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I am trying to solve a problem in which I have to prove that for a particle with mass $m$ in a potential $U=A|x|^n$, has a period of oscillation given by

$$\tau =\frac{2}{n} \sqrt\frac{2\pi m}{E} \left ( \frac{E}{A}\right)^{1/n} \frac{\Gamma(1/n)}{\Gamma(1/2+1/n)}$$

What I have tried is to write down the second law by using $F = -\frac{d}{dx}U$, and I got the form

$$\frac{d^2 x}{dt^2}+\frac{A}{m}n x|x|^{n-2} = 0 $$

At this point I am thinking in taking cases, for example when $n$ is an even positive integer with $n>2$ we can solve this, I used the Laplace transformation knowing the transformation

$$ L\{x^{n-1}\} = \Gamma(n)s^{-n} $$

also it's inverse have a similar form, but as a result I got a non-periodic function

$$x(t) = x(0)-\frac{A}{m}n\Gamma(n) \frac{t^{n+1}}{\Gamma(n+2)} $$

and I am kind of confused because I wasn't expecting to get this, I would like to know if this approach is correct or someone knows a betters way to tackle this problem.

Adán González
  • 33

2 Answers2

5

Assume the total energy of the particle is $E$, a conservative quantity: $$ E = \frac{1}{2}m v^2 + A|x|^n. $$

We can obtain the velocity as function of position: $$ v = \pm \sqrt{\frac{2(E-A|x|^n)}{m}} $$

Fot the periodic motion, the particle will run back and forth in between two turning points,

$$ x_r = \pm \left( \frac{E}{A}\right)^{1/n}. $$

The integral for the period $\tau$:

$$ \tau = 4 \int_0^{x_r} \frac{dx}{v(x)} = 4 \int_0^{x_r} \frac{\sqrt{m} dx}{\sqrt{2(E-A x^n)}} $$ The integral is over postive $x$, thus the absolute symbole may be removed.

Simplify the integrand: $$ \tau = 4 \frac{\sqrt{m}}{\sqrt{2E}} \int_0^{x_r} \frac{dx}{\sqrt{1-\frac{A}{E} x^n}} = 2 \frac{\sqrt{2m}}{\sqrt{E}} \left(\frac{E}{A}\right)^{1/n} \int_0^{1} \frac{d\xi}{\sqrt{1-\xi^n}} $$

WHere $\xi = \left(\frac{A}{E}\right)^{1/n} x$. It should have little problem to go from the last equation to the answer.

ytlu
  • 4,354
2

I think that you can obtain analytical solution for the period $~T$

follow the solution from @ytlu

with A and m equal 1 and omit the abs() function

the energy E is:

$$E=\frac{1}{2}\,v^2+x^n$$ with: $$E_0(v=0~,x=x_0)=E(v~,x)$$ you obtain $$v=\sqrt{2(\,x_0^n-x^n)}$$ thus for $~v=0~,x_T=x_0$ and the period T is: $$\frac{dx}{dt}=v(x)$$ $\Rightarrow$ $$T=4\,\int_{0}^{x_0}\left(\frac{1}{v(x)}\right)\,dx\tag 1$$

you can choose the initial condition arbitrary $~x_0=1~$

the evaluation of equation (1) : \begin{align*} &n=2 \qquad T=\sqrt(2)\,\pi\\ &n=3 \qquad T=\frac{2}{3}\sqrt{2}\,\mathcal{B}\left(\frac{1}{3}~,\frac{1}{2}\right)\\ &n=4 \qquad T=\frac{1}{2}\sqrt{2}\,\mathcal{B}\left(\frac{1}{4}~,\frac{1}{2}\right)\\ &n=5 \qquad T=\frac{2}{5}\sqrt{2}\,\mathcal{B}\left(\frac{1}{5}~,\frac{1}{2}\right)\\ &n=6 \qquad T=\frac{2}{6}\sqrt{2}\,\mathcal{B}\left(\frac{1}{6}~,\frac{1}{2}\right)\\ \end{align*} from here you can deduce that \begin{align*} &\text{if $n~$ is even}\qquad, T=\frac{1}{2}\sqrt{2}\,\mathcal{B}\left(\frac{1}{n}~,\frac{1}{2}\right)\\ &\text{if $n~$ is odd}\qquad, T=\frac{2}{n}\sqrt{2}\,\mathcal{B}\left(\frac{1}{n}~,\frac{1}{2}\right)\\ &\text{where}\\ &\mathcal{B}(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \end{align*}

Eli
  • 13,829