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I am trying to solve a problem in which I have to prove that for a particule 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

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

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.

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.

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

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

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

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

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.

I am trying to solve a problem in which I have to prove that for a particule 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

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

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.

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

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

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}{dx^2}{x}+\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.

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

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

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

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

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.