Difference between revisions of "Forest UCM Energy Line1D"

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: <math>t = \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx = \int dt  </math>
 
: <math>t = \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx = \int dt  </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-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>
  
  
  
 
[[Forest_UCM_Energy#Energy_for_Linear_1-D_systems]]
 
[[Forest_UCM_Energy#Energy_for_Linear_1-D_systems]]

Revision as of 12:20, 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.

T+U(x)= cosntant E
T=EU(x)
12m˙x2=EU(x)
˙x=±2(EU(x))m
±m2(EU(x))dx=dt=tti=t

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.

F=kx

The potential is given by

U(x)=F(x)dx=12kx2
t=±m2(EU(x))dx=dt
=m2xx0(EU(x))12dx
=m2xx0(E12kx2)12dx


Forest_UCM_Energy#Energy_for_Linear_1-D_systems