Anharmonic Oscillation

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Homework Statement



Assume that the potential is symmetric with respect to zero and the system has amplitude ##a## suppose that ##V(x)=x^4## and the mass of the particle is ##m=1##. Write a java function that calculates the period of the oscillator for given amplitude ##a## using Gaussian quadrature with ##N=20## points.

Homework Equations



##E = \frac12 m(\frac{dx}{dt})^2+V(x)##
##T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}.##

The Attempt at a Solution



I don't have much experience with numerical methods so I am having difficulty understanding the question and how to proceed.

Is the question asking instead of using ##T##, i.e the period given by : ##T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}##, we instead must use the Gaussian quadrature of ##T##? Also, will ##T=\sqrt{8m}\int^a_0\frac{dx}{\sqrt{V(a)-V(x)}}## work for a nonquadratic function ##V(x)##?
 

Answers and Replies

  • #2
stevendaryl
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The method of Gaussian quadratures is (in my opinion) a really difficult way to efficiently compute integrals numerically. I think it's too complicated to explain in a message. The basic idea is simple enough: You can approximate an integral

[itex]\int_{a}^b f(x) dx = \sum_{i=1}^N w_i f(x_i)[/itex] where [itex]x_1, x_2, ...[/itex] are points in the interval [itex][a,b][/itex] and [itex]w_i[/itex] is a weighting function. The simplest approach is the "rectangle rule", which uses [itex]w_i = \frac{b-a}{N}[/itex] and [itex]x_i = a + i \cdot \frac{b-a}{N}[/itex]. Gaussian quadrature is a way to get a much more accurate approximation by choosing [itex]w_i[/itex] and [itex]x_i[/itex] in a more sophisticated way, but it's beyond me to explain how to do it in a post.

As to the second question, if a particle goes from [itex]a[/itex] to [itex]b[/itex] and back, then the time taken for a full period is twice the time to go from [itex]a[/itex] to [itex]b[/itex]. To compute that time, you use:

[itex]T = 2 \int dt = \int_a^b \frac{dt}{dx} dx = \int_a^b \frac{1}{v} dx[/itex] where [itex]v = \frac{dx}{dt}[/itex]. To compute [itex]v[/itex], you use conservation of energy:

[itex]E = \frac{1}{2} m v^2 + V[/itex]
[itex]v = \sqrt{\frac{2}{m} (E - V)}[/itex]

So [itex]T = 2 \int_a^b \sqrt{\frac{m}{2(E-V)}} dx[/itex]

If the potential is symmetric about x=0, then you can just do half the integral and double it:

[itex]T = 4 \int_0^b \sqrt{\frac{m}{2(E-V)}} dx[/itex]

So it works no matter what the potential [itex]V[/itex] is--it doesn't have to be quadratic.
 
  • #3
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The method of Gaussian quadratures is (in my opinion) a really difficult way to efficiently compute integrals numerically. I think it's too complicated to explain in a message. The basic idea is simple enough: You can approximate an integral

[itex]\int_{a}^b f(x) dx = \sum_{i=1}^N w_i f(x_i)[/itex] where [itex]x_1, x_2, ...[/itex] are points in the interval [itex][a,b][/itex] and [itex]w_i[/itex] is a weighting function. The simplest approach is the "rectangle rule", which uses [itex]w_i = \frac{b-a}{N}[/itex] and [itex]x_i = a + i \cdot \frac{b-a}{N}[/itex]. Gaussian quadrature is a way to get a much more accurate approximation by choosing [itex]w_i[/itex] and [itex]x_i[/itex] in a more sophisticated way, but it's beyond me to explain how to do it in a post.

As to the second question, if a particle goes from [itex]a[/itex] to [itex]b[/itex] and back, then the time taken for a full period is twice the time to go from [itex]a[/itex] to [itex]b[/itex]. To compute that time, you use:

[itex]T = 2 \int dt = \int_a^b \frac{dt}{dx} dx = \int_a^b \frac{1}{v} dx[/itex] where [itex]v = \frac{dx}{dt}[/itex]. To compute [itex]v[/itex], you use conservation of energy:

[itex]E = \frac{1}{2} m v^2 + V[/itex]
[itex]v = \sqrt{\frac{2}{m} (E - V)}[/itex]

So [itex]T = 2 \int_a^b \sqrt{\frac{m}{2(E-V)}} dx[/itex]

If the potential is symmetric about x=0, then you can just do half the integral and double it:

[itex]T = 4 \int_0^b \sqrt{\frac{m}{2(E-V)}} dx[/itex]

So it works no matter what the potential [itex]V[/itex] is--it doesn't have to be quadratic.
I see, thank you very much!

So our weighted function for this particular question will be ##w_i = \frac{b-a}{N} = \frac{a-0}{20}## which implies ##
\int_{a}^b f(x) dx = \sum_{i=1}^N w_i f(x_i) = \sum_{i=1}^{20} \frac{a-0}{20}\sqrt{8 \dot\ (1)}\frac{1}{\sqrt{V(a)-V(x_i = 0 +i\frac{a-0}{N})}}?##
 
  • #4
BvU
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Wouldn't count as a gaussian quadrature in my book. (a/20 is equal weights) But I wouldn't know how to do it more sophisticated, sorry.
 
  • #5
stevendaryl
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I see, thank you very much!

So our weighted function for this particular question will be ##w_i = \frac{b-a}{N} = \frac{a-0}{20}## which implies ##
\int_{a}^b f(x) dx = \sum_{i=1}^N w_i f(x_i) = \sum_{i=1}^{20} \frac{a-0}{20}\sqrt{8 \dot\ (1)}\frac{1}{\sqrt{V(a)-V(x_i = 0 +i\frac{a-0}{N})}}?##
That's NOT using quadratures. That's using the much simpler "rectangle rule", also known as Newton's method (I think).
 

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