(1 point) A 10 kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeter

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1 Point A 10 Kilogram Object Suspended From The End Of A Vertically Hanging Spring Stretches The Spring 9 8 Centimeter 1 (56.37 KiB) Viewed 42 times

(1 point) A 10 kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeters. At time t = 0, the resulting mass-spring system is disturbed from its rest state by the force F(t) = 160 cos(8t). The force F(t) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds. a. Determine the spring constant k. k= Newtons / meter b. Formulate the initial value problem for y(t), where y(t) is the displacement of the object from its equilibrium rest state, measured positive in the downward direction. (Give your answer in terms of y, y', y",t. Differential equation: help (equations) Initial conditions: y(0) : = and y'(0) = help (numbers) c. Solve the initial value problem for y(t). y() = help (formulas) d. Plot the solution and determine the maximum excursion from equilibrium made by the object on the time interval 0 < t < oo. If there is no such maximum, enter NONE. maximum excursion= meters help (numbers)