Period of Anharmonic Oscillator

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).
  • #1
Vuldoraq
272
1

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



[tex]U(x)=(cx^4)/4[/tex], 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 [tex](s)[/tex]

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

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

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

in terms of the above equations this is,

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

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

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

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

However this is incorrect, my answer is wrong by a multiplicative factor. Please could someone show me where I have gone wrong?
 
Last edited:
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes Steve4Physics and Delta2
  • #3
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 $$m\frac{dv}{dt}=-cx^3\Rightarrow m\frac{dv}{dx}\frac{dx}{dt}=-cx^3\Rightarrow m\frac{dv}{dx}v=-cx^3$$
 
  • #4
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 $$\frac{x^4}{c^{1/2}A^2}+\frac{x^6}{10c^{3/2}A^6}$$
 
  • Like
Likes Steve4Physics and malawi_glenn
  • #5
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...
 
  • Like
Likes Delta2
  • #6
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.
 
  • #7
and a Taylor series always have a point which you expand around, I guess you took x=0.
 
  • Like
Likes Delta2
  • #9
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 ;)
 
  • Love
Likes Delta2
  • #10
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}##
 
  • Love
  • Like
Likes malawi_glenn and Delta2
  • #11
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 :oldsmile:
1657206972319.png
 
  • Like
Likes TSny

Related to Period of Anharmonic Oscillator

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.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
734
  • Introductory Physics Homework Help
Replies
2
Views
522
  • Introductory Physics Homework Help
Replies
14
Views
559
  • Introductory Physics Homework Help
Replies
13
Views
736
  • Introductory Physics Homework Help
Replies
7
Views
758
  • Introductory Physics Homework Help
Replies
16
Views
495
Replies
31
Views
810
  • Introductory Physics Homework Help
Replies
6
Views
413
  • Introductory Physics Homework Help
Replies
4
Views
920
  • Introductory Physics Homework Help
Replies
9
Views
827
Back
Top