# Simple Harmonic Motion

## 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.

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

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?

Last edited:
Doc Al
Mentor

God, that is embarrassing ha.
Thanks very much.