Question

6. A 32-pound object stretches a spring 8 feet to reach equilibrium. The object is then pulled to a point that is I foot below the equilibrium position and released from with an upward velocity of 6 feet per second Assume a damping force of 5. What is the position of the object when it attains its maximum displacement above the equilibrium position? Express your answer to 4 decimal places

Answer 1

A 32 round object stiches a spring of feet reach equilibrium. From Hooks law at equilibrium position. Mg = ks we are given m = 32, s = 8 k = mg/s k = 32 times 32.174/8 k = 128.7 the differential equation governing given problem is md^2 x/dt^2 + beta dx/dt + kx = 0 32 d^2 x/dt^2 + 5 dx/dt + 128.7x = 0 d^2 x/dt^2 + 0.156 dx/dt + 4.02x = 0 d^2 x/dt^2 + 0.156 dx/dt + 4.02x = 0 auxiliary equation is \r\n m^2 + 0.156 m + 4.02 = 0 Roots are m_1 = -0.078 + j2: m_2 = -0.078 - j2 solution is x(t) = e^- 0.078 t (c_1 cos 2t + c_2 sin 2t) c_1 & c_2 are determined from initial condition x(theta) = 1 c-1 = 1: x(t) = e^- 0.078 t (cos 2t + c_2