A block of mass M is suspended from a ceiling by a spring with spring stiffness constant k. A penny of mass m is placed on top of the block.

Question

A block of mass M is suspended from a ceiling by a spring with spring stiffness constant k. A penny of mass m is placed on top of the block. What is the maximum amplitude of oscillations that will allow the penny to just stay on top of the block

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Ladonna 4 years 2021-08-26T06:30:28+00:00 1 Answers 38 views 0

Answers ( )

  1. Philomena
    0
    2021-08-26T06:31:40+00:00
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    Answer:

    Explanation:

    Given that:

    Mass of block M

    Mass of penny m

    spring stiffness constant k

    The frequency of oscillation of the block f=\frac{1}{2\pi}\sqrt{\frac{k}{m} }

    The angular velocity is \omega =2\pi f

    =\sqrt{k/m}

    when the penny is resting on the block

    The acceleration of the penny = acceleration of the block

    If R is the reaction of the block on the penny

    R-mg=a_{max}m\\\\=-\omega^2A_{max}m\\\\R=mg-\omega^2A_{max}m

    The penny will leave the block if R = 0

    mg=\omega^2A_{max}m\\\\g=\omega^2A_{max}\\\\A_{max}=\frac{g}{\omega^2} \\\\=\frac{g}{(k/M)} \\\\A_{max}=gM/k

    Therefore the amplitude A_{max}<gM/k for the penny to remain on the block

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