A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring \frac{128}{401} feet.
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A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring \frac{128}{401} feet.

[From: ] [author: ] [Date: 12-03-31] [Hit: ]
The IVP you have to solve,4y + 4y + 401y = 0, y(0) = 0,Solving this should be straight forward.......
A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring 128/401 feet.

The ball is started in motion from the equilibrium position with a downward velocity of 5 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 postiive direction for y is down.)

y=?????

Take as the gravitational acceleration 32 feet per second per second.

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Mass is weight/(gravitational acceleration), so

m = 128/32 = 4 slugs.

The spring constant, assuming Hooke's Law, is determined by the given displacement

F = 128lb = k(128/401) ft ==> k = 401 lb/ft.

The damping constant is given as 4 (slug/sec). The IVP you have to solve, with the "down is positive" orientation is

4y'' + 4y' + 401y = 0, y(0) = 0, y'(0) = 5

Solving this should be straight forward.
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