- #1
Emspak
- 240
- 1
Homework Statement
Given: \frac{1}{2}m (\dot x)^2 + mg x = E
Gravitational force is mg.
We need to show that by solving this DE that we can confirm that the conservation of energy correctly describes one-dimensional motion (the motion in a uniform field). That is, that the same motion is obtained as predicted by Newton' equation of motion.
The Attempt at a Solution
My attempt was as follows:
\frac{1}{2}m\dot x + mg x = E so by moving things around a bit I can reduce this to
(\dot x)^2 = \frac{2(E - mgx)}{m}
and from there I can solve the DE:
\dot x = \sqrt{\frac{2(E - mgx)}{m}} \Rightarrow \int dx =\int \sqrt{\frac{2(E - mgx)}{m}}dt
from there I can move some variables some more:
\frac{\sqrt{m}}{\sqrt{E-mgx}} \int dx = \int \sqrt{2}dt \Rightarrow \sqrt{m} \int \frac{dx}{{\sqrt{E-mgx}}} = \int \sqrt{2}dt
trying a u substitution where u=\sqrt{E-mgx} and du = -mgdx I should have \frac{\sqrt{m}}{mg} \int \frac{du}{{\sqrt{u}}} = \int \sqrt{2}dt which gets me to \frac{\sqrt{m}}{mg} \sqrt{u} + c = \sqrt{2}t
going back to what I substituted for u I have:
\frac{\sqrt{m}}{mg} \sqrt{E-mgx} + c = \sqrt{2}t
and when I do the algebra I get (after moving the c over and squaring both sides):
\frac{m}{m^2g^2}{E-mgx} = {2}t^2 - 2\sqrt{2}c+ c^2
\frac{E}{mg^2}-\frac{x}{mg} = {2}t^2 - 2\sqrt{2}c+ c^2
-\frac{x}{mg} =-\frac{E}{mg^2}+ {2}t^2 - 2\sqrt{2}ct+ c^2
x(t) =\frac{E}{g}- 2mgt^2 + 2\sqrt{2}mgct- c^2mg
I am unsure of the last step. I see something that looks like an equation of motion there -- and since gt = v = \dot x I can see that term, and gt^2 = x. But I am not quite clear on what to do with the energy term.
I know, I know, don't just plug in formulas. But I presume that the prof gave us the instructions he did for a reason and it wasn't just to confuse us, though he has succeeded in that regard. :-)
Anyhow, any assistance is appreciated.
