- #1

Istiak

- 158

- 12

- 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?

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?