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

[LoveForGauss]

Homework Statement

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

Homework Equations

$T(E)=\sqrt{2m}\int_{x_1(E)}^{x_2(E)}\frac{dx}{ \sqrt{E-U(x)} }$

The Attempt at a Solution

I rewrite the integral, due to symmetry, as
$T(E)=2\sqrt{2m}\int_{0}^{x_0}\frac{dx}{ \sqrt{E-U(x)} }$
and find the $x_0$ that satisfies $E=U(x)$:
$x_0=\frac{1}{\alpha}\cosh^{-1}(\sqrt{-U_0/E})$

*For sake of brevity I won't show some intermediate steps, unless it is requested of me.*

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

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

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

Which I have no idea how to simplify. I know the answer is supposed to be $\frac{\pi}{\alpha}\sqrt{2m/E}$ which my answer somewhat resembles, but I don't 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 $T$ on $E$ 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:
$U=A|x|^n\rightarrow T=\frac{2}{n}\sqrt{2\pi m/E}\cdot(E/A)^{1/n}\frac{\Gamma(1/n)}{\Gamma(1/n+1/2)}$. But i don't really understand MS so i couldn't use it.

BIO: I am a high school 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?

 

[mark726]

Hello there! I am a scientist and I would be happy to help you with this problem. It seems like you have made a good attempt at solving it so far. Let's take a look at your solution and see if we can simplify it.

First, let's rewrite the integral as you did, but using the variable u instead of x. This will make it easier to see the simplifications later on.

T(E) = 2√(2m) ∫0x0 du/ √(E-U(u))

Next, we can use the substitution u = (1/α)arccosh(√(-U0/E)) to simplify the integral. This will also change the limits of integration to 0 and u0.

T(E) = 2√(2m) ∫0u0 du/ √(E-U(u)) = 2√(2m) ∫0u0 du/ √(E+U0sech^2(αu))

Now, using the identity sech^2(x) = 1/cosh^2(x), we can rewrite the denominator as √(E+U0/cosh^2(αu)). This will allow us to use the trigonometric substitution cosh(αu) = sec(θ), which will simplify the integral further.

T(E) = 2√(2m) ∫0u0 du/ √(E+U0/cosh^2(αu)) = 2√(2m) ∫0θ0 dθ/ √(E+U0sec^2(θ))

Now, using the identity sec^2(x) = 1+tan^2(x), we can rewrite the denominator as √(E+U0(1+tan^2(θ))). This will simplify the integral even further.

T(E) = 2√(2m) ∫0θ0 dθ/ √(E+U0(1+tan^2(θ))) = 2√(2m) ∫0θ0 dθ/ √(E+U0(1+tan^2(θ))) = 2√(2m) ∫0θ0 dθ/ √(E+U0(1+tan^2(θ)))

Using the