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A 40.0 N force stretches a vertical spring 0.250 m.
(A) What mass must be suspended from the spring so that the system will oscillate with a period of 1.0 s?
(B) If the amplitude of the motion is 0.050 m and the period is as in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position moving downwards?
(C) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

  • Physics - ,

    From the force and deflecion information, the spring constant is k = F/X = 160 N/m

    (A) Use the equation
    P (period) = 2 pi sqrt(M/k)
    (Solve for M)

    (B) The total energy (kinetic + potential) = (1/2) k (Amplitude)^2

    If t=0 is the equilibrium position,
    X = -A sin (2 pi t/P)
    A = 0.050 m
    Substitute t = -0.35 s
    That is more than 1/4 period past the t=0 postion. It should be at negatime deflection but on the way back.

    (C) X = -0.030 m. Multiply than by k for the force. The force wil be up.

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