A 10 kilogram object suspended from the end of a vertically

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) = 120cos(10f). The force F(t) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds. Determine the spring constant k k = 1000. Newtons/meter 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: 10y\"M+1000y=120cos(10t) help (equations) Initial conditions: y(0) = Q and y\'(0) = Q help (numbers) Solve the initial value problem for y(f ). y(t) = help (formulas) Plot the solution and determine the maximum excursion from equilibrium made by the object on the time interval 0

Solution

1) let spring constant be k

therefore kx = Mg

k x 0.098 = 10x9.8 => k = 1000

2) at any point

F - Ky = Ma

F = Ma + ky

120 cos(10t) = 10 y\'\' + 1000 y ( as we know acceleration is double derivative of displacement with respect to time so a= y\'\')

3)

at t=0 body was at equilibrium position so y (0) = 0 that is no displace ment from equilibrium position

body was at rest so velocity at equilibrium is 0 there fore y\'(0) = 0