A block of mass m slides on a horizontal frictionless surface. The block is attached to a spring with a spring constant K. At the instant shown below, the spring is stretched and the block is traveling to the right. Neglect air resistance. Equilibrium position

a) Draw the free body diagram for the block.
b) Apply Newton's Second Law, Fnetma, to the block for the x direction. Put into your equation the fact that the spring force magnitude, Fopring Kx, and solve for the acceleration of the bloc.
c) Start with the position equation x x cos( t+%) and differentiate twice to get dt d2x d2x
d)Sct a% from part b) equal to dtx from part c)toshow that ω-ν e) Write the frequency f in terms of K and m. f Write the period T in terms of K and m. g) Write the block velocity V as a function of time. h) Write the block acceleration 'a' as a function of time.

b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e)  T = 2π √m / k

h)   a = - A w² cos (wt+Ф)

Explanation:

a) see free body diagram in the attachment

b) We write Newton's second law

          Fe = m a

          -k x = ma

           a = -k / m x

c) the acceleration is

         a = d²x / dt²

      If x = A cos wt

        v = dx / dt = -A w sin (wt +Ф)

        a = d²x / dt² = - A w² cos (wt+Ф)

d) we substitute in Newton's second law

        d²x / dt² = -k / m x

We call

       w² = k / m

e) substitute to find w

     -A w² cos (wt+Ф) = -k / m A cos (wt+Ф)

      w² = k / m

Angular velocity and frequency are related

       w = 2π f

       f = 1 / T

 We substitute

      T = 2π / w

      T = 2π √m / k

g)    v= - A w sin (wt+Ф)

h) acceleration is

       a = - A w² cos (wt+Ф)

a and b sometimes have a positive value

the bal will not change its velocity.