An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s. Calculate the dista

Question

An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s. Calculate the distance d by which the spring stretches from its unstrained length when the object is allowed to hang stationary from it.

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bonexptip 3 years 2021-08-12T21:41:04+00:00 1 Answers 8 views 0

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  1. Thunguyet
    0
    2021-08-12T21:42:30+00:00
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    Answer:

    The distance by which the spring stretches is 1.48 cm.

    Explanation:

    Given that,

    An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s, \omega=25.7\ rad/s

    We know that angular frequency in SHM is given by :

    \omega=\sqrt{\dfrac{k}{m}} \\\\\omega^2=\dfrac{k}{m}\\\\\dfrac{k}{m}=(25.7)^2.............(1)

    When the object is allowed to hang stationary from it, the force due to spring is balanced by its weight. Such that :

    kd=mg\\\\d=\dfrac{g}{(k/m)}

    From equation (1) :

    d=\dfrac{9.8}{(25.7)^2}\\\\d=0.0148\ m\\\\d=1.48\ cm

    So, the distance by which the spring stretches is 1.48 cm.

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