A spring with an mm-kg mass and a damping constant 9 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 2.5 new

Question

A spring with an mm-kg mass and a damping constant 9 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 2.5 newtons. If the spring is stretched 1 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.

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bonexptip 3 months 2021-08-09T11:58:19+00:00 1 Answers 0 views 0

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    2021-08-09T11:59:50+00:00

    Answer:

    4.05 kg

    Explanation:

    From Hooke’s law, force required to stretch the spring is  represented as follows:

    k (0.5) = 2.5

    Spring Constant, k = 5

    For critical damping, c² – 4mk =0

    m = c² / 4 k

    c= damping constant

    m = Mass to produce critical damping

    There fore, m = (9²/4*5)

    = 81/20

    = 4.05 kg

    Therefore, the mass that would produce critical damping. = 4.05 kg

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