmate Date Page .. u(t) e - 10.027344) + [A-cos + B sin (6.549+7+) 0.5ft ult=0) > A=0.5 & ult=o) = 40.027344) u(t) = 10.027344) e Acos (6.94977 +) +B sin (6.949776) (0.027344 € te A(6.94977)-sting (6.9497747 +876.94917) cos(5.94977477 ulo) = -0.027344 [A] + [6.94977 B] to = B = 0.001967 -0.027344)t i ult) = e 0.001967 sij(6-94977) [(6-94977)+7 +0.5 cos [(5.9497774