Because this question is long I\'m giving extra points. A brick of mass 2 kg han

Question

Because this question is long I'm giving extra points. A brick of mass 2 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 2 cm and released with a downward force of F(t)=2.59259cos(2t) N acts on it. Assume there is no air resistance. Note that the acceleration due to gravity, g, is g=980 cm/s^2 a) Find the spring constant. b) Set up a differential equation that describes this system. Let y(t) to denote the displacement, in cm, of the brick from its equilibrium position, and give your anser in terms of y, y', y''. Assume that positive displacement means the mass below the equilibrium position (when the spring stretched 8820 cm) c) Solve the differential equation with initial conditions describing the motion/the displacement y(t) of the mass from its equilibrium position.

Explanation / Answer

a) at equilibrium; m*g = k*x k = m*g/x = 2*9.8/88.2 = 2/9 N/m b) my'' + ky = F(t) 2*y'' + (2/9)*y = 2.59259cos(2t) c) Initial conditions: At t = 0 , y = 2cm and t = o , y' = 0 cm/sec solution of the differential equation = Complementary function + particular integral Complementary function = A cos(wt) + B sin(wt) , where A and B are constant from initial conditions we get A = 0 and B = 2 particular integral = F*cos(2t)/k*(r^2 -1) where r = natural frequency/ forced frequency = 6 natural frequency = w = sqrt(k/m) and forced frequency = 2 (given) therefore particular integral = 0.33 cos(2t) solution of the differential equation = 2 sin(0.33t) + 0.33 cos(2t)

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