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Period of an Oscillating Particle

  1. Jun 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle oscillates with amplitude A in a one-dimensional potential that is symmetric about x=0. Meaning U(x)=U(-x)

    First find velocity at displacement x in terms of U(A), U(x), and m.

    Then show that the period is given by ##4\sqrt{\frac{m}{2U(A)}}\int_0^A \frac{1}{1-\frac{U(x)}{U(A)}}dx##
    (hint: ##dt=\frac{dx}{v}##)
    2. Relevant equations

    Total energy = ##U(x) + \frac{1}{2}mv^2##
    Period = time it takes to complete an oscillation

    I found velocity by noting that at x=A, all the energy is potential, so total energy of the system (which doesn't change) is U(A), so we can set up the equation:

    ##U(A)=U(x)+\frac{1}{2}mv^2## and solving for v we get:

    ##v=x'=\sqrt{\frac{2(U(A)-U(x))}{m}}##.

    3. The attempt at a solution

    Now I was thinking we could multiply the velocity at x=0 by Δt, then do v(0)Δt+v(Δt)Δt, and then v(0)Δt+v(Δt)Δt+v(2Δt)Δt and so on and so forth... this will help us find how far the particle has traveled at any time t. But obviously the right way to do the above process is by integrating our ##v## equation with respect to time, which will give us distance traveled over that time which will let us know how long it takes to travel A distance (1/4 of an oscillation).

    ##\int_0^t v dt = \int_0^t \sqrt{\frac{2(U(A)-U(x))}{m}}dt##

    But we know ##\frac{dx}{dt}=v## so we can write ##dt=\frac{dx}{v}##...


    umm... I'm stuck...
     
    Last edited: Jun 11, 2013
  2. jcsd
  3. Jun 11, 2013 #2
    You have a differential equation in the form ## \frac {dx} {dt} = f(x) ##. How do you integrate it?

    If unclear, try these simple examples: ## f(x) = a + x ##, ## f(x) = x^2 ##.
     
  4. Jun 11, 2013 #3
    Since the input into our velocity equation we found is x not t, I should edit:

    ... to say:

    See how as we make Δt smaller and smaller, the above becomes a better and better approximation of displacement x as a function of time? But I don't know how to turn that method into an integral?





    As for integrating ##\frac{dx}{dt}=f(x)##, if we do ##\int \frac{dx}{dt}dt=\int f(x)dt##, we get ##x(t)=\int f(x)dt##... (don't know how we could evaluate that integral.)

    Whereas if we do it ##\int \frac{dx}{dt}dx = \int f(x)dx## I don't know what the left side would be and we'd have to evaluate the right side.

    How are either of these useful to me?
     
  5. Jun 11, 2013 #4
    Have you studied differential equations? Separation of variables sound familiar?
     
  6. Jun 11, 2013 #5
    ##\frac{dx}{dt}=f(x)##
    ##\frac{1}{f(x)}dx=dt##
    ##\int f(x)^{-1} dx = \int 1 dt=t##

    is that right?

    That would give us time as a function of velocity which can give us velocity as a function of time?

    Then we can integrate that with respect to t to find displacement as a function of time?



    thanks for the quick response btw
     
    Last edited: Jun 11, 2013
  7. Jun 11, 2013 #6
    What would that have to do with velocity? Is x velocity?
     
  8. Jun 11, 2013 #7
    oh whoops the integral of velocity is displacement so we would have x(t). So we could solve x(?)=A and then multiply it by 4 to get the period?
     
  9. Jun 11, 2013 #8
    You will have t(x) for sure. Whether you can convert that to x(t) depends on U(x).

    But if you just need to get the period, you don't need x(t). t(x) is exactly what you need in this case.
     
  10. Jun 11, 2013 #9
    I'm afraid I don't get what it means to take ##\frac{dx}{dt}=f(x)## and turn it into ##\frac{1}{f(x)}dx=dt.

    And thus I don't get what it means to then take the integral of both sides of that...


    can you help me make sense of what those procedures mean?
     
  11. Jun 11, 2013 #10
    In #2, I suggested a couple of examples. Try them.
     
  12. Jun 11, 2013 #11
    If ##\frac{dx}{dt}=a+x## then ##\frac{1}{a+x}dx=dt## (whatever that means), so ##\int \frac{1}{a+x} dx = \int dt = t##. So ##ln(a+x)=t##.

    If ##\frac{dx}{dt}=x^2## then ##\frac{1}{x^2}dx=dt## (once again idk what exactly i'm doing there), so ##\int \frac{1}{x^2}dx = \int dt##. So ##-\frac{1}{x}=t##.



    But I still don't get how to solve the original problem...
     
  13. Jun 11, 2013 #12
    What is a "period" and how much of it is spent when the particle moves from 0 to A?
     
  14. Jun 11, 2013 #13
    "Period" is the time it takes for the particle to make one cycle. The particle traveling from 0 to A takes 1/4 of the time it would take to travel the whole period.

    But I don't see how we can find the time it takes the particle to travel from 0 to A? We have this equation that gives us time:

    ... but I don't get what the left hand side is a function of? Like how would I plug in A to the LHS without having to evaluate that integral?

    Apparently the answer might be to make it a definite integral we would have to make both sides of ##\int \sqrt{\frac{m}{2(U(A)-U(x))}}dx = \int dt## definite integrals and I don't understand what the boundaries on both sides would be... Would we have to make them match up somehow? But the right hand side is a function of x and the LHS is a function of t...?



    thanks for sticking with me
     
  15. Jun 11, 2013 #14
    What you really have is ## \frac {dt} {dx} = f(x) ##. Which pretty much defines ## t(x) = F(x) ##, where ## F(x) = \int f(x) dx ##. You could also use definite integrals, then you would have, by the fundamental theorem of calculus, ## t(b) - t(a) = \int_a^b f(x) dx ##.
     
  16. Jun 11, 2013 #15
    Sorry, how do we have ## \frac {dt} {dx} = f(x) ##?
     
  17. Jun 12, 2013 #16
    Poor choice of symbols o n my part. Originally I wrote ## \frac {dx} {dt} = f(x) ##, which is equivalent to ## \frac {dt} {dx} = g(x) = \frac 1 {fx(x) } ##.

    Then ## t(x) = \int g(x) dx ##, and ## t(b) - t(a) = \int_a^b g(x) dx ##.
     
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