At what distance from equilibrium is the energy of a sho half kinetic and half potential?

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At what distance from equilibrium is the energy of a sho half kinetic and half potential?

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Thu Nguyệt 4 years 2021-07-26T08:42:50+00:00 1 Answers 102 views 0

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  1. thnahdat
    0
    2021-07-26T08:44:02+00:00
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    Answer:

    Distance from equilibrium x = 0.707A

    Where A is the amplitude of the SHO

    Explanation:

    all the energy of a SHO is a potential energy at

    PE = 0.5kA^2

    Where A is the amplitude of the SHO

    Half that energy means

    PE = 0.25kA^2

    The energy of a SHO is half KE and half PE at a displacement of

    PE = 0.5kx^2

    Where x is the displacement the displacement from equilibrium when the energy is half potential energy and half kinetic energy.

    Equating both equations and solving, we have,

    0.25kA^2 = 0.5kx^2

    We solve for x

    x^2 = (0.25kA^2)/(0.5kx^2)

    x^2 = 0.5A^2

    x = 0.707A.

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