Simple harmonic motion and static equilibrium

[oceanwalk]

Homework Statement

In static equilibrium a spring is stretched 0.25 m by a hanging mass. If the system is initially moved up to the unstretched position and released with zero velocity, determine position of the mass (in m) after 4.2s. Take the equilibrium position to be zero and up the positive direction.

Homework Equations

at equilibrium
-kx+mg = 0
m(x double dot (t)) + kx(t) = 0

The Attempt at a Solution

 

[Mark44]

Homework Statement

In static equilibrium a spring is stretched 0.25 m by a hanging mass. If the system is initially moved up to the unstretched position and released with zero velocity, determine position of the mass (in m) after 4.2s. Take the equilibrium position to be zero and up the positive direction.

Homework Equations

at equilibrium
-kx+mg = 0
m(x double dot (t)) + kx(t) = 0

Do you know how to solve a 2nd-order, constant coefficient, homogeneous differential equation?
This is the initial value problem:
x''(t) + (k/m)x(t) = 0, with initial conditios x(0) = 0, x'(0) = 0

 

[oceanwalk]

Do you know how to solve a 2nd-order, constant coefficient, homogeneous differential equation?
This is the initial value problem:
x''(t) + (k/m)x(t) = 0, with initial conditios x(0) = 0, x'(0) = 0

Ok, I must not know how to do it correctly because I'm getting it to equal zero for all time.

Solving the equation you gave me:
x(t) = C2 sin(sqrt(a) t) + C1 cos(sqrt(a) t) = 0
using the initial conditions
C2 = 0 , C1 = 0

as such, x(t) = 0 for all time

 

[Mark44]

Try it with one change: x(0) = 0.25. This means that we are taking the origin to be the point where the spring is in static equilibrium. Raising it .25 m. and then letting it go means x(0) = .25 and x'(0) = 0.