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.

[math]T + U(x) =[/math] constant [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.

Free fall

Consider a rock dropped at t=0 from a tower of height h.

The potential energy stored in the rock at any instant is given by

[math]U(x) = -mgx[/math]

Note

The potential is highest at x=0 and becomes negative as x increases

The initial total energy is

[math]E_{tot} = T + U = 0 -0 = 0[/math]

[math]t = \int \pm \sqrt{\frac{m}{2\left (E-U(x) \right )}} dx [/math]

[math] = \int \pm \sqrt{\frac{m}{2\left (0-(-mgx) \right )}} dx [/math]

[math] = \int \pm \sqrt{\frac{1}{2gx}} dx = \int \pm (2gx)^{-\frac{1}{2}}dx [/math]

[math] = \pm (2g)^{-\frac{1}{2}} 2\sqrt x = \sqrt{\frac{2x}{g} } [/math]

or

[math]x = \frac{1}{2} gt^2[/math]

spring example (problem 2.8)

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]

[math] \theta = \omega t + \theta_0 [/math]

[math] \sin \theta = \sin {\left (\omega t + \theta_0 \right )}[/math]

[math]\sqrt{\frac{2E}{m}} \sin \theta = \sqrt{\frac{2E}{m}} \sin {\left (\omega t + \theta_0 \right )}[/math]

[math]x = \sqrt{\frac{2E}{m}} \sin {\left (\omega t + \theta_0 \right )}[/math]

[math]x = A \sin {\left (\omega t + \theta_0 \right )}[/math]

[math]A = \sqrt{\frac{2E}{m}}[/math] = amplitude of oscillating motion

[math]U(x) = \frac{1}{2} k x^2 = \frac{1}{2} kA^2 \sin^2 {\left (\omega t + \theta_0 \right )}[/math]

[math]E = T + U(x) = \frac{1}{2}kA^2[/math]

Forest_UCM_Energy#Energy_for_Linear_1-D_systems