Find the values of A, B, and C such that the action is a minimum

AI Thread Summary
The discussion focuses on finding the constants A, B, and C in the motion equation of a particle subjected to a potential V(x) = -Fx. Participants emphasize the need to minimize the action rather than simply integrating the equations of motion. There is confusion regarding the treatment of A, B, and C as functions of time instead of constants. The correct approach involves computing the action integral and applying the principles of Lagrangian mechanics. Ultimately, the initial attempts to solve the problem are criticized for not adhering to the requirements stated in the question.
Istiak
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Homework Statement
A particle is subjected to the potential V (x) = −F x, where F is a constant. The
particle travels from x = 0 to x = a in a time interval t0 . Assume the motion of the
particle can be expressed in the form ##x(t) = A + B t + C t^2## . Find the values of A, B,
and C such that the action is a minimum.
Relevant Equations
Lagrangian
> A particle is subjected to the potential V (x) = −F x, where F is a constant. The
particle travels from x = 0 to x = a in a time interval t0 . Assume the motion of the
particle can be expressed in the form ##x(t) = A + B t + C t^2## . Find the values of A, B,
and C such that the action is a minimum.

I was thinking it can solved using Lagrangian rather than Hamilton. There's no frictional force.

$$L=\frac{1}{2}m\dot{x}^2+Fx$$
$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}=0$$
$$m\ddot{x}=F$$
$$\ddot{x}=\frac{F}{m}$$
Differentiate ##x(t)## twice. $$2C=\frac{F}{m}=>C=\frac{F}{2m}$$

For finding B I was thinking to integrate ##\ddot{x}## once. $$\dot{x}=\int \ddot{x} \mathrm dt$$
$$=\ddot{x}t$$
initial position is 0 so, not writing constant.

$$\dot{x}=\frac{F}{m}$$
Differentiate ##x(t)## once.
$$B+2Ct=\frac{F}{m}$$
$$B=\frac{F}{m}-\frac{2Ft}{2m}$$
$$=-\frac{Ft}{2m}$$

Again, going to integrate ##\ddot{x}## twice.
$$x=\int \int \ddot{x} dt dt$$
$$=\frac{\ddot{x}t^2}{2}$$

initial velocity and initial position is 0.

$$x=\frac{Ft^2}{2m}$$
$$A+Bt+Ct^2=\frac{Ft^2}{2m}$$
$$A=\frac{Ft^2+Ft-F}{2m}$$

According to my, I think that C is the minimum (I think B is cause, B is negative; negative is less than positive). And, A is maximum. Did I do any mistake?
 
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Where to begin?

First, you were supposed to compute the action and minimise it.

Everything you did from integrating ##\ddot x## got pretty wild. You definitely cannot integrate ##\ddot x## as though it were constant.

##A, B, C## were supposed to be constants, not functions of ##t##.
 
PeroK said:
Where to begin?

First, you were supposed to compute the action and minimise it.

Everything you did from integrating ##\ddot x## got pretty wild. You definitely cannot integrate ##\ddot x## as though it were constant.

##A, B, C## were supposed to be constants, not functions of ##t##.
Umm, I had found ##F=m\ddot{x}## 🤔. couldn't get you... then started differentiating ##x## function.
 
Istiakshovon said:
Umm, I had found ##F=m\ddot{x}## 🤔. couldn't get you... then started differentiating ##x## function.
You ignored most things in the question:

It asked you to minimise the action; it told you the particle moved from ##0## to ##a## in time ##t_0##; it gave you the equation of the trajectory.

You didn't do the problem that was asked.
 
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