# Object oscillates such that displacement is x = (.222)sin(314t)

1. May 13, 2013

### jorgegalvan93

1. The problem statement, all variables and given/known data
Object is oscillates such that displacement is x = (.222)sin(314t), where t is in seconds.
In one period, the objects moves what distance?

2. Relevant equations
What is the relationship between displacement, amplitude and distance?

3. The attempt at a solution
The amplitude is the 0.222 m, so the displacement is four times that, as it moves from the centre to the edge, back to the centre, to the other edge and then back to the centre again, so the object moves 0.888 m. but WHY?????

2. May 13, 2013

### Staff: Mentor

I think you explained it quite well (I bolded your solution). Is there something about that explanation that you are not getting?

3. May 14, 2013

### rock.freak667

If you consider the graph of x=0.222sin(314t) for one period.

Then if you consider the distances covered in going from 0 to max, max to 0 and back, then you will get 4*Amplitude.

Which is exactly what you explained, but I don't know if you wanted to know with respect to the graph/equation.