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๕ ブ Ο ώ Owebworks2.csu hio edu etwork2/CSU MTH286 Fal 18 HwS Sections61 62Rod 2 effecti el ser-24543558 displayModesMathlax&sho OldAns erm ☆ 左久这 HW5-Sections6.1-6.2Rod: Problem 2 Prev Up Next (1 pt) A spring with a 8-kg mass and a damping constant 9 can be held stretched 0.5 meters beyond its natural length by a force of 2.5 newtons. Suppose the spring is stretched 1 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c24mk? t seconds. Your answer should be a function of the variable t with the general form cleat cos(β) + c2c" sin(6) α- βー Find the position of the mass after Note: You can earn partial credit on this problem. Preview Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining Email instructor O Type here to search 10/15/2018

Answer 1

From Hooke's law F = kx implies k = F/x = 25/0.5 = 5 N/m from c^2 - 4mk = 9^2 - 4 times 8 times 5 = -79 the differential equation md^2x /dt^2 + c dx/dt + kx = 0 8 d^2x/dt^2 + 9dx/dt + 5x = 0 the auxiliary equation is 8m^2 + 9m + 5 = 0 m = -0.5625 plusminus 0.5555 the general solution is x(t) = c_1e^-0.5625t cos (0.5555 t) + c_2 e^-0.5625t sin/0.5555t) at t = 0 implies x/0) = lm implies 1 = c_1 + 0 implies c_1 = 1 x^'(t) = 0.5628 c_1 e605625t cos (0.5555t) - 0.5555 c_1e^0.5625t sin (0.5555t) - 0.5625 c_2 e^0 .54625t sin theta 0.5555t) + 0.5555 c_2 e^-0.5625 cos (0.5555t) \r\n at t = 0 implies x^1(0) = 0 -0.5625 c_1 + 0 .5555 c_2 = 0 c_2 = 1.01258 x(t) = e^-0.5625t cos (0.5555t) + 1.012358 e^05625t sin (0.5555t) therefore = 0.5625 beta = 0.5555 r = 0.56256 = 0.5555 c_1 = 1 c_2 = 1.01258