A 3.72-kg iron box is suspended from a spring with a spring constant of k =...
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A 3.72-kg iron box is suspended from a spring with a spring constant of k = 18 N/m. The mass is pulled 0.35 m downward from its equilibrium position and allowed to oscillate.
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What will the angular velocity be for this iron box?
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With what period will the iron box oscillate?
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Homework Answers
Angular velocity = sqroot( k/m)
Or angular velocity = sqroot(18/3.72) = 2.1997 rad / sec
Now omega = 2×pi/T
So time period T = 2×3.14/2.1997 = 2.855 sec
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