A 32lb weight is attached to a spring suspended from the cei

A 32-lb weight is attached to a spring suspended from the ceiling, stretching the spring by 1.6 ft before coming to rest. At time t=0 an external force of f(t)=20cos(2t) is applied to the system. Assume that the mass is aacted upon by a damping force of 4 du/dt, where du/dt is the instantaneous velocity in feet per second. Find the displacement of the weight.

Solution

Given:
   W = 32 lb
   Dx = 1.6ft
but F = kDx;
or W = k Dx
or k = W/ Dx = 32 lb / 1.6 ft = 20 lb/ft -- (1)

now, Damping Force = 4 Du / Dt
and F(t) = 20cos(2t) = F\'sin(w\' t)

or F ( total ) = F(t) - (kx + 4 * Du / Dt)
but Du / Dt = instantaneous velocity in feet per second = Dx / Dt
so, F ( total) = F(t) - (kx + 4 *Dx / Dt)

But F (total) = m * a = m * D2x/Dx2

so m * D2x/Dt2 = 20cos(2t) - kx - 4 *Dx/Dt

compare this with D2z/Dt2 + 2 * c * Dz/Dt + w2Z = F(t)/m

Here, c = 2 / m -- (2)
w^2 = k / m -- (3)

the solution to the above problem is given by :
z(t) = [ F\' sin(w\' t + phi) ]/m*Zm * w\'
Here F\' = 20
w\' = 2
m = 32lb/g = 1
w = sqroot(k/m) = sqroot(20/1) = 4.47
phi = arctan([2* w * w\']/[w*w - w\'*w\']) - pi/2 = arctan(17.88/[4 - 19.98]) = -179 deg
c = 2 / m = 2
Zm = sqroot([2*w*c]^2 + [[w*w - w\'*w\']^2]/w\'*w\') = sqroot(64 + [[4-19.98]^2]/19.98) = 7.94

z(t) = [20 sin(4.47*t -179)]/1*7.94*4.47 = 0.56 sin (4.47t - 179)