# Period of Anharmonic Oscillator

• Vuldoraq
In summary, the homework statement is that 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).
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?

Last edited:
Delta2
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.

Steve4Physics and 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 $$m\frac{dv}{dt}=-cx^3\Rightarrow m\frac{dv}{dx}\frac{dx}{dt}=-cx^3\Rightarrow m\frac{dv}{dx}v=-cx^3$$

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 $$\frac{x^4}{c^{1/2}A^2}+\frac{x^6}{10c^{3/2}A^6}$$

Steve4Physics and malawi_glenn
Delta2 said:
Taylor series approximation of the integral

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

Delta2
drmalawi said:
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.

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

Delta2
Delta2 said:
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 ;)

Delta2
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 and Delta2
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

TSny

## 1. What is an anharmonic oscillator?

An anharmonic oscillator is a type of physical system that exhibits oscillatory motion, or a back-and-forth movement, that is not a perfect sinusoidal wave. This means that the restoring force acting on the system is not directly proportional to the displacement from equilibrium, unlike in a harmonic oscillator.

## 2. How is the period of an anharmonic oscillator different from a harmonic oscillator?

In a harmonic oscillator, the period is constant and independent of the amplitude of the oscillation. However, in an anharmonic oscillator, the period increases as the amplitude increases. This means that larger amplitudes result in longer periods of oscillation.

## 3. What is the significance of the period in an anharmonic oscillator?

The period of an anharmonic oscillator is an important parameter that determines the behavior of the system. As the period increases, the oscillation becomes slower and less regular. This can lead to interesting phenomena, such as chaos and bifurcations, in the system.

## 4. How is the period of an anharmonic oscillator affected by the potential energy function?

The potential energy function of an anharmonic oscillator plays a crucial role in determining the period of oscillation. Different types of potential energy functions result in different periods of oscillation, with some functions leading to longer or shorter periods compared to others.

## 5. Can the period of an anharmonic oscillator be measured experimentally?

Yes, the period of an anharmonic oscillator can be measured experimentally by observing the oscillatory motion of the system and recording the time it takes to complete one full cycle. This can then be compared to theoretical predictions to validate the behavior of the system.

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