Simple Harmonic Motion Equation with Period and Initial Conditions

[Ed Aboud]

Homework Statement

A particle moves with simple harmonic motion of period \frac{\pi}{2}. Initially it is 8cm from the centre of motion and moving away from the centre with a speed of 4 \sqrt{2} cm/s.
Find an equation for the position of the particle in time t second.

Homework Equations

x = A \cos{ \omega t + \epsilon}
v^2 = \omega^2 (A^2 - x^2)
T = \frac{2 \pi}{\omega}

The Attempt at a Solution

T = \frac{2 \pi}{\omega}
\omega = 4 rad s^-1
v^2 = \omega^2 (A^2 - x^2)
32 = 16(A^2 - 64)
A = \sqrt{66}
x = A \cos( \omega t + \epsilon)
x = \sqrt{66}\cos(4t + \epsilon)

The answer in the book is:
x = \sqrt{66}\cos(4t + .175)

I don't understand where the .175 comes from.
Thanks for any help.

 

[Doc Al]

Make use of the initial conditions at t = 0 to solve for the phase.

 

[Ed Aboud]

Thanks for helping.
So at t=0 x=8

8 = \sqrt{66}\cos(\epsilon)
\cos(\epsilon) = \frac{8}{\sqrt{66}}
\epsilon = \arccos\frac{8}{\sqrt{66}}
\epsilon = 10.02498786

Have I made a mistake somewhere?

 

[Doc Al]

Use radians, not degrees.

 

[Ed Aboud]

God, that is embarrassing ha.
Thanks very much.