Question

(1 point) A mass weighing 8 lb stretches a spring 3 in. Suppose the mass is displaced an additional 11 in in the positive (downward) direction and then released with an initial upward velocity of 2 ft/s. The mass is in a medium, that exerts a viscuouse resistance of 1 lb when the mass has a velocity of 4 ft/s. Assume g 32 ft/s is the gravitational acceleration (a) Find the mass m (in lb.s/ft) (b) Find the damping coefficient c (in lb - s/ft) (c) Find the spring contant k (in 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 Z,T ,,T ,,. (e) Enter the initial conditions r(0) - : ft l ft/s (f) Is this system under damped, over damped, or critically damped??

Answer 1

"(a) = mass = w/g = 8/32 lb/ft/s^2 = 0.25 lb s^2/ft (b) 'c' damping coefficient = Resistance/velocity = 1 lb/2 ft/s = 0.50 lb s/ft (c) spring constant k = weight/initial displacement = 8 lb/0.2 ft = 32 lb/ft (d) The equation is given by the formula if x denote the displacement and f(t) the external from mx"" + cx' + kx = f(t) Now in dues f(t) = 0 there is no external value of m c k f(t) well get 0.25 x"" + 0.50 x' + 32x = 0 x"" + 2x' + 128x = 0 (e) The initial conditions x(0) = 11/12 = 0.9166 ft (f) The spring mass system is in the equation mx"" + cx' + kx = 0 over temp is c^2 - 4mk = 0 under defined is c^2 - 4 mk < 0 critical denoted if c^2 - 4mk = 0 Now in two case"