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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

Respuesta :

Answer:

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.

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Cricetus

The time at which the mass passes will be "0.1 s".

Equilibrium

According to the question,

Mass weighing, w = 4 pounds

Spring constant, k = 2 lb/ft

Gravitational constant, g = 10 m/s² = 32 ft/s²

Point above equilibrium, β = 1

Velocity = 14 ft/s

By using equation of motion,

→     x(t) =  (-1 + gt)

By substituting the values,

         0 =  (-1 + 10t)

-1 + 10t = 0

By adding "1" both sides, we get

-1 + 10t + 1 = 1

           10t = 1

               t = [tex]\frac{1}{10}[/tex]

                 = 0.1 s

Thus the above answer is right.

Find out more information about equilibrium here:

https://brainly.com/question/517289

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