Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Stuck on the last step of a problem. Help please.

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data
    This is a problem from L.D. Landau and E.M. Lifgarbagez "Mechanics (3 ed): Course of Theoretical Physics, Volume 1" [itex]\S 11.[/itex] Motion in one dimension:
    Problem 2b. Determine the period of oscillation, as a function of energy, when a particle of mass [itex]m[/itex] moves in the firld for which the potential energy is
    [itex]U=-U_0/\cosh^2\alpha x,\quad -U_0<E<0[/itex]


    2. Relevant equations
    [itex]T(E)=\sqrt{2m}\int_{x_1(E)}^{x_2(E)}\frac{dx}{ \sqrt{E-U(x)} }[/itex]


    3. The attempt at a solution
    I rewrite the integral, due to symmetry, as
    [itex]T(E)=2\sqrt{2m}\int_{0}^{x_0}\frac{dx}{ \sqrt{E-U(x)} }[/itex]
    and find the [itex]x_0[/itex] that satisfies [itex]E=U(x)[/itex]:
    [itex]x_0=\frac{1}{\alpha}\cosh^{-1}(\sqrt{-U_0/E})[/itex]

    *For sake of brevity I wont show some intermediate steps, unless it is requested of me.*

    Simplifying: *Note:[itex]\sech\psi=\sqrt{-U_0/E}\sech\alpha x[/itex]
    [itex]\frac{dx}{\sqrt{E+U_0\sech^2\alpha x}}=\frac{1}{\alpha\sqrt{-U_0}}\frac{d\psi}{\sqrt{1+\sech^2\psi}}=\frac{d\psi}{\alpha\sqrt{-U_0}\tanh\psi}[/itex]

    Integration leads to:
    [itex] T=\frac{2}{\alpha}\sqrt{-2m/U_0}[\ln|\sinh\psi|]_0^{\psi_0}=2\sqrt{2m/E}[\ln|\sqrt{-U_0/E}\sinh\alpha x|]_0^{x_0} [/itex]

    Which gives me something messy:
    [itex] \frac{2}{\alpha}\sqrt{-2m/U_0}\ln[\sinh(\frac{1}{\alpha}\cosh^{-1}(\sqrt{-U_0/E}))] [/itex]

    Which I have no idea how to simplify. I know the answer is supposed to be [itex] \frac{\pi}{\alpha}\sqrt{2m/E} [/itex] which my answer somewhat resembles, but I dont know how to simplify that Natural log. I tried writing sinh in terms of exponents, and I tried using Mechanical Similarity (Landau made a vauge comment alluding to MS: right before stating the answer to part (b), to which he shows no work) he states:
    "The dependence of [itex] T [/itex] on [itex] E [/itex] is in accordance with the law of mechanical similarity (10.2), (10.3)." I tried using the result from part (a), which i easliy enough found using beta functions:
    [itex] U=A|x|^n\rightarrow T=\frac{2}{n}\sqrt{2\pi m/E}\cdot(E/A)^{1/n}\frac{\Gamma(1/n)}{\Gamma(1/n+1/2)} [/itex]. But i don't really understand MS so i couldnt use it.

    BIO: Im a highschool student who is self taught, and i have nobody to ask for guidence on problems, any help is appreciated...

    PS: is there any way to just type in LaTeX language, rather than use the HTML like headers?
     
    Last edited: Aug 11, 2011
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted