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.

Question

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?

in progress 0
Linh Đan 2 months 2021-07-24T13:55:31+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-07-24T13:57:17+00:00

    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 :

    kx=mg

    k is spring constant

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

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

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

    The period of oscillation is :

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

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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )