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

[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.

 

[quasar987]

Well, now you got to solve

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

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

 

[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?

 

[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.

 

[Tonyt88]

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