A spring is stretched 6 in by a mass weighing 8 lb The dampi

A spring is stretched 6 in by a mass weighing 8 lb. The damping force measures 1 lb at a velocity of 4 ft/see, and the mass is acted on by an external force of 4 cos 8t lbs. Calculate k, tau, and m. Remember that pounds are not the correct units for mass and that gamma > 0. Set up the differential equation for u(t), the position of the mass at time t. Do not solve for u(t). Solve for the steady state response U(t).

Solution

a) 1 lb = 4 ft/s * (gamma)
c = 0.25 lb s /ft

mg = kx
8 lb = k * 6 in
k = 1.33 lb/in = 15.96 lb/ft

mg = 8 lb
m = 0.25 lb s^2 / ft

gamma = c/2m = 0.25/2*0.25 lb s ft/lb s^2 ft = 0.5/s
b) Applying force balance:

0.25 lb s^2/ft * u\'(t) + 0.25 lb s/ft * u(t) + 1.33 lb/in [integration of u(t)dt] = Fcos8t lbs

c) Steady state solution is given by
u(t) = A/w (sin (wt - phi))
where A = Fo/m /sqroot[(Wo^2 - W^2)^2 + 4 (gamma)^2 (w)^2]
phi = arctan[c*w/(k - mw^2)]
w = 8, c = 0.25, k = 15.96, m = 0.25
phi = arctan(-50) = -88.85 degree
A = 4/0.25 / sqroot([15.96/0.25 - 64]^2 + 4*0.5^2*64) = 16 / sqroot(64.0256 ) = 2
u(t) = 0.25(sin(8t + 88.85))