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An object with mass 3.0 kg is attached to a spring with spring stiffness constant k = 300 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 0.55 m/s.

(a) Calculate the amplitude of the motion.


m
(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]
m/s

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5 answers
  1. Total energy (which is conserved)
    = (1/2)M*V^2 + (1/2)k X^2
    = 1.5*0.55^2 + 150*0.02^2 = 0.5138 J

    (a) when V = 0 you get the maximum deflection Xmax. That is the amplitude.
    0.5138 = (300/2)*Xmax^2
    Xmax = 0.0585 m

    (b) when X = 0 you have maximum speed:
    0.5138 = 1.5*Vmax^2
    Vmax = 0.5853 m/s

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