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 1/2 feet. The ball is started in motion from the equilibrium position with a downward velocity of 6 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

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 1/2 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=

Answer

mg-k(x+y)-4v=ma and mg=kx k=4*32/9 k=14.22 so -ky-4v=md^2y/dt^2 -ky-4dy/dt = 4d^2y/dt^2 4d^2y/dt^2 +14.22y+4dy/dt=0 integrate and you will get the value

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