F(t) as it relates to v(t) and x(t)

1. Oct 11, 2006

Tonyt88

Starting at t = 0, an object of mass m is subject to a force

F(t) = F nought cos omega t

If the initial speed and position are v nought and x nought, find v(t) and x(t), any help would be greatly appreciated.

2. Oct 11, 2006

quasar987

Well, now you gotta solve

$$\frac{dv}{dt}=\frac{F_0}{m}\cos(\omega t)$$

That's easy. What is the function whose derivative is cos?

3. Oct 11, 2006

Tonyt88

Okay, so this is how it worked out for me:

dv = (F0/m) cos(ωt) dt

v = (F0/mω) sin(ωt) + C
v(t) = (F0/mω) sin(ωt) + v0
dx =[ (F0/mω) sin(ωt) + v0 ] dt
x = -(F0/mω2) cos(ωt) + v0t + x0

x(t) = -(F0/mω2) cos(ωt) + v0t + x0

Does that make sense?

4. Oct 11, 2006

quasar987

You went a little too fast. After integrating

dx =[ (F0/mω) sin(ωt) + v0 ] dt

you get

x = -(F0/mω²) cos(ωt) + v0t + C

But plugging x(0)= x_0 does not give C=x_0 because cos(0)=1.

5. Oct 11, 2006

Tonyt88

Oh, oops, so then C = x_0 + (F_0/mω²) ?