A block of mass 0.464 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.

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A block of mass 0.464 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.646 m. Find the spring constant. N/m The block is then pulled down an additional 0.338 m and released from rest. Assuming no damping, what is its period of oscillation? s How high above the point of release does the block reach as it oscillates?

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  1. Explanation:

    Mass, m = 0.464 kg

    Compression in the spring, x = 0.646 m

    (a) The net force acting on the spring is given by :

    [tex]kx=mg[/tex]

    k is spring constant

    [tex]k=\dfrac{mg}{x}\\\\k=\dfrac{0.464\times 9.8}{0.646 }\\\\k=7.03\ N/m[/tex]

    (b) The angular frequency of the spring mass system is given by :

    [tex]\omega=\sqrt{\dfrac{k}{m}} \\\\\omega=\sqrt{\dfrac{7.03}{0.464 }} \\\\\omega=3.89\ rad/s[/tex]

    The period of oscillation is :

    [tex]T=\dfrac{2\pi}{\omega}\\\\T=\dfrac{2\pi}{3.89}\\\\T=1.61\ m/s[/tex]

    (c) As the spring oscillates, its will reach to a height of 2(0.338) = 0.676 m

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