A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless track. At time t = 0.00 s, the mass is released from rest at x = 10.0 cm. (That is, the spring is stretched by 10.0 cm.)Express the displacement x as a function of time t.

x = 0.100 cos (3.65 t)

Explanation:

We must solve Newton's second law for this case

          F = m a

the spring force is given by Hooke's law

          F = -kx

and the definition of acceleration is

         a =

let's substitute

         - kx = m \frac{d^2x}{dt^2}

          \frac{d^2x}{dt^2} + x = 0

this equation has as a solution

           x = A cos (wt + Ф)

where

           w =

let's calculate

           w =

           w = 3.65 rad / s

to find the constant fi let's use the condition t = 0 x = 10.0 cm

         x = A cos (0 +Фi)

         0.100 = A

therefore Ф = 0

the complete solution is

      x = 0.100 cos (3.65 t)

this solution is in meters