A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude is 0.45 m, w

Question

A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude is 0.45 m, what is the frequency of the oscillation?

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Minh Khuê 5 months 2021-09-05T13:53:28+00:00 1 Answers 0 views 0

Answers ( )

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    2021-09-05T13:54:33+00:00

    Answer:

    The frequency of the oscillation is 2.45 Hz.

    Explanation:

    Given;

    mass of the spring, m = 0.5 kg

    total mechanical energy of the spring, E = 12 J

    Determine the spring constant, k as follows;

    E = ¹/₂kA²

    kA² = 2E

    k = (2E) / (A²)

    k = (2 x 12) / (0.45²)

    k = 118.519 N/m

    Determine the angular frequency, ω;

    \omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

    Determine the frequency of the oscillation;

    ω = 2πf

    f = (ω) / (2π)

    f = (15.396) / (2π)

    f = 2.45 Hz

    Therefore, the frequency of the oscillation is 2.45 Hz.

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