Forest UCM Energy Line1D

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 $\equiv E$

$\Rightarrow T = E - U(x)$

$\frac{1}{2} m \dot {x}^2 = E -U(x)$

$\dot x = \pm \sqrt{\frac{2\left (E-U(x) \right )}{m}}$

$\int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx = \int dt = t-t_i =t$

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

$U(x) = - \int F(x) dx = \frac{1}{2} k x^2$

$t = \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx = \int dt$

$= \sqrt{\frac{m}{2}} \int_{x_0}^x\left (E-U(x) \right )^{-\frac{1}{2}} dx$

$= \sqrt{\frac{m}{2}} \int_{x_0}^x\left (E-\frac{1}{2}kx^2 \right )^{-\frac{1}{2}} dx$

$= \sqrt{\frac{m}{2}} \int_{x_0}^x E \left (1-\left ( x \sqrt{\frac{k}{2E}}\right ) ^2 \right )^{-\frac{1}{2}} dx$

let

$\sin \theta = x \sqrt{\frac{k}{2E}}$ and $\omega = \sqrt{\frac{k}{m}}$

then

t = \frac{1}{\omega} \int_{\theta_0}^{\theta} d \theta

Forest_UCM_Energy#Energy_for_Linear_1-D_systems