A steel bail weighing 128 pounds is suspended from a spring. This stretches the

Question

A steel bail weighing 128 pounds is suspended from a spring. This stretches the spring 128/17 feet. The ball is started in motion from the equilibrium position with a downward velocity of 9 feet per second. The air resistance (in pounds) of the moving bail numerically equals 4 times its velocity (in feet per second). Suppose that after t seconds the bail is y feet below its rest position. Find y in terms of t. (Note that this means that the positive direction for y is down.) Take as the gravitational acceleration 32 feet per second per second.

Explanation / Answer

Solution:

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(128/17) ft ==> k = 17 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' + 17y = 0, y(0) = 0, y'(0) = 9

Solving this should be straight forward in this equation

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