# (1 point) A brick of mass 8 kg

(1 point) A brick of mass 8 kg hangs from the end of a spring.When the brick is at rest, the spring is stretched by 3920 cm. Thespring is then stretched an additional 2 cm and released with adownward force of F(t)=143cos(6t) NF(t)=143cos(6t) N acts on it.Assume there is no air resistance. Note that the acceleration dueto gravity, gg, is g=980g=980 cm/s22.

- Find the spring constant N/cm
- Set up a differential equation that describes this system. Lety(t)y(t) to denote the displacement, in centimeters, of the brickfrom its equilibrium position, and give your answer in terms ofy,y′,y′′y,y′,y″. Assume that positive displacement means the massbelow the equilibrium position (when the spring stretched 3920cm).
- Solve the differential equation with initial conditionsdescribing the motion/the displacement y(t)y(t) of the mass fromits equilibrium position.

Answer:

mass is

kg

.

a spring stretches 3920 cm

so x=3920

from the Hooke’s law, spring constant k is

**…………………..springconstant**

.

.

there is no air resistance. so damping constant is

.

force is

DE is given by

………………………..differentialequation

.

find roots

for complex roots general solution is

………………..(1)

.

here we have

so assume that a particular solution is

.

put all values in DE

compare coefficient boh sides

put both constant in a particular solution

.

.

general solution is

The spring is then stretched an additional 2 cm

so y(0)=2

.

take the derivative of a general solution

there is no initial velocity so y'(0)=0

put both constant in general solution