Question

Help please and show all work to the answer. If written please have work legible. Will up vote for correct answer.

(1 point) A mass weighing 10 lb stretches a spring 6 in. Suppose the mass is displaced an additional 11 in in the positive (downward) direction and then released with an initial upward velocity of 3 ft/s. The mass is in a medium, that exerts a viscuouse resistance of 3 lb when the mass has a velocity of 2 ft/s. Assume g 32 ft/s2 is the gravitational acceleration. (a) Find the mass m (in lb s /ft) (b) Find the damping coefficient c (in Ib s/ft). (c) Find the spring contant k (n lb/ft) (d) Set up a differential equation that describes this system. Let x to denote the displacement, in meters, of the mass from its equilibrium position, and give your answer in terms of x,x'.x" (e) Enter the initial conditions: EE ft, x' (0) E ft/s ( Is this system under damped, over damped, or critically damped?

Answer 1

At B weight is balance by spring force w = kx, w = 10lb x = 6inch = 1/2 feet 10 = k(1/2) k = 10 times 2 k = 20 lb/ft rightarrow spring constant \r\n weight rightarrow w = m.g so, m = w/g m = 10/32 mass m = 0.3125 lb s^2/ft here viscous resistance = c v(t) 3 = c dx/dt 3 = c(2) rightarrow velocity is 2ft/s given c = 3/2 c = 1.5 damping coefficients \r\n the de can be written as mx''+ cx'+ kx = 0 (0.3125)x''+(1.5)x'+ 20x = 0 or we can multiply by 3.2 both sides x''+ 4.8x'+ 64x = 0 spring constant we have already find k = 20 at t = 0 mass was at position c x = 11 inch v= -3 ft/seconds negative sign indicates up so, x(0) = 11 and x'(0) = -3 c^2 = (1.5)^2 = 2.25 4mk = 4(0.3125) times 20 = 25 c62 < 4mk so, system is under damped