Question

A spring with a 6-kg mass and a damping constant 5 can be stretched 1.5 meters beyond its natural length and held at rest there by a force of 7.5 newtons. Suppose the spring is stretched 3 meters below spring-mass equilibrium and then released with zero velocity.

Find the position of the mass after t seconds.

y(t)=?

Please show all the work.

Answer 1

mass is

kg

.

a force of 7.5 N, stretches a spring 1.5 m

so x=1.5

from the Hooke's law, spring constant k is

.

.

damping constant is

.

DE is given by

find roots

for complex roots general solution is

........................(1)

.

here mass is initially released from rest from a point 3 m  above the equilibrium position.

so y(0)=3

................put it back in equation 1

.

.......................(2)

take derivative

here initial velocity is zero so y'(0)=0

................put it back in the equation 2

.