Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
Find the values of A, B, and C such that the action is a minimum
Reply to thread
Message
[QUOTE="Istiak, post: 6537960, member: 690416"] [B]Homework Statement:[/B] 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. [B]Relevant Equations:[/B] 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? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
Find the values of A, B, and C such that the action is a minimum
Back
Top