.A mass of 1.61 kg stretches a vertical spring 0.313 m. If the spring is stretched an additional 0.132 m and released, how long does it take to reach the (new) equilibrium position again?

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  1. Assuming this is a physics question, and not a differential equation problem, I will just apply the established formulas:
    y(t)=A sin(ωt)
    A=maximum displacement from equilibrium position, = 0.132m
    m=mass = 1.61 kg
    k=stiffness of string=m/d=1.61/0.313

    From the extreme lowest point to the equilibrium position, it is one-quarter of the cycle, or (1/4)*2π=(1/2)π.

    Thus, solve for t in

    I get ω=1.787
    and consequently t=0.88 s.

    Check my calculations

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