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) = 190cos(10t). 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. 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. Solve the initial value problem for y(t). Plot the solution and determine the maximum excursion from equilibrium made by the object on the time interval 0 less than or equal to t less than or equal to infinity. If there is no such maximum, enter NONE.

Solution

[a] At t= sec

F[t=0sec]= 190 cos[0]= 190N

F= Kx----------Hook\'s law

190= K*9.8*10^-2

K= 190/9.8*10^-2 = 1938.8N/m

Spring constant, K=1938.8N/m

[b]

F[t]= ma

a= F[t]/m= 190 cos[10t]/10= 19cos[10t]

y\'\'=19cos[10t]--------------------------------acceleration

y\'= integral[ 19 cos [10t]+y\'[0]= 19sin[10t]/10+y\'[0]= 1.9sin[10t]+y\'[0]

y\'[t]=1.9sin[10t]+y\'[0]----------------------------------velocity

y=integral1.9sin[10t]= -1.9cos[10t]/10+y[0]= -0.19cos[10t]+y[0]

y[t]=-0.19cos[10t]+y[0]-------------------------------displacement

At initial condition t=0 sec; y\'=0+y\'[0]

Hence y\'[0]= y

y=-0.19+y[0]

y[0]= y+0.19

[c]

y[t]=-0.19cos[10t]+y[0]-------------------------------displacement