A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 16 ft/s. Determine the time (in s) at which the mass passes through the equilibrium position. (Use g

0.125 s

Explanation:

Given that

Mass of spring is 4lb

Spring constant = 2 lb/ft

Speed of the mass = -16 ft/s

See attachment for the calculations, and here's the final answer

The mass passes through the equilibrium position when x(t) = 0, thus

e^-4t [-1 + 8t] = 0

-1 + 8t = 0

8t = 1

t = 1/8

t = 0.125 s

Therefore, the time at which the mass passes through equilibrium position is 0.125 s

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