A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring 12865 feet. The ball is started in motion from the equilibrium position with a downward velocity of 8 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second). Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that this means that the positive direction for y is down.)

Answer and Step-by-step explanation::

Downward force mg = upward force kδg

Spring stiffness k = Load/extension = 128 / (128/257) = 257 lb/ft

At instant t when spring extension from equilibrium position is y, downward force on the ball is weight mg and upward forces are (k (y+δ)) g and air resistance.

But air resistance = (4v) g. Hence, by Newton's second law, my'' = mg - k (y+δ) g - 4vg

Further, v = dy/dt = y' and mg = kδg

So, my'' = - kyg - 4y'g

or, (m/g) y'' + 4y' + ky = 0,

Boundary conditions: y (0) = 0, y' (0) = 8 ft/s

Characteristic equation is (m/g) r^2 + 4r + k = 0

r = (g/2m) [-4 ± √ (16-4mk/g) ]

r = (32.2/2*128) [-4 ± √ (16-4*128*257/32.2) ] = (-0.503125 ± 8.024 i)

Roots are imaginary

So, y = e^ (-0.503125*t) [C1*Cos (8.024*t) + C2*Sin (8.024*t) ]

Putting y (0) = 0 we get, C1 = 0.

So, y' = C2*e^ (-0.503125*t) * [8.024*Cos (8.024*t) + Sin (8.024*t) * (-0.503125) ]

Putting y' (0) = 8 ft/s, we get C2 = 0.9969

Thus, y = 0.9969*e^ (-0.503125*t) Sin (8.024*t)