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]\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.

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]\displaystyle{ U(x) = -mgx }[/math]

Note

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

The initial total energy is

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

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

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

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

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

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{m}{2E}} \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]

then

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

[math]\displaystyle{ = \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{dx}{ \cos \theta } }[/math]

[math]\displaystyle{ = \sqrt{\frac{m}{2E}} \int_{x_0}^x \frac{\cos \theta d\theta}{ \cos \theta } \sqrt{\frac{2E}{k}} }[/math]

[math]\displaystyle{ = \sqrt{\frac{m}{k}} \int_{x_0}^x d\theta }[/math]

[math]\displaystyle{ = \frac{1}{\omega} \int_{\theta_0}^{\theta} d \theta }[/math]

[math]\displaystyle{ \theta = \omega t + \theta_0 }[/math]

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

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

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

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

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

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

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

Forest_UCM_Energy#Energy_for_Linear_1-D_systems