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

we need answer choices

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)