A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that numerically equal to the instantaneous velocity. The mass is initially released from a point 1 ft above the equilibrium position with a downward velocity of 14 feet/second. Determine the time at which the mass passes through the equilibrium position

the time at which it passes through the equilibrum position is:

t = 0.1 second

Explanation:

given

w= 4pounds

k(spring constant) = 2lb/ft

g(gravitational constant) = 10m/s² = 32ft/s²

β(initial point above equilibrum) = 1

velocity = 14ft/s

attached is an image showing the calculations, because some of the parameters aren't convenient to type.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s