- #1

Emspak

- 243

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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. I had this up as a thread before -- (thanks to tiny-tim!) but I discovered that there might have been some bad errors in my original set-up. So I wanted to see if I corrected those.

## 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, in my initial attempt I left the [itex]\sqrt{2}[/itex] on the right but this time I will move it over:

[itex]\frac{\sqrt{m}}{\sqrt{2(E-mgx)}} \int dx = \int dt \Rightarrow \frac{\sqrt{m}}{\sqrt{2}} \int \frac{dx}{{\sqrt{E-mgx}}} = \int 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 \sqrt{2}} \int \frac{du}{{\sqrt{u}}} = \int dt[/itex] which gets me to [itex]\frac{2\sqrt{m}}{mg\sqrt{2}} \sqrt{u} + c = t[/itex]

going back to what I substituted for u I have:

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

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

[itex]\frac{4m}{2m^2g^2}({E-mgx}) = t^2 - 2ct+ c^2[/itex]

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

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

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

From here I plug in the equation for kinetic energy. [itex]E = (1/2) mv^2[/itex] which in this case would be [itex]\frac{E}{mg} = \frac{(1/2) mg^2t^2}{mg} = \frac{1}{2}gt^2[/itex]

which changes my x(t) expression to:

[itex]x(t) =-\frac{1}{2}gt^2 + cgt- \frac{c^2g}{2}[/itex]

That looks like an equation of motion to me. And if we take a derivative of the original expression we get:

[itex]\frac{1}{2}m \dot x \ddot x + \frac{1}{2}m \dot x \ddot x + mg \dot x = 0 \Rightarrow m \dot x \ddot x + mg \dot x = 0[/itex]

and that shows that [itex]g=\ddot x [/itex]. Maybe that's what we were shooting for in the first place.

and looking at the x(t) expression I got, we can replace g with [itex]\ddot x[/itex] and that makes x(t) into:

[itex]x(t) =-\frac{1}{2}x + c\dot x- \frac{c^2}{2}\ddot x[/itex]

I feel like I am close here, but not quite there. So I am hoping someone can tell me where I messed up.