A brick of mass 8 kg hangs from the end of a spring. When the brick is at rest,

Question

A brick of mass 8 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 8820 cm. The spring is then stretched an additional 4 cm and released with a downward force of F(t)=95.7037cos(6t) N acts on it. Assume there is no air resistance. Note that the acceleration due to gravity, g, is g=980 cm/s2. Find the spring constant N/cm Set up a differential equation that describes this system. Let y(t) to denote the displacement, in centimeters, of the brick from its equilibrium position, and give your answer in terms of y,y?,y??. Assume that positive displacement means the mass below the equilibrium position (when the spring stretched 8820 cm). Solve the differential equation with initial conditions describing the motion/the displacement y(t) of the mass from its equilibrium position. y(t)= cm

Explanation / Answer

8 kg x 9.81 m/s^2 = 78.48 N
78.48 N = k x .09m
k = 872 N/m
y(t) = 1/2 x 872 N/m x .04m^2
y(t) = .6976m

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