Period of Anharmonic Oscillator

[Vuldoraq]

Homework Statement

Hi,

For a certain oscillator the net force on a body, with mass m, is given by F=-cx^3.

One quarter of a period is the time taken for the body to move from x=0 to x=A (where A is the amplitude of the oscillation). Calculate this time and hence the period.

Homework Equations

U(x)=(cx^4)/4, where U(x) represents the potential energy of the body.

The Attempt at a Solution

In order to solve this I used a homogeneity of units argument as follows,

Units of time are (s)

Units of potential energy are (kg*m^2)/(s^2)

In order to get from the potential energy units to the time units,

(s)=\sqrt{((kg*m^2)/(s^2))}

in terms of the above equations this is,

\sqrt{(m*x^2/U(x))}=\sqrt{((4*m*x^2)/(c*x^4))}

let x=A and the equation =T/4,

T/4=\sqrt{((4*m)/(cA^2))}

hence, T=4*\sqrt{((4*m)/(cA^2))}

However this is incorrect, my answer is wrong by a multiplicative factor. Please could someone show me where I have gone wrong?

 

[malawi_glenn]

You can not use dimensional analysis to find multiplicative factors.

There is no closed form analytical solution to this question (?)

Use conservation of energy $\dfrac{m(x')^2}{2} + U(x) = U(A)$ where $m$ is the mass of a weight attached to the oscillator. With some algebra, you should end up with this $\dfrac{d x}{ dt } =\sqrt{ \dfrac{c(A^4 - x^4) }{ 2m }}$ which is a separable differential equation. Integrate t from 0 to T/4 and x from 0 to A.

 

[Delta2]

Well the ODE as a function of time doesn't have closed form solution indeed. Instead the ODE as a function of x (and velocity v as unknown function) has a closed form solution $$m\frac{dv}{dt}=-cx^3\Rightarrow m\frac{dv}{dx}\frac{dx}{dt}=-cx^3\Rightarrow m\frac{dv}{dx}v=-cx^3$$

 

[Delta2]

Nvm what I said in post #3, doesn't help us find the period, seems one way to approximately solve the ode of post #2 is to find a Taylor series approximation of the integral $$\frac{T}{4}=\sqrt{2m}\int_0^A \frac{dx}{\sqrt{c(A^4-x^4)}}$$. Wolfram gives the first two terms of this approximation as $$\frac{x^4}{c^{1/2}A^2}+\frac{x^6}{10c^{3/2}A^6}$$

 

[malawi_glenn]

Taylor series approximation of the integral

Don't you mean "Tayloer series of the integrand"?
The integral is not a function of x...

 

[Delta2]

Don't you mean "Tayloer series of the integrand"?
The integral is not a function of x...

Yes well, I mean the respective indefinite integral.

 

[malawi_glenn]

and a Taylor series always have a point which you expand around, I guess you took x=0.

 

[Delta2]

here is what I get from wolfram, judge for yourself
https://www.wolframalpha.com/input?i=integral+dx/sqrt(aA^4-ax^4)

 

[malawi_glenn]

here is what I get from wolfram, judge for yourself

I am not questioning whether it is correct or not, I am just nitpicking about nomenclature ;)

 

[TSny]

Since this is a very old thread, I think it would OK to continue on from @drmalawi 's setup in post #2 and finish it:

$T = \large \frac{4}{A} \sqrt{\frac{m}{2c}} \int_0^1 \frac{du}{\sqrt{1-u^4}}$

Mathematica expresses the value of the integral in terms of the gamma function:

$\large \int_0^1 \frac{du}{\sqrt{1-u^4}} = \sqrt \pi \frac{\Gamma(5/4)}{\Gamma(3/4)}$ $\approx 1.311$

Then $T \approx 3.708 \large \frac{\sqrt{m/c}}{A}$

 

[malawi_glenn]

Yeah I was hesitaded not to give away too much.

One can also integrate numerically from 0 to 1-ε where ε is a small positive number

Or, just plug it into wolfram alpha