Physics

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A 2kg object is attached to a spring and undergoes simple harmonic motion. At t=0 the object starts from rest, 10 cm from the equilibrium position. If the force constant of the spring k is equal to 75 N/m, calculate
a. the max speed
b. the max acceleration of the object. Fine the velocity of the object at t=5s

  • Physics -

    10 cm = 0.1 m

    x = .1 cos 2 pi t/T

    v = -(2 pi/T) .1 sin 2 pi t/T

    a = - (2 pi/T)^2 .1 cos 2 pi t/T = -(2 pi/T)^2 x

    F = - k x

    F = m a = - 2 (2 pi/T)^2 x
    k= 8 pi^2/T^2 = 75

    T ^2 = 8 pi^2/75
    T = 1.03 s
    2pi /T = 6.12
    (2 pi/T)^2 = 37.5

    so
    if v = -(2 pi/T) .1 sin 2 pi t/T
    then v max = (2 pi/T) .1
    =.612
    and
    if a = - (2 pi/T)^2 .1 cos 2 pi t/T
    then a max = .1(2 pi/T)^2
    =3.75

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