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Question about computing period of a function

Ghunho Min

Ghunho Min, high school student

Posted 3 years ago

Hello there! I'm stuck at how to calculate the period of a function in Mathematica. y[t] and m[t] are functions I made from different differential equations, and they should be periodic function. But I am not sure how to get the period. Also, I want to know how to compute x intercepts of my graphs, as in that way I can calculate the period as well. Thank you so much. edit: this is my code. g = 9.81 L = 0.5 y =. m =. y = First[ y /. NDSolve[{y''[t] == -(g/L)Sin[y[t]], y[0] == Pi/2, y'[0] == 0}, y, {t, 0, 10}]] m = First[ m /. NDSolve[{m''[t] == -(g/L)m[t], m[0] == Pi/2, m'[0] == 0}, m, {t, 0, 10}]] Plot[{m[t], y[t]}, {t, 0, 10}, PlotLegends -> "Expressions",] FunctionPeriod[{m[t], y[t]}, {t, 0, 10}]

POSTED BY: Ghunho Min

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Ghunho Min

Ghunho Min, high school student

Posted 3 years ago

Oh wow, you are my savior, thank you so much. And it works for m(t) too! And I am getting expected answers! But I'd like to ask why the peak difference is not consistent and slightly varies every time. m(t) should be clearly trigonometric function, and its period should be consistent. And I am sure y(t) should also have consistent period. could you explain me the reason why there's small difference between each peak y values? Thank you! I really appreciate your help.

POSTED BY: Ghunho Min

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 3 years ago

Some responses to a very similar question appear at this Mathematica StackExchange thread. They include location of successive peaks as one method. As for inconsistent periods, sure, it is quite likely the two do not have commensurate periods (that is to say, their ratio is irrational).

POSTED BY: Daniel Lichtblau

Rohit Namjoshi

Posted 3 years ago

Hi Ghunho, For `m` there is an analytic solution so we can get an exact answer using `DSolve`. g = Rationalize@9.81; L = Rationalize@0.5; y =.; m =.; solm = DSolve[{m''[t] == -(g/L)m[t], m[0] == Pi/2, m'[0] == 0}, m, {t, 0, 10}] fm = solm[[1, 1, 2]] ( Function[{t}, 1/2 π Cos[3/5 Sqrt[109/2] t]] ) FunctionPeriod[fm[t], t] ( 10/3 Sqrt[2/109] π ) N@% ( 1.4185 *) The numerical solutions will always have some variability due to the precision of the computations.You can control those in WL to arrive at a more precise computation of the period. For `NDSolve` you can try increasing `AccuracyGoal` and `PrecisionGoal`. For `Plot` you can try increasing `PlotPoints`.

POSTED BY: Rohit Namjoshi

Rohit Namjoshi

Posted 3 years ago

Hi Ghunho, Here is one rather inelegant way. The idea is to use the data from the plot to identify the peaks and use their location to get the period. g = 9.81; L = 0.5; y =.; m =.; soly = NDSolve[{y''[t] == -(g/L)Sin[y[t]], y[0] == Pi/2, y'[0] == 0},y, {t, 0, 10}] y = y /. soly // First py = Plot[y[t], {t, 0, 10}, PlotLegends -> "Expressions"] points = py // Cases[#, Line[x_] :> x, All] & // First; yValues = points[[All, 2]]; yPeaks = FindPeaks[yValues] yPeakIndexes = yPeaks[[All, 1]] Table[points[[i, 1]], {i, yPeakIndexes}] // Differences ( {1.67438, 1.6757, 1.67197, 1.6763, 1.67339, 1.62825} *) The peaks are separated by ~1.67. Try the above procedure on `m`.

Hi Ghunho,

Here is one rather inelegant way. The idea is to use the data from the plot to identify the peaks and use their location to get the period.

g = 9.81; L = 0.5; y =.; m =.;
soly = NDSolve[{y''[t] == -(g/L)*Sin[y[t]], y[0] == Pi/2, y'[0] == 0},y, {t, 0, 10}]
y = y /. soly // First

py = Plot[y[t], {t, 0, 10}, PlotLegends -> "Expressions"]

points = py // Cases[#, Line[x_] :> x, All] & // First;
yValues = points[[All, 2]];

yPeaks = FindPeaks[yValues]
yPeakIndexes = yPeaks[[All, 1]]

Table[points[[i, 1]], {i, yPeakIndexes}] // Differences
(* {1.67438, 1.6757, 1.67197, 1.6763, 1.67339, 1.62825} *)

The peaks are separated by ~1.67. Try the above procedure on m.

POSTED BY: Rohit Namjoshi