Question

4. (15 pts) A mass weighing 8 pounds, attached to the end of a spring, stretches it 6 inches. Initially, the mass is released from a point 2 inches above the equilibrium position. Find the equation of motion.

Answer 1

The Newton's second low of motion for the system is -kx m d²x at² -11) Or dt2 Given : 8 eb :. g the spring m d’x + kx = 0 Mass weighs 8 Pounds ne W = 816 32 ft/s2 1 sluge This 8 pound weight stretches 6 inches 1/2 ft flooke's law, spring constant 16 Ablft 1/2.34 Thus, from ean I Id²x By W k= 16 = + T 64% =0 + w2x = 0 is + or d²z at solution to d²x ata W2-64 => W=8 x(t) = c, coswt + Ca sinut thus Xit) = C, cos 8t + Ca sin 87 -M Hence

the initial conditious are: 20 = -2 inch. = 0 ft/s & xe) from egin TIT, X(t) = c, cos 8t +Cg sin 8t x'(t) = -84 sin&t+ 8 (2 eszt 3 20) = -2 = G, COS 0° + Casino -2 = C. 1 + 20 G = -2 X'10) = 0 = -86, sin 0 + 8 Cą COSO 0 = 8 cz C2 = 0 Answer. - 2 cos cosot 22(t) = :.