k = 20 N/.098 m = 204 N/m
m = 20/9.81 = 2.04 kg
omega = w = sqrt (k/m) radians/second
so
w = sqrt (204/2.04) = 10 rad/s
if x is distance from equilibrium position, positive up
x = -.05 cos 10 t
The object weighing 20 N is hung on the bottom of the spiral spring , so that stretch as far as 9.8 cm after a state of balance , the weight pulled down as far as 5 cm , measured from idle after burdened . Find the equation of motion generated assuming no friction and air resistance
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