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I'm studying Landau, Lifshitz - Mechanics. Could someone help me with this problem ? =)

Problem 2 (Page 27 3rd Edition) Determine the period of oscillation, as a function of the energy, when a particle of mass $m$ moves in fields for which the potential energy is

$(b)$ $U=-U_{0}/\cosh^{2}\alpha x$


The period is given by

$ T=4 \sqrt{\frac{m}{2}}\int\frac{dx}{\sqrt{E+\frac{U_{0}}{\cosh^{2}\alpha x}}}$

How can I evaluate this integral? I know that the answer is $T=(\pi/\alpha)\sqrt{2m/|E|}$

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closed as too localized by Manishearth Jan 26 '13 at 8:26

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I removed my last comment, because it was incomplete. You should start with two things: 1) determine the domain of integration and 2) rewrite the integrand in the most compact way possible (i.e. use only dimensionless parameters). One way to see that you should not brute-force the integral is that $U_0$ does not appear in the final answer. – Vibert Jan 24 '13 at 23:56
Welcome to Physics! Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) – Manishearth Jan 26 '13 at 8:26

Some idea to evaluate the integral: $$\int\frac{dx}{\sqrt{E+\frac{U_{0}}{\cosh^{2}\alpha x}}}=\int \cosh\alpha x\frac{dx}{\sqrt{E \cosh^{2}\alpha x+U_{0}}} = \int \frac{d(\sinh\alpha x)/\alpha}{\sqrt{(E+U_0) + E \sinh^{2}\alpha }}$$ The last equality use the identity $\cosh^2 x - \sinh^2 x = 1$. Using $y=\sinh \alpha x$, you should then able see it as a standard integral of the form $\int dy/\sqrt{a^2+y^2} = \sinh^{-1}(y/a)$

As what Vibert said, to get the correct answer, you need the correct domain of integration

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Hints following Vibert's advices:

  1. From energy conservation $$\frac{1}{2}m\dot{x}^2 ~=~ E-U,$$ with symmetric potential $$U ~=~ - \frac{U_0}{c^2},\qquad c~:=~\cosh(\alpha x), \qquad 0<-E<U_0,$$ one gets the quarter period $$\frac{T}{4}~=~ \sqrt{\frac{m}{2}} \int_0^{x_1}\frac{dx}{\sqrt{E-U}},$$ where $x_1>0$ is the upper turning point determined by the condition $E=U(x_1)$.

  2. Prove that $$\int_0^{x_1}\frac{dx}{\sqrt{E-U}}~=~\frac{1}{\alpha\sqrt{|E|}}\int_0^{\sqrt{a}}\frac{ds}{\sqrt{(a-s^2)}}, $$ where $$ s~:=~\sinh(\alpha x),\qquad ~a:=~ -U_0/E-1~>~0.$$

  3. Show that the last integral does not depend on $a>0$. Choose $a=1$, and perform a substitution $s=\sin(t)$.

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Remark: By a symmetric potential is meant an even potential $U(x)=U(-x)$. – Qmechanic Jan 27 '13 at 11:49

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