(1 point) A frictionless spring with a 6-kg mass can be held stretched 0.2 meters beyond its natural length by a force of 50 newtons. If the

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(1 point) A frictionless spring with a 6-kg mass can be held stretched 0.2 meters beyond its natural length by a force of 50 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 0.5 m/sec, find the position of the mass after tt seconds.

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Thông Đạt 3 years 2021-07-17T11:57:46+00:00 1 Answers 9 views 0

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  1. Neala
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    2021-07-17T11:59:24+00:00
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    Answer:

    x(t) = 0.077cos(6.455t)

    Explanation:

    If the spring can be stretched 0.2 m by a force of 50 N, then the spring constant is:

    k = 50 / 0.2 = 250 N/m

    The equation of simple harmonic motion is as the following:

    x(t) = Acos(\omega t - \phi)

    where \omega = \sqrt{k/m} = \sqrt{250 / 6} = 6.455

    We also know that the initial velocity is 0.5 m/s, which is also the maximum speed at the equilibrium:

    v_{max} = A\omega

    A = v_{max}/\omega = 0.5 / 6.455 = 0.077 m

    \phi = 0 is the initial phase

    Therefore, the position of the mass after t seconds is

    x(t) = 0.077cos(6.455t)

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