Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 decade ago

An object with mass 3.0 kg is attached to to a spring with spring stiffness constant k = 280 N/m and is?

executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of speed of 0.55 m/s (a) calculate the amplitude of the motion. (b) calculate the maximum velocity attained by the object?.

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  • 1 decade ago
    Best Answer

    for spring restorative force

    F = - k x

    m dv / dt = - k x

    dv = - k/m x dt

    dv = - k/m x {dx/v}

    v dv = - k/m dx integrating

    (v^2/2) = - k/m (x^2/2) + c

    (v^2) = - k/m (x^2) + c1

    ----------------------

    at x = a (amplitude)

    v = 0 (spring retreats after reaching amplitude

    -------------------------

    0 = - k/m (a^2) + c1 >>> c1 put back

    (v^2) = - k/m (x^2) + k/m (a^2)

    v^2 = k/m [a^2 - x^2]

    -----------------------------

    given x = 0.02, v = 0.55

    0.55^2 = k/m [a^2 - 0.02^2]

    a^2 = [(m/k) * 0.55^2] + 0.02^2

    calculate (a) amplitude

    --------------------------------

    v^2 = k/m [a^2 - x^2]

    velocity will be maximum when x = 0 ie at mean position

    by seeing v^2 formula you can know that (- x^2) term is trying to decrease the v^2 so if x=0 this term will not reduce v^2

    hence vmax

    (vmax)^2 = k/m * a^2

    just found value of a is known and others are known

    vmax = a √(k/m)

    Source(s): hope you will settle these questions
  • kasee
    Lv 4
    3 years ago

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