(1 point) A brick of mass 8 kg hangs from the end of a spring. When...

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(1 point) A brick of mass 8 kg hangs from the end of a spring.When the brick is at rest, the spring is stretched by 3920 cm. Thespring is then stretched an additional 2 cm and released with adownward force of F(t)=143cos(6t) NF(t)=143cos?(6t) N acts on it.Assume there is no air resistance. Note that the acceleration dueto gravity, gg, is g=980g=980 cm/s22.

  1. Find the spring constant  N/cm
  2. Set up a differential equation that describes this system. Lety(t)y(t) to denote the displacement, in centimeters, of the brickfrom its equilibrium position, and give your answer in terms ofy,y?,y??y,y?,y?. Assume that positive displacement means the massbelow the equilibrium position (when the spring stretched 3920cm).
  3. Solve the differential equation with initial conditionsdescribing the motion/the displacement y(t)y(t) of the mass fromits equilibrium position.

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4.1 Ratings (746 Votes)
mass is kga spring stretches 3920 cmso x3920from the Hookes law spring constant k isspringconstantthere is no air    See Answer
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