# Simple harmonic motion of a particle

1. Feb 13, 2008

### ~christina~

1. The problem statement, all variables and given/known data

A particle moves along x axis It is initially at 0.270m moving with velocity of 0.140m/s
and acceleration of -0.320m/s^2
Suppose it moves with constant acceleration for 4.50s.

Find it's:
a) position and velocity at end of time interval

next assume it moves with simple harmonic motin for 4.50s and x= 0 is it's equillibrium position.

Find it's:

b) position and velocity at end of this time interval

2. Relevant equations
$$x(t)= A cos (\omega t + \phi )$$
$$v(t)= - \omega A sin (\omega t + \phi )$$
$$a(t)= - \omega ^2 A cos (\omega t + \phi)$$

3. The attempt at a solution

well I know for the first situation that:

$$x_o$$= 0.270
$$v_o$$= 0.140m/s
$$a_o$$= -0.320m/s^2

and the acceration remain constant for 4.50s

to find a) the postiion and velocity at end of time interval....

I was thinking that I would take the period (T) to equal T= 4.50s?
and would I use 4.51 seconds and use the velocity as a function of time equation

$$v(t)= - \omega A sin (\omega t + \phi )$$

though I need the $$\phi$$....hm..

and I think I'm lost here.

Last edited: Feb 13, 2008
2. Feb 13, 2008

### Littlepig

so, a) it's no problem as is an 1 dimension problem, with movement equations...

about b), it's a whole new problem:
you will need to know the position and velocity at $$t_0=4.50$$
so, you can write: $$t_1=0$$ at $$x_1=0$$(equilibrium point)
by a), you calculated velocity at $$t_0$$, so, it's the particles velocity at $$t_1$$

I guess this should help...

Last edited: Feb 13, 2008