- #1

Emspak

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## Homework Statement

Given: [itex]\frac{1}{2}m (\dot x)^2 + mg x = E[/itex]

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:

[itex]\frac{1}{2}m\dot x + mg x = E[/itex] so by moving things around a bit I can reduce this to

[itex](\dot x)^2 = \frac{2(E - mgx)}{m}[/itex]

and from there I can solve the DE:

[itex]\dot x = \sqrt{\frac{2(E - mgx)}{m}} \Rightarrow \int dx =\int \sqrt{\frac{2(E - mgx)}{m}}dt[/itex]

from there I can move some variables some more:

[itex]\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[/itex]

trying a u substitution where [itex]u=\sqrt{E-mgx}[/itex] and [itex]du = -mgdx[/itex] I should have [itex]\frac{\sqrt{m}}{mg} \int \frac{du}{{\sqrt{u}}} = \int \sqrt{2}dt[/itex] which gets me to [itex]\frac{\sqrt{m}}{mg} \sqrt{u} + c = \sqrt{2}t[/itex]

going back to what I substituted for u I have:

[itex]\frac{\sqrt{m}}{mg} \sqrt{E-mgx} + c = \sqrt{2}t[/itex]

and when I do the algebra I get (after moving the c over and squaring both sides):

[itex]\frac{m}{m^2g^2}{E-mgx} = {2}t^2 - 2\sqrt{2}c+ c^2[/itex]

[itex]\frac{E}{mg^2}-\frac{x}{mg} = {2}t^2 - 2\sqrt{2}c+ c^2[/itex]

[itex]-\frac{x}{mg} =-\frac{E}{mg^2}+ {2}t^2 - 2\sqrt{2}ct+ c^2[/itex]

[itex]x(t) =\frac{E}{g}- 2mgt^2 + 2\sqrt{2}mgct- c^2mg[/itex]

I am unsure of the last step. I see something that looks like an equation of motion there -- and since [itex]gt = v = \dot x[/itex] I can see that term, and [itex]gt^2 = x[/itex]. 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.