Difference between revisions of "Forest UCM Energy Line1D"

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: <math>t = \sqrt{\frac{m}{2E}} \int_{x_0}^x \left (1-\sin^2 \theta \right )^{-\frac{1}{2}} dx </math>
 
: <math>t = \sqrt{\frac{m}{2E}} \int_{x_0}^x \left (1-\sin^2 \theta \right )^{-\frac{1}{2}} dx </math>
:: <math>= \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{d\theta}{ \cos \theta } \sqrt{\frac{2E}{k}}</math>
 
:: <math>= \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{dx}{ \cos \theta } </math>
 
 
:: <math>= \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{dx}{ \cos \theta } </math>
 
:: <math>= \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{dx}{ \cos \theta } </math>
 +
:: <math>= \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{\cos  \theta d\theta}{ \cos \theta } \sqrt{\frac{2E}{k}}</math>
 +
:: <math>= \sqrt{\frac{m}{k}} \int_{x_0}^x d\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 12:36, 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]T + U(x) =[/math] cosntant [math]\equiv E[/math]
[math]\Rightarrow T = E - U(x)[/math]
[math] \frac{1}{2} m \dot {x}^2 = E -U(x)[/math]
[math]\dot x = \pm \sqrt{\frac{2\left (E-U(x) \right )}{m}}[/math]
[math]\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] F = -kx[/math]

The potential is given by

[math]U(x) = - \int F(x) dx = \frac{1}{2} k x^2[/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-\frac{1}{2}kx^2 \right )^{-\frac{1}{2}} dx [/math]
[math] = \sqrt{\frac{m}{2E}} \int_{x_0}^x \left (1-\left ( x \sqrt{\frac{k}{2E}}\right ) ^2 \right )^{-\frac{1}{2}} dx [/math]

let

[math]\sin \theta = x \sqrt{\frac{k}{2E}}[/math] and [math] \omega = \sqrt{\frac{k}{m}}[/math]
[math]\cos \theta d \theta = dx \sqrt{\frac{k}{2E}}[/math]

then

[math]t = \sqrt{\frac{m}{2E}} \int_{x_0}^x \left (1-\sin^2 \theta \right )^{-\frac{1}{2}} dx [/math]
[math]= \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{dx}{ \cos \theta } [/math]
[math]= \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{\cos \theta d\theta}{ \cos \theta } \sqrt{\frac{2E}{k}}[/math]
[math]= \sqrt{\frac{m}{k}} \int_{x_0}^x d\theta[/math]
[math]= \frac{1}{\omega} \int_{\theta_0}^{\theta} d \theta[/math]

Forest_UCM_Energy#Energy_for_Linear_1-D_systems