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 9 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. y= _________________________?

Explanation / Answer

mu'' + gamma*u' + ku = 0

where m is weight/acceleration due to gravity (4/32) or 1/8 gamma is the damping coefficient (it's given) and k = spring constant or weight/displacement = 4/(12) =1/3

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