Question

This problem is an example of critically damped harmonic motion.

A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 1818 feet. The ball is started in motion from the equilibrium position with a downward velocity of 99 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.

Take as the gravitational acceleration 32 feet per second per second. (Note that the positive y direction is down in this problem.)

Answer 1

Spring constant = mg/l = (432)/1818 = 0.07 pound-force/ft

angular frequency is given by

velocity at equilibrium position = A = 99 ft/s

where A is amplitude of oscillation, A = 99/0.133 = 744 ft

damping coefficient b is given by

b velocity = damping force

If damping force in pound-force, b = 4 32 = 128 pound-force / (ft/s)

displacement after t seconds from initial time = A exp( -bt / (2m) ) = 744 exp[ -128t / 8 ]

displacement after t seconds from initial time = 744 exp[ -16t ]