Question

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

Answer 1

Answer:

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

Answer 2

The time at which the mass passes will be "0.125 s".

Equilibrium

According to the question,

Spring's mass = 4 lb

Spring constant = 2 lb/ft

Speed of mass = -16 ft/s

By using equation of motion,

→   x(t) = $e^{-4t}$ (-1 + gt)

By substituting the values,

      0 = $e^{-4t}$ (-1 + 8t)

-1 + 8t = 0

By adding "1" both sides, we get

-1 + 8t + 1 = 1

          8t = 1

            t = $\frac{1}{8}$

              = 0.125 s

Thus the response above is appropriate.

Find out more information about equilibrium here:

https://brainly.com/question/517289