A 10 kilogram object suspended from the end of a vertically hanging spring stret

Question

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)=40cos(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.

k= ____?____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:________?________ Initial conditions: y(0)= _____?_____ and y'(0)= _____?_____

Solve the initial value problem for y(t). y(t)=______?______ Plot the solution and determine the maximum excursion from equilibrium made by the object on the time interval 0 </= t < inf. If there is no such maximum, enter NONE.

maximum excursion =_____?_____ meters

Explanation / Answer

Its given that,

mass of the object = 10 kg

strech of the spring = x = 9.8 cm = 0.098 m

F(t) = 40 Cos (10t) at time t= 0

The gravitational and the Hook's forces are balancing each other, in order to keep the mass anged vertically from the spring. So,

F(grav) = F(hooks)

m g = - K (x) this gives us

K = m g / x = 10 x 9.8 / 0.098 = 1000 N/m

Hence Spring constant K = 1000 N/m

The diffrential equation will be as follows :

m x d2y/dt2 + K dy /dt = 40 Cos (10 t)

10 d2y/dt2 + 1000 dy/dt = 40 Cos (10 t)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.