Forest UCM Energy Line1D: Difference between revisions
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::<math> = \sqrt{\frac{m}{2}} \int_{x_0}^x\left (E-U(x) \right )^{-\frac{1}{2}} dx </math> | ::<math> = \sqrt{\frac{m}{2}} \int_{x_0}^x\left (E-U(x) \right )^{-\frac{1}{2}} dx </math> | ||
::<math> = \sqrt{\frac{m}{2}} \int_{x_0}^x\left (E-\frac{1}{2}kx^2 \right )^{-\frac{1}{2}} dx </math> | ::<math> = \sqrt{\frac{m}{2}} \int_{x_0}^x\left (E-\frac{1}{2}kx^2 \right )^{-\frac{1}{2}} dx </math> | ||
::<math> = \sqrt{\frac{ | ::<math> = \sqrt{\frac{mE}{2}} \int_{x_0}^x \left (1-\left ( x \sqrt{\frac{k}{2E}}\right ) ^2 \right )^{-\frac{1}{2}} dx </math> | ||
let | let | ||
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then | then | ||
: <math>t = \sqrt{\frac{ | : <math>t = \sqrt{\frac{mE}{2}} \int_{x_0}^x \left (1-\sin^2 \theta \right )^{-\frac{1}{2}} dx </math> | ||
: <math>= \sqrt{\frac{ | : <math>= \sqrt{\frac{mE}{2}} \int_{x_0}^x \frac{dx}{ \cos \theta } </math> | ||
:: <math>= \frac{1}{\omega} \int_{\theta_0}^{\theta} d \theta</math> | :: <math>= \frac{1}{\omega} \int_{\theta_0}^{\theta} d \theta</math> | ||
[[Forest_UCM_Energy#Energy_for_Linear_1-D_systems]] | [[Forest_UCM_Energy#Energy_for_Linear_1-D_systems]] | ||
Revision as of 12:32, 26 September 2014
The equation of motion for a system restricted to 1-D is readily solved from conservation of energy when the force is conservative.
- [math]\displaystyle{ T + U(x) = }[/math] cosntant [math]\displaystyle{ \equiv E }[/math]
- [math]\displaystyle{ \Rightarrow T = E - U(x) }[/math]
- [math]\displaystyle{ \frac{1}{2} m \dot {x}^2 = E -U(x) }[/math]
- [math]\displaystyle{ \dot x = \pm \sqrt{\frac{2\left (E-U(x) \right )}{m}} }[/math]
- [math]\displaystyle{ \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx = \int dt = t-t_i =t }[/math]
The ambiguity in the sign of the above relation, due to the square root operation, is easily resolved in one dimension by inspection and more difficult to resolve in 3-D.
The velocity can change direction (signs) during the motion. In such cases it is best to separte the inegral into a part for one direction of the velocity and a second integral for the case of a negative velocity.
spring example
Consider the problem of a mass attached to a spring in 1-D.
- [math]\displaystyle{ F = -kx }[/math]
The potential is given by
- [math]\displaystyle{ U(x) = - \int F(x) dx = \frac{1}{2} k x^2 }[/math]
- [math]\displaystyle{ t = \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx = \int dt }[/math]
- [math]\displaystyle{ = \sqrt{\frac{m}{2}} \int_{x_0}^x\left (E-U(x) \right )^{-\frac{1}{2}} dx }[/math]
- [math]\displaystyle{ = \sqrt{\frac{m}{2}} \int_{x_0}^x\left (E-\frac{1}{2}kx^2 \right )^{-\frac{1}{2}} dx }[/math]
- [math]\displaystyle{ = \sqrt{\frac{mE}{2}} \int_{x_0}^x \left (1-\left ( x \sqrt{\frac{k}{2E}}\right ) ^2 \right )^{-\frac{1}{2}} dx }[/math]
let
- [math]\displaystyle{ \sin \theta = x \sqrt{\frac{k}{2E}} }[/math] and [math]\displaystyle{ \omega = \sqrt{\frac{k}{m}} }[/math]
- [math]\displaystyle{ \cos \theta d \theta = dx \sqrt{\frac{k}{2E}} }[/math] </math>
then
- [math]\displaystyle{ t = \sqrt{\frac{mE}{2}} \int_{x_0}^x \left (1-\sin^2 \theta \right )^{-\frac{1}{2}} dx }[/math]
- [math]\displaystyle{ = \sqrt{\frac{mE}{2}} \int_{x_0}^x \frac{dx}{ \cos \theta } }[/math]
- [math]\displaystyle{ = \frac{1}{\omega} \int_{\theta_0}^{\theta} d \theta }[/math]