Question

(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. The spring is then stretched an additional 2 cm and released with a downward force of F(t)=143cos(6t) NF(t)=143cos⁡(6t) N acts on it. Assume there is no air resistance. Note that the acceleration due to gravity, gg, is g=980g=980 cm/s22.

1. Find the spring constant  N/cm
   
2. Set up a differential equation that describes this system. Let y(t)y(t) to denote the displacement, in centimeters, of the brick from its equilibrium position, and give your answer in terms of y,y′,y′′y,y′,y″. Assume that positive displacement means the mass below the equilibrium position (when the spring stretched 3920 cm).
3. Solve the differential equation with initial conditions describing the motion/the displacement y(t)y(t) of the mass from its equilibrium position.

Answer 1

mass is

kg

.

a spring stretches 3920 cm

so x=3920

from the Hooke's law, spring constant k is

.......................spring constant

.

.

there is no air resistance. so damping constant is

.

force is

DE is given by

.............................differential equation

.

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