201902200839 Exercise 5 Chapter 2

A particle is subjected to the potential $V(x)=-Fx$ , where $F$ is a constant. The particle travels from $x=0$ to $x=a$ in a time interval $t_0$ . Assume the motion of the particle can be expressed in the form $x(t)=A+Bt+Ct^2$ . Find the values of $A$ , $B$ , and $C$ such that the action is a minimum.

Solution.

The solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

1D-case:

$\mathcal{L}=T-V=\displaystyle{\frac{1}{2}}m\dot{x}^2+Fx$ .

Euler-Lagrange (E-L) equation:

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{x}}}\bigg) =\frac{\partial \mathcal{L}}{\partial x}$

gives the path over which the action is stationary.

That is,

$\begin{aligned} \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}( m\dot{x}) & = F \\ \Rightarrow \ddot{x} & =\displaystyle{\frac{F}{m}}\\ \end{aligned}$ .