# Simple Harmonic Motion + Regular Kinematics

1. Jan 23, 2012

### fnaarfnaar

1. The problem statement, all variables and given/known data
A particle moves along the x axis. It is initially at the position 0.250 m,
moving with velocity 0.200 m/s and
acceleration -0.290 m/s2.
Suppose it moves with constant acceleration for 5.80 s.

Position of Particle after 5.80 Seconds - -3.4678 m.
Velocity of Particle After 5.80 Seconds -1.482 m/s

Next, assume it moves with simple harmonic motion for 5.80 s and x = 0 is its equilibrium position.

What is its position at the end of this time interval?
What is its velocity at the end of this time interval?

2. Relevant equations

Attempts :

I attempted to use the formula X(t)= A cos (wt), with A being the amplitude 0.250 (because a spring can't stretch farther than its intial position) and w = angular velocity which I used 0.2 / .25.

For the velocity I used the formula- v(t)=w A sin (t)

From this point I was completely lost because I could not find a way to incorporate the constant acceleration.

Simple Harmonic Motion equations

if you could leave an explanation, it's be greatly appreciated!!

Last edited: Jan 23, 2012
2. Jan 23, 2012

### vela

Staff Emeritus
You need to make an attempt at solving the problem yourself and showing us what you did before you can receive assistant, per the forum rules.