Simple Oscillator Problem: Spring and a block

Problem: A 0.200 Kg mass is attached to a spring of constant k = 5.6 N/m and set into oscillation with an amplitude of 0.25m. Solve the following circumstances:

A) Determine the frequency of the motion         

Trying to solve for F (frequency). By using the free body diagram we sum the forces in the x direction (horizontal in the example).

               

By summing forces in the X axis (positive to the right) we get: Sum of Forces in x = Force of the spring. Steps shown below:

Sum of Forces in x = (mass)(acceleration)
Showing the only force in the X direction is the Force of the spring in the negative direction.
A force of a spring is the spring constant times the distance from rest.
Substituting the force of the spring in we get  (mass)(accell.) = -(spring const.)(distance)

We are looking to get this into the form to where we can show what the second derivative of x (the position) is equal to a so we can extract omega from it. Acceleration (a) is the second derivative of position (x). Therefore we can solve for (a) from our forces above. Steps:

Start with our equation from our sum of forces above.
Solving for a we get:   a = -(k/m)x
We need to find that second derivative of position equals negative omega times the position (second derivative of x = -(w^2)a). Acceleration is the second derivative of position. We replace a with (d^2/dt^2)x to show that. Notice that k/m is omega squared.
Because we got our equation in the correct form, we show that k/m = w^2. Solve for w (omega) we get sqrt(k/m) = w.
We know that w (omega) equals (2)(Pi)(Frequency). So we can easily solve for F to find out answer.
Plugging in the given values of the problem we can get a numerical answer which is approximately .842169 hertz.

B) Determine the Period

Frequency is defined as 1 divided by the period, T. So we use F=1/T for our equation to work with.
Solving for T we get T=1/F. So T equals 1/.842169. Therefore F equals 1.18741 seconds.

C) Determine Maximum Velocity

Using the velocity function of a simple oscillator we get this equation. From this, wx(not), is the maximum velocity (amplitude of the wave form)
Solving for Vmax we get (.842169)(.25) where .25 is the amplitude of the x function (given). Therefore we get .210542 m/s

D) Determine the Maximum Force

From our free-body diagram, we know our only force is the spring. Which the force of the spring = -kx.
So the max force would be k times the amplitude of x. So we have (5.6)(.25) which equals 1.4 Newtons.