A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 115 N/m an

Question

A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 115 N/m and negligible mass. At time t = 0 the mass is released from rest at a distance d = 0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by y(t) = A cos(ωt – φ). The positive y-axis points upward. show answer Correct Answer 17% Part (a) Find the angular frequency of oscillation, in radians per second. ω = 10.22 ✔ Correct! show answer Incorrect Answer 17% Part (b) Determine the value of the coefficient A, in meters.

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Sigridomena 7 months 2021-07-14T08:05:32+00:00 1 Answers 7 views 0

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    2021-07-14T08:06:49+00:00

    Answer:

    a) = 10.22 rad/s

    b) = 0.35 m

    Explanation:

    Given

    Mass of the particle, m = 1.1 kg

    Force constant of the spring, k = 115 N/m

    Distance at which the mass is released, d = 0.35 m

    According to the differential equation of s Simple Harmonic Motion,

    ω² = k / m, where

    ω = angular frequency in rad/s

    k = force constant in N/m

    m = mass in kg

    So,

    ω² = 115 / 1.1

    ω² = 104.55

    ω = √104.55

    ω = 10.22 rad/s

    If y(0) = -0.35 m and we want our A to be positive, then suffice to say,

    The value of coefficient A in meters is 0.35 m

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