Ge108: Homework 3 Solution

Transcription

1 Ge8: Homework 3 Solution Due: Wednesday, Oct. In class October 7, 5 General Comments In general, people are making some very beautiful plots in mathematica with all kinds of colors and properties. Keep it up! It makes your plots easier to read. Comment on your important results/plots, even when we forget to ask you to do so. Basically, this demonstrates to me that you ve briefly done the critical thinking about what your plot/result/calculation actually means. You should consider this as practice for figures/calculations that you ll make for papers and posters, which you will certainly write brief critical thoughts on. A sentence or two generally suffices, and it can state the obvious but try to take it to the next step. e.g. The plot of the motion of a pendulum looks like a sine wave for small initial displacements. This indicates that a pendulum acts like a simple harmonic oscillator if the oscillations are small. Presumably the fact that your plot is a sine wave is obvious, but its nice to show me you re making connections. Please label your axes!! Its easy to forget on these problem sets, but it really is a bad habit. I ve let it slide so far, but I m going to get pickier about it. Someday you ll look back at these plots and you ll want to be able to understand them, or at the very least recall how to label plots when its really critical that you do (like in your orals!!). It can be a bit of an organizational nightmare to get all the mathematica plots and your handwritten stuff pulled together in a manner that makes sense. Many of you are directing me to look at plots/comments which are attached at the back of your homework, which works wonderfully for me. Also most are doing a good job of labeling sections. This makes finding things much easier, and I really appreciate this. Keep it up! Part Derive the equation for the Pendulum Below is a graphic of how the pendulum system works. Using the force balance equation, mlẍ = Restoring Force + External Force (.) The L is on the LHS because x in an angle. The angular velocity, y, given by y= ẋ. Restoring force is the force that opposes the motion of the pendulum, in this case its the component of the gravitational force that

2 is antiparallel to the motion. This gives ml dy dt = mg sin(x) + F ext dy dt = g L sin(x) + F ext (.) In the case that x, then sin(x) x, which you get from the Taylor expansion. If F ext =, then eq.. reduces to dy dt = g L x (.3) This is the equation for the simple harmonic oscillator (SHO), since ẋ=y. The oscillator has a frequency, ω = g L. The general solution to the SHO equation is x(t) = A cos(ω t + ) y(t) = ω A sin(ω t + ) (.4) Using the initial conditions that x()= and y()=, you get the following set of equations x() =A cos( ) = (.5) y() =ω A sin( ) = (.6) You know that A and ω aren t, so this means that sin( )=, or that = and therefore A=. We re told that ω is actually π in this problem. Therefore final solution is x(t) = cos(πt) (.7) L X F ext y=dx/dt m mg Sin(x) mg Figure : The figure for part a. The relevant variables are labeled. Forces are in blue.

3 sol3_.nb Problem Set 4 WARNING: There is a long integration in this part 5 of this notebook, so don't evaluate this whole notebook at once. ü Part In[86]:= Clear@x, t, yd ü Analytical Solution This is the plot of the analytical solution that you just found for the SHO. I've plotted x(t) in red and the y(t) is in blue. In[86]:= x@t_d := Cos@p * td In[86]:= y@t_d := -p * Sin@p * td In[863]:= p = Plot@8x@tD, y@td<, 8t,, <, PlotStyle Ø 88AbsoluteThickness@D, Hue@D<, 8GrayLevel@.5D, AbsoluteThickness@D, Hue@.7D<<, Frame Ø True, FrameLabel Ø 8"", " & yhtl"<, RotateLabel Ø True D 3 & yhtl Out[863]= In[864]:= Clear@x, t, yd ü Numerical Solution This is the numerical result for the SHO. Here the x(t) is in black and y(t) is grey. In[45]:= sol = NDSolve@8x'@tD ã y@td, y '@td ã -p ^ * x@td, x@d ã., y@d ã <, 8x, y<, 8t,, <D Out[45]= 88x Ø InterpolatingFunction@88.,.<<, <>D, y Ø InterpolatingFunction@88.,.<<, <>D<<

4 sol3_.nb In[866]:= p = Plot@8x@tD ê. First@solD, y@td ê. First@solD<, 8t,, <, PlotStyle Ø 88AbsoluteThickness@D, Dashing@8.5,.3<D<, 8GrayLevel@.5D, AbsoluteThickness@D, Dashing@8.5,.3<D<<, Frame Ø True, FrameLabel Ø 8"", " & yhtl"<, RotateLabel Ø TrueD 3 & yhtl Out[866]= ü Comparing the Analytical and Numerical Solution: Here I've compared the two plots by plotting them on the same graph. As you can see they're exactly the same. In[867]:= Show@p, pd 3 & yhtl Out[867]= ü The Parametric Plot This is plot of the x vs. t, as you can see, its a closed ellipse

5 sol3_.nb 3 In[93]:= ParametricPlot@8Evaluate@x@tD ê. First@solDD, Evaluate@y@tDD ê. First@solD<, 8t,, 4<, PlotRange Ø All, Frame Ø True, FrameLabel Ø 8"", "yhtl"<, RotateLabel Ø TrueD 3 yhtl Out[93]= Here, I've just set the keyword AspectRatio-> so that I get a nice circle In[94]:= pp = ParametricPlot@8Evaluate@x@tD ê. First@solDD, Evaluate@y@tDD ê. First@solD<, 8t,, 4<, PlotRange Ø All, AspectRatio Ø, Frame Ø True, FrameLabel Ø 8"", "yhtl"<, RotateLabel Ø TrueD 3 yhtl Out[94]=

6 sol3_.nb 4 ü Part Here I've found the numerical solution for the full pendulum equation. The wording of this part was a bit confusing, but Andy wanted you to compare the numerical solution of the full pendulum to the analytical solution for the SHO. In some sense there is no "analytical" solution to the the full pendulum equation. If you try using DSolve in Mathematica, you get a funky solution that involves "elliptical functions". Elliptical functions are basically defined to be the solution to the pendulum differential equation and other similar equations. Once you have your elliptical function, you figure out what inputs you want and look up the output on a table (or rather, have Mathematica do this). Basically there is no "closed form" solution to the full pendulum equation. Here I picked a value of 7/8p as the initial x-so this is a pretty large amplitude. As you can see, the solution doesn't look like the SHO. The black plot is x(t) and y(t) is in grey. In[334]:= sol3 = NDSolve@ 8x'@tD ã y@td, y'@td ã -p ^ * Sin@x@tDD, x@d ã 7 * p ê 8, y@d ã <, 8x, y<, 8t,, <D Out[334]= 88x Ø InterpolatingFunction@88.,.<<, <>D, y Ø InterpolatingFunction@88.,.<<, <>D<< In[868]:= p3 = Plot@8x@tD ê. First@sol3D, y@td ê. First@sol3D<, 8t,, <, PlotStyle Ø 88AbsoluteThickness@D<, 8GrayLevel@.5D, AbsoluteThickness@D<<, Frame Ø True, FrameLabel Ø 8"", " & yhtl"<, RotateLabel Ø TrueD 6 4 & yhtl Out[868]= This is the comparison of the analytical x & y (red and blue respectively) for the SHO to the numerical solution to the pendulum. As you can see, when x gets large, the shape of the curves can get strange. Also, the period of the oscillation is no longer p.

7 sol3_.nb 5 In[869]:= Show@p, p3d 6 4 & yhtl Out[869]= Here is the parametric plot, x vs. y. As you can see, it starts to look boxy, but its still a closed form In[95]:= pp = ParametricPlot@8Evaluate@x@tD ê. First@sol3DD, Evaluate@y@tDD ê. First@sol3D<, 8t,, <, PlotRange Ø All, AspectRatio Ø, Frame Ø True, FrameLabel Ø 8"", "yhtl"<, RotateLabel Ø TrueD 6 4 yhtl Out[95]=

8 sol3_.nb 6 ü Part 3a The pendulum can go over the top if it has total energy > Lmg, which is the potential energy that is has at the top of its motion. Any energy above this will go into kinetic energy, causing the pendulum to swing around. If it starts with x()=, then the minimum velocity is given by: ÅÅÅÅ mv = Lmg Øv= è!!!!!!!!! 4 Lg Since Ly=v, Then you can use the fact that è!!!!!!!!! g ê L = p and find that the minimum velocity is given by y()=p In[857]:= sol4 = NDSolve@8x'@tD ã y@td, y '@td ã -p ^ * Sin@x@tDD, x@d ã, y@d ã * p *.<, 8x, y<, 8t, -3 p, 3 p<d Out[857]= 88x Ø InterpolatingFunction@ , <<, <>D, y Ø InterpolatingFunction@ , <<, <>D<< Here I've picked a value for the initial velocity that will definitely get the pendulum around. If you pick exactly p, then you'll find that the pendulum gets stuck at the top, but eventually falls down due to fluctuations in the numerical answer. In[87]:= p4 = Plot@8x@tD ê. First@sol4D, y@td ê. First@sol4D<, 8t,, 3 * p<, PlotStyle Ø 88AbsoluteThickness@D<, 8GrayLevel@.5D, AbsoluteThickness@D<<, Frame Ø True, FrameLabel Ø 8"", " & yhtl"<, RotateLabel Ø TrueD 5 & yhtl Out[87]= Here is the parametric plot

9 sol3_.nb 7 In[96]:= pp = ParametricPlot@8Evaluate@x@tD ê. First@sol4DD, Evaluate@y@tDD ê. First@sol4D<, 8t, -p, p<, PlotRange Ø All, AspectRatio Ø, Frame Ø True, FrameLabel Ø 8"", "yhtl"<, RotateLabel Ø TrueD 6 5 yhtl 4 3 Out[96]=

10 sol3_.nb 8 ü Part 3b-Optional ü Code for part 3b ü Plot of constant energy contours for the pendulum The code for this plot is hidden above because its long. Ive just show the result below, but all the plots are similar to what you've already made with just slightly different initial conditions. You can download the notebook and look at the code for yourself if you're interested. The figure below is essentially a plot of constant energy contours. You initial energy will dictate where on the plot you fall and then you're constrained to move along the contour. The pendulum will never cross contours in the case of no damping and no forcing. You can see the two regimes- one where the pendulum is not going over the top and making closed ellipses. The other regime is where the pendulum goes over the top and continues oscillating in the x-y plane. Note that there are also two kinds of equilibirum points- One is in the center of the circles, which is a blue dot on the graph below, and corresponds to the pendulum at rest at the bottom. The other equilibrium points are between the circles and are marked by red dots and corresponds to the pendulum at rest, but at the top of the arc, balanced over the pin. The blue equilibrium point is stable, meaning if the system recieves a small perturbation away from this point, it stays nearby. The red equilibrium points are unstable, since a small perturbation will result in large motions away from the equilibrium point- also, if damping is present, then the pendulum will eventually end up oscillating around the stable equilibrium point. In[464]:= a = Graphics@8Hue@.65D, AbsolutePointSize@7D, Point@8, <D, Hue@D, AbsolutePointSize@7D, Point@8-p, <D, Hue@D, AbsolutePointSize@7D, Point@8p, <D<D; In[97]:= Show@pp, pp, pp, pp3, pp4, pp5, pp6, pp7, pp8, pp9, ad yhtl Out[97]=

11 sol3_.nb 9 ü Part 4a This is the numerical result for the damped, forced harmonic oscillator. In[883]:= g =.5; c =.; w = p *.5; In[886]:= sol4a = NDSolve@8x'@tD ã y@td, y '@td ã -p ^ * x@td - g * y@td + c * Cos@w * td, x@d ã, y@d ã <, 8x, y<, 8t,, 5 * p<d Out[886]= 88x Ø InterpolatingFunction@88., <<, <>D, y Ø InterpolatingFunction@88., <<, <>D<< In class we learned that the amplitude of the residual solution is given by In[877]:= a@wa_, wb_, Amp_, g_d := Amp ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Hwa^ - wb^l^ + 4 * Hg ê L^ * wb^ So the amplitude of steady state oscillation in this case is analytically given as: In[878]:= a@p, w, c, gd Out[878]=.3434 Below is the plot of the numerical solution and you can see after all the transients have died away that the amplitude is indeed.3. In[887]:= p4a = Plot@8x@tD ê. First@sol4aD, a@p, w, c, gd, -a@p, w, c, gd<, 8t,, * p<, PlotStyle Ø 88AbsoluteThickness@D<, 8RGBColor@,, D<, 8RGBColor@,, D<<, PlotRange Ø 8-.,.<, Frame Ø True, FrameLabel Ø 8"", ""<, RotateLabel Ø TrueD Out[887]=

12 sol3_.nb Here is for a case above the natural frequency In[888]:= g =.5; c =.; w = p *.5; In[89]:= sol4b = NDSolve@8x'@tD ã y@td, y '@td ã -p ^ * x@td - g * y@td + c * Cos@w * td, x@d ã, y@d ã <, 8x, y<, 8t,, 5 * p<d; The amplitude from the analytical solution: In[65]:= a@p, w, c, gd Out[65]=.6988 In[893]:= p4b = Plot@8x@tD ê. First@sol4bD, a@p, w, c, gd, -a@p, w, c, gd<, 8t, * p, 5 * p<, PlotStyle Ø 88AbsoluteThickness@D<, 8RGBColor@,, D<, 8RGBColor@,, D<<, PlotRange Ø 8-.6,.6<, Frame Ø True, FrameLabel Ø 8"", ""<, RotateLabel Ø TrueD; And for good measure, lets look at what happens when you get really close to the resonant frequency. In[95]:= g =.5; c =.; w3 = p * ; In[897]:= sol4c = NDSolve@8x'@tD ã y@td, y '@td ã -p ^ * x@td - g * y@td + c * Cos@w3 * td, x@d ã, y@d ã <, 8x, y<, 8t,, * p<d; In[898]:= a@p, w3, c, gd Out[898]=.6366

13 sol3_.nb In[9]:= p4c = Plot@8x@tD ê. First@sol4cD, a@p, w3, c, gd, -a@p, w3, c, gd<, 8t, * p, 5 * p<, PlotStyle Ø 88AbsoluteThickness@D<, 8RGBColor@,, D<, 8RGBColor@,, D<<, PlotRange Ø 8-, <, Frame Ø True, FrameLabel Ø 8"", ""<, RotateLabel Ø TrueD; ü Part 4b (Optional) Here is the x-y plot for the damped, forced oscillator. Specifically, this is for the last case shown above. As you can see it sprirals towards a single solution, but it never quite a closed ellipse. In[99]:= pp4 = ParametricPlot@8Evaluate@x@tD ê. First@sol4cDD, Evaluate@y@tDD ê. First@sol4cD<, 8t,, 5 * p<, PlotRange Ø All, AspectRatio Ø, Frame Ø True, FrameLabel Ø 8"", "yhtl"<, RotateLabel Ø TrueD yhtl Out[99]= Here I've created an array of values t which satisfy the equation wt=np, where n=,,4 etc.

14 sol3_.nb In[65]:= s = Table@n * p ê w3, 8n,,, <D; This is x & y evaluated at those times: In[66]:= x4b = x@sd ê. First@sol4cD; In[67]:= y4b = y@sd ê. First@sol4cD; The plot shows that the points approac a single solution. They aren't exactly piled up on each other because the "transient" is an exponential, and always contributes some. In[9]:= ListPlot@Transpose@8x4b, y4b<d, Prolog Ø AbsolutePointSize@5D, Frame Ø True, FrameLabel Ø 8"xHT=npêwL", "yht=npêwl"<, RotateLabel Ø TrueD.99 yht=npêwl Out[9]= xht=npêwl ü Part 5a In this section, I've replaced x by sin(x). I'm using the same parameters shown below. I found that I got chaotic behavior at C=, so instead I've used c=5., to show that you do get nice behaviour for a small enough forcing frequency. In[9]:= g =.5; c = 5.; w4 =.5; sol5b = NDSolve@8x'@tD ã y@td, y '@td ã -p ^ * Sin@x@tDD - g * y@td + c * Cos@w4 * td, x@d ã, y@d ã <, 8x, y<, 8t,, 3 * p<, MaxSteps Ø D Out[698]= 88x Ø InterpolatingFunction@88., <<, <>D, y Ø InterpolatingFunction@88., <<, <>D<<

15 sol3_.nb 3 In[95]:= p5c = Plot@8x@tD ê. First@sol5bD<, 8t, * p, * p<, PlotStyle Ø 88AbsoluteThickness@D, RGBColor@,, D<, 8RGBColor@,, D<, 8RGBColor@,, D<<, Frame Ø True, FrameLabel Ø 8"", ""<, RotateLabel Ø TrueD Out[95]= Here we're examining the chaotic behavior. I set c= as Andy suggested, and I integrated for a really long time so make the chaotic behavior more obvious. WARNING: Notice I had to bump up the MaxSteps to,, and made the integration really long. I recommend upping your step sizes gradually so you can figure out how many stepsizes your computer can handle before crashing. In[945]:= g =.5; c =.; w4 =.5; In[969]:= sol5c = NDSolve@8x'@tD ã y@td, y '@td ã -p ^ * Sin@x@tDD - g * y@td + c * Cos@w4 * td, x@d ã, y@d ã <, 8x, y<, 8t,, * p<, MaxSteps Ø D Out[969]= 88x Ø InterpolatingFunction@88., 345.9<<, <>D, y Ø InterpolatingFunction@88., 345.9<<, <>D<<

16 sol3_.nb 4 In[986]:= p5c = Plot@8x@tD ê. First@sol5cD<, 8t, * p, * p<, PlotStyle Ø 88AbsoluteThickness@D, RGBColor@,, D<, 8RGBColor@,, D<, 8RGBColor@,, D<<, Frame Ø True, FrameLabel Ø 8"", ""<, RotateLabel Ø TrueD; As you can see, the amplitude of the oscillation is regular, but x(t) wanders in an unpredictable fashion. In fact, if you start with slightly different initial conditions, you find that you get a completely different plot. In[975]:= g =.5; c =.; w4 =.5; In[978]:= sol5d = NDSolve@ 8x'@tD ã y@td, y'@td ã -p ^ * Sin@x@tDD - g * y@td + c * Cos@w4 * td, x@d ã, y@d ã.<, 8x, y<, 8t,, 3 * p<, Method Ø ExplicitRungeKutta, MaxSteps Ø D Out[978]= 88x Ø InterpolatingFunction@88., <<, <>D, y Ø InterpolatingFunction@88., <<, <>D<<

17 sol3_.nb 5 In[987]:= p5d = Plot@8x@tD ê. First@sol5dD<, 8t, * p, * p<, PlotStyle Ø 88AbsoluteThickness@D, RGBColor@,, D<, 8RGBColor@,, D<, 8RGBColor@,, D<<, Frame Ø True, FrameLabel Ø 8"", ""<, RotateLabel Ø TrueD; In[988]:= Show@p5c, p5dd; ü Part 5b (Optional) The parametric makes this wandering more obvious. Below I've plotted the x vs. y for the case just above. The kinks are real and not a numerical effect.

18 sol3_.nb 6 In[989]:= pp5a = ParametricPlot@8Evaluate@x@tD ê. First@sol5cDD, Evaluate@y@tDD ê. First@sol5cD<, 8t,, * p<, PlotRange Ø All, AspectRatio Ø, Frame Ø True, FrameLabel Ø 8"", "yhtl"<, RotateLabel Ø TrueD 5 5 yhtl Out[989]= 5 5 In[97]:= s = Table@n * p ê w4, 8n,, 5, <D; In[97]:= s@@5dd Out[97]= This is x & y evaluated at those times-note I've put this into modulus p using Mod: In[97]:= x5b = Mod@x@sD, pd ê. First@sol5cD; In[973]:= y5b = Mod@y@sD, pd ê. First@sol5cD; The plot shows that the points don't approach a single value, although they do appear to have some structure to them. This isn't exactly the plot that was shown on the second website, but its of the same flavor. I may not have been able to integrate to have enough points or my initial conditions may be the issue.

19 sol3_.nb 7 In[974]:= ListPlot@Transpose@8x5b, y5b<d, Prolog Ø AbsolutePointSize@D, Frame Ø True, FrameLabel Ø 8"xHT=npêwL", "yht=npêwl"<, RotateLabel Ø TrueD 3.5 yht=npêwl xht=npêwl Out[974]=