A 16 pound weight attached to a spring exhibits simple harmonic motion. Determine the equation of motion if the spring constant is 8 lb/ft a

Question

A 16 pound weight attached to a spring exhibits simple harmonic motion. Determine the equation of motion if the spring constant is 8 lb/ft and if the weight is released 6 inches below the equilibrium position with a downward velocity of 1 ft/s.

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Ngọc Diệp 4 years 2021-08-15T23:12:19+00:00 1 Answers 28 views 0

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  1. ThanhThu
    0
    2021-08-15T23:13:52+00:00
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    Answer:

    T = 2.82π s

    x = Acos(0.71t)

    Explanation:

    This problem can be solved by using the expressions

    T = 2\pi \sqrt{\frac{m}{k} }    ( 1 )

    x=Acos(\omega t) = Acos(\frac{2\pi }{T}t )   ( 2 )

    where T is the period of oscillation of the system, m is the mass of the object attached to the spring, k is the spring constant and x is the position of the object.

    By replacing in the expression (1):

    T=2\pi \sqrt{\frac{16 lb}{8lb/ft}} = 2.82\pi s

    Taking 6 inches as the amplitude of the motion, we have

    x=6cos(\frac{2\pi }{2.82\pi }t ) = 6cos(0.71t)

    I hope this is useful for you

    Regards

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