A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring so that its

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A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f?

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Eirian 2 months 2021-07-22T13:43:08+00:00 1 Answers 5 views 0

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    2021-07-22T13:44:47+00:00

    Answer:

    COMPLETE QUESTION

    A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

    Explanation:

    Given that,

    Extension of spring

    x = 0.0208m

    Mass attached m = 3.39kg

    Additional mass to have a frequency f

    Let the additional mass be m

    Using Hooke’s law

    F= kx

    Where F = W = mg = 3.39 ×9.81

    F = 33.26N

    Then,

    F = kx

    k = F/x

    k = 33.26/0.0208

    k = 1598.84 N/m

    The frequency is given as

    f = ½π√k/m

    Make m subject of formula

    f² = ¼π² •(k/m

    4π²f² = k/m

    Then, m4π²f² = k

    So, m = k/(4π²f²)

    So, this is the general formula,

    Then let use the frequency above

    f = 3Hz

    m = 1598.84/(4×π²×3²)

    m = 4.5 kg

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