- #1
Istiak
- 158
- 10
- 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?