A 2 kg mass connected to a spring with spring constant k = 10 N/m oscillates in simple harmonic motion with an amplitude of A = 0.1 m. What

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A 2 kg mass connected to a spring with spring constant k = 10 N/m oscillates in simple harmonic motion with an amplitude of A = 0.1 m. What is the kinetic energy of the mass when its position is at x = 0.05 m? g

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Diễm Kiều 5 years 2021-08-12T21:37:29+00:00 1 Answers 13 views 0

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  1. kimchi2
    0
    2021-08-12T21:38:33+00:00
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    Answer:

    Kinetic energy at 0.05 m is 0.037 J

    Explanation:

    Given:

    Mass, m = 2 kg

    Spring constant, k = 10 N/m

    Amplitude, A = 0.1 m

    Angular frequency, ω = √k/m

    Substitute the suitable values in the above equation.

    \omega = \sqrt\frac{10}{2}

    ω = 2.24 s⁻¹

    Simple harmonic equation is represent by the equation:

    x = A cos ωt

    Substitute 0.05 m for x, 0.1 m for A and 2.24 s⁻¹ for ω in the above equation.

    0.05 = 0.1\cos(2.24t)

    t = 0.47 s

    Kinetic energy at x = 0.05 is determine by the relation:

    E=\frac{1}{2} kA^{2}\sin^{2}(\omega t)

    Substitute the suitable values in the above equation.

    E=\frac{1}{2} \times 10 \times 0.1^{2} \sin^{2}(2.24\times0.47)

    E = 0.037 J

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