Fallen Seraph
Jan9-07, 01:23 PM
Sorry for the undescriptive title, but I couldn't think of a better one.
My question is essentially this: is the following procedure correct?
a=-kx and we want v.
Multiply both sides by v to get:
\frac {d^2x} {dt^2} * \frac {dx} {dt} = -kx *\frac {dx} {dt}
Now, by dt, and some rearranging :
(\frac {d^2x} {dt^2} dt) * \frac {dx} {dt} = -kx *dx
And the step which I'm not convinced can be done:
( \int \frac {d^2x} {dt^2} dt) * \frac {dx} {dt} = -k \int x *dx
and so we end up with v^2 = -(kx^2)/2 +c which is but an errant 1/2 away from what I need.
I just don't trust integrating a wrt t, while having a v just sitting there as if it were a constant. The only reason that I haven't already disregarded it is that it checks out if you analyse the dimensions...
My question is essentially this: is the following procedure correct?
a=-kx and we want v.
Multiply both sides by v to get:
\frac {d^2x} {dt^2} * \frac {dx} {dt} = -kx *\frac {dx} {dt}
Now, by dt, and some rearranging :
(\frac {d^2x} {dt^2} dt) * \frac {dx} {dt} = -kx *dx
And the step which I'm not convinced can be done:
( \int \frac {d^2x} {dt^2} dt) * \frac {dx} {dt} = -k \int x *dx
and so we end up with v^2 = -(kx^2)/2 +c which is but an errant 1/2 away from what I need.
I just don't trust integrating a wrt t, while having a v just sitting there as if it were a constant. The only reason that I haven't already disregarded it is that it checks out if you analyse the dimensions...