# Simple Harmonic Motion Question

1. Aug 2, 2010

### coffeem

1. The problem statement, all variables and given/known data
A particle undergoes simple harmonic with amplitude of 10cm and frequency 5Hz. If it passes through the equilibrium position (x=0) at time t = 0, calculate the position, velocity and acceleration of the particle at t = 3.5m5 ms.

3. The attempt at a solution

Given that:

A = 0.1m
f = 5Hz
therefroe w = 2pif = 10pi
phase difference = 0.

So using the standard: x = Acos(wt+PD)

where PD = phase difference.

I get x = 0.1m.
differentiating this I get:
v = -0.34 m/s
and again differentiating I get:
a = -98.0 m/s^2.

However these answers appear to be wrong. Any advice on where I am messing up?

2. Aug 2, 2010

### rock.freak667

Your attempt at getting 'A' and 'ω' are correct. But the particle starts at x=0, t=0. So your general equation should be sine and not cosine.

If it said x=A at t=0, then you'd use cosine.

3. Aug 2, 2010

### coffeem

Thanks - that makes sense. I just read the position from the book without thinking...

So:

x = 0.1*sin(10pi*t)
v = 0.1*10pi*cos(10pi*t)
a = -0.1*100*pi^2*sin(10pi*t)

would that be correct? thanks

4. Aug 2, 2010

### rock.freak667

Yes that should be correct now.

5. Aug 2, 2010

### coffeem

Thank you very much - i appreciate the hepl.