A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches

Question

A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 12 feet. The ball is started in motion from the equilibrium position with a downward velocity of 8 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 the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second.

Explanation / Answer

Starting from chuckt's eqn. derived by him taking 4 lb as weight 4lbforce, k = 28 lbf/ft and b = 6 lbf/(ft/s),
(1/8)y'' + 4y' + 28u = 0
=> y" + 32y' + 224y = 0

The solution of this
y = Ae^(-bt/2m) sin(?' t + ?),
where A and ? are constants and ?' = ?[k/m - b/(2m)^2] with k/m > (b/2m)^2, i.e., k > b^2/4m
This condition of k > b^2/4m is not satisfied.
In a repost of this question, you have changed the value of elongation to 1/8 ft
=> k = 32 lbf/ft.
With b = 4 and m = 1/8, b^2/4m = 32
This also does not satisfy the condition k/m > (b/2m)^2

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