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Position-Dependent Potential

  1. Sep 17, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moving in one dimension has potential energy [itex]U(x)=U_0[2(\frac{x}{a})^2-(\frac{x}{a})^4][/itex], where [itex]U_0[/itex] and a are positive constants.

    What is the function x(t) for the position of the particle when it starts at the origin with a velocity in the positive direction equal to the escape velocity?

    2. Relevant equations
    The escape velocity is [itex]$\sqrt{\frac{2U_0}{m}}$[/itex] which escapes at x=a.


    3. The attempt at a solution
    I was able to determine the function for the force easily, but I ran into many problems with logs when trying to integrate it twice and ended up with logarithms that could not be paired up so that I could actually solve for x.

    Having a fair amount of trouble getting the latex code to work on these forums properly... Looks like it's because of Chrome.
     
    Last edited: Sep 17, 2010
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  3. Sep 17, 2010 #2

    vela

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    What exactly is the question?
     
  4. Sep 17, 2010 #3
    Sorry, I had been having so much trouble when trying to get it done with Chrome (restarting the post multiple times) that I forgot to include that part. It's there now.
     
  5. Sep 17, 2010 #4

    vela

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    Could you show us what you tried and got? It's hard to see where the problem is if all you do is give a vague description of what you tried.
     
  6. Sep 17, 2010 #5
    [itex]$(\frac{1}{a^2})(x \ln (x)-x-\frac{1}{2}(x\ln(x^2-a^2)+a\ln(x^2-a^2)-2x))+C$[/itex] all over ma^2 I believe. It's difficult getting this stuff to show up correctly on the forums :s But basically something that I can't really continue with.
     
  7. Sep 17, 2010 #6

    gabbagabbahey

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    What is this expression supposed to represent (position, velocity, force, the numer of apples in a carton of width [itex]x[/itex])? And how did you get it?

    Posting an expression without any work shown or description of what the expression is and where it came from doesn't help us to see where you are going wrong.

    In any case, can you think of a conserved quantity that will help you get an expression for velocity? Can you determine position as a function of time from that expression?
     
  8. Sep 17, 2010 #7
    I figured out what I was doing wrong. The correct method was the second I tried, but I had represented something in a way that prevented me from seeing the ease with which I could have solved it. Basically the process would use E=1/2 mv^2 + U(r), which I can simplify so that I get a first order differential equation that requires some simple partial fractions to integrate. Thanks though.
     
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