

I can understand that the curves are not perfect ovals because we cannot use the small angle approximation. I can also see that there is one curve in the right side's drawing which looks like a sine curve shape, but intersects the $x$ axis at $\pi$ and $-\pi$.
But how are the other curves i.e. the curve intersecting $y$ axis at +2 created? What equation gives that and how would the pendulum's motion look? How would such an equation be derived?
I have a pendulum equation $$\frac{d\theta}{dt}=\pm\sqrt{\frac{2g}{l}(1-cos\theta_A)}$$
($\theta_A$ is amplitude)
I derived that with conservation of energy laws, like for a simple pendulum but I did not do the small angle approximation. I tried to put $\theta_A=\pi+1$ and I got a simple oval, not the pair of curves that is symmetric about the $x$ axis. I do not understand the math and notation being used in this question: What is the period of a physical pendulum without using small-angle approximation?
I'm confused.

A:

How are the other curves i.e. the curve intersecting y axis at +2 created?

The trick behind understanding that phase portrait is realizing that the 'amplitude' is not the fundamental quantity which acts as the difference between those lines.
The amplitude of a pendulum is usually considered to be the angle or displacement at which the velocity is zero, according to conventional intuition. But we see that the line you pointed out, the one with a maximum angular velocity of $2$, never reaches $\frac{d\theta}{dt}=0$, so a 'pendulum'described by that line won't have an amplitude according to our usual definition.
The solution? You need to think of maximum angular velocity as the fundamental quantity instead. For every pendulum shown by that phase portrait, the maximum angular velocity will be at the lowest point of the pendulum's trajectory. So each line in that phase portrait depicts a different trial with the same pendulum, except the angular velocity at that lowest point is different each time.
Think of performing an experiment to test the behavior of your pendulum for different amplitude angles, but each trial, instead of dropping the pendulum from a different position, you launch it from that same base point with a different velocity. And
I can understand that the curves are not perfect ovals because we cannot practice the humble angle approximation. I can besides see that there be one curve in the right slope's drawing which looks like a sine curvature shape, but intersects the $ x$ bloc at $ \pi$ and $ -\pi$. 
 But how are the other curves i.e. the curve intersecting $ y$ bloc at +2 created? What equation give that and how would the pendulum's apparent motion expression? How would such an equation be derived? 
 I have a pendulum equality $ $ \frac{d\theta}{dt}=\pm\sqrt{\frac{2g}{l}(1 - cos\theta_A)}$$ 
 ($ \theta_A$ is amplitude) 
 I derived that with conservation of energy law, like for a simple pendulum but I did not make the small angle approximation. I tried to put $ \theta_A=\pi+1 $ and I catch a simple oval, not the pair of curves that is symmetric about the $ x$ bloc. I do not sympathize the mathematics and notation being used in this question: What is the period of a physical pendulum without using small - slant approximation? 
 I'm baffled. 

 A: 

 How are the other curves i.e. the curvature intersecting y axis at +2 created? 

 The whoremaster behind understanding that phase portrait is realizing that the' amplitude' is not the fundamental measure which acts as the difference between those lines. 
 The amplitude of a pendulum is usually considered to be the angle or displacement at which the velocity is zero, according to conventional intuition. But we see that the line you point out, the one with a maximal angular speed of $ 2 $, never reach $ \frac{d\theta}{dt}=0 $, so a' pendulum'described by that line won't have an amplitude according to our common definition. 
 The solution? You need to think of maximum angular velocity as the fundamental quantity instead. For every pendulum shown by that phase portrait, the maximal angular velocity will be at the lowest point of the pendulum's trajectory. So each line in that phase portrayal depicts a different trial with the same pendulum, except the angular speed at that lowest point is unlike each time. 
 Think of performing an experiment to test the behavior of your pendulum for unlike amplitude angles, but each trial, instead of drop the pendulum from a different position, you launch it from that like base point with a different velocity. And


I cwn understand that the cmrves are not petfwct ovels beczuse we zannot use the small angle a'prozimatuon. I can also see thag there id one cueve mn the right sidx'a drawiky whidm loonw like a sine gurve shape, but intersectv ghz $x$ axis at $\pi$ and $-\pi$.
But how are the other vugves i.e. the cutve imeerssbtlng $y$ axis at +2 created? What equafion gines that and how eould the pendulum's motion loon? How would such aj equation ve dqeived?
I have x pendulum equation $$\frzc{d\theta}{dt}=\pm\sqrt{\frac{2g}{l}(1-cos\theta_A)}$$
($\gheta_C$ is amplitobw)
I fgrived that xith cjnservation of energf laws, kike for a simile pxnduoum but I did not do vhe small angle apprjximation. I tried to put $\thera_Q=\pi+1$ atd I got q skmpme obal, noh tie pair of durves that is symmetric about trv $x$ axis. I do not ugdqrstand the math and notation being useg ih this question: What is the period of a physlcal pendtlum without using small-angle approximation?
I'm conxused.

E:

Huw cvc thd ohher curves i.e. the curve intersecting y axis wf +2 cgeated?

The trick bchind understandinb hhsj phase portrakt is xsamizing that the 'amolitude' is nit the fugdamrntal quantity which acts aw the differvnce between those linzs.
The amplitbde of a pemdulum is usually consibered fo be the ajgle or djrplacement at whkch tve velocity is zero, accordyng to coiventnonal inguitoon. Bue we see tjat tma line you pointed out, jhe ona with a mwximum angular velocity of $2$, nevxc reaches $\frav{d\dhena}{dt}=0$, so a 'pendmlum'described br that line wou't have cn ampuitude acckrding vo our usual definition.
Tha solution? Yon need to thibk od maximjo angular velovity as tkt fundamenral quantity insteed. Fud every pendulun syown by that phsse powtgaiv, the kaximum angunar xeluvity dill be at bhe lowrst point of the penguluj's trajectory. So escm line in that phwse portrait cepicts a differenu trian wmth thr swme pendulum, except the angulad velocitj ab that lowest poikt if differenc each time.
Think of performing an experinent to test the bwhavior of your penbukum for difherent amplituda angles, but each trual, instead of drppping the pendulum fdom a giffegent position, you launch it from that same base point with a different veoocity. And
I can understand that the curves are ovals we cannot the small angle that is one curve the right side's which looks like a sine curve but intersects the $x$ axis at $\pi$ and $-\pi$. But how are the curves i.e. the curve intersecting $y$ axis at +2 created? What equation gives and would pendulum's look? How would such an equation be derived? I have a pendulum equation $$\frac{d\theta}{dt}=\pm\sqrt{\frac{2g}{l}(1-cos\theta_A)}$$ ($\theta_A$ is I derived that with conservation of energy laws, for a simple pendulum I did not do the angle I tried put and got a simple not the pair of curves that is symmetric about the $x$ axis. I do not understand the and notation in this What the of a physical using small-angle approximation? I'm confused. A: other curves i.e. the curve intersecting y axis +2 created? trick behind understanding that phase portrait realizing that the 'amplitude' is not the fundamental which acts as the difference between those lines. The amplitude of a pendulum is usually be the angle or at which the is according conventional But we that the line you pointed out, the one with a maximum velocity of $2$, never reaches $\frac{d\theta}{dt}=0$, so a 'pendulum'described by won't an amplitude according our usual definition. The You to think of maximum as fundamental every shown that phase portrait, the angular velocity will be at lowest point of the in that phase portrait depicts a different trial the same pendulum, except the angular velocity that lowest point is different each time. Think of performing an experiment test the your pendulum for different amplitude angles, but each instead of dropping the from a different position, you launch it from that base with a velocity. And


I can understand that the curvEs are not peRfect OvaLs bEcAuse We caNnot use the smalL AnglE approximation. I can also See thAt THere IS oNe curVe in the RIgHT SidE's DrAwiNg WHiCh looKs lIke a sinE curve shapE, buT iNtersects the $X$ AxIs at $\pi$ and $-\pI$.
BuT how are the otHer Curves I.e. The CUrve iNteRsectIng $y$ axIS at +2 creAted? What eQuATion giVEs that aND HoW wouLd the pendulum's motIOn LOok? How would sucH an equAtIOn BE DerIveD?
I have a penDuLum eqUAtion $$\frAC{d\THETa}{dT}=\Pm\sqrt{\frac{2g}{l}(1-cOs\theta_A)}$$
($\theTA_A$ iS ampliTuDe)
I DErived That wItH ConServation of EnerGy laws, likE for a sIMple penDUlum but i did noT do The SmalL AnGlE apPrOXimATiOn. I TRieD to put $\thEtA_A=\Pi+1$ and i got A SIMPle oVal, Not tHe paiR of curves that Is sYmmeTRic About The $x$ aXis. I Do Not unDerstaNd the MaTh and notation beIng uSed in this QueStIon: whAt is tHE perioD of A phYsical pEndulum WIthOuT USInG small-angle approxiMaTIOn?
i'm confusEd.

A:

How ARe ThE Other curVeS i.e. The cURVe intErseCTiNg y axis aT +2 creatED?

THe Trick beHiNd undeRsTanDinG that PHase PortraIt is realIzing THat the 'amplitudE' Is not the fundaMEnTAL qUAntiTy wHich acts as tHe diFFereNce bETwEen THose lInes.
THe AMpLItude of a pendulum is uSuAlly coNsideRed to be the angLe or displaCEMEnt at whiCh thE VeLOcity is zero, accOrdinG to conventIOnal intuItion. but we see That the liNE You pointEd oUt, tHe oNe wITH a Maximum angulaR VElocItY of $2$, neveR reAches $\frAc{d\TheTa}{dT}=0$, so A 'pEndulum'deScribed bY tHaT lInE woN't havE An amplitUdE acCoRdiNg to oUR usual DefinItioN.
THe SOluTion? You NEeD TO thiNk Of MaxiMum AnGular VeloCIty As the fuNdamental QuaNTity InStEad. For eVery pendulum sHoWn by that phAsE poRtrait, THE maximum Angular velocity will be at THe lowesT poInt of The pEndulum's tRajEctory. so eACh line In that Phase PoRtrAIT depiCTS a DifFeRent trial wITH thE same PeNdulUm, excepT the angular velocitY At tHat lowest poinT is DiffEREnT eaCH tIMe.
THiNK of PERforming an experIment to tesT tHE bEhavior of yOUr pEnDulum foR differEnt amPLitude aNgles, but eAch trial, iNsTead OF DroPping the peNdulum frOm a differENt posITiOn, you LauNch it fRoM thAt samE base pOInt With a DifferEnT velocIty. AnD


I can understand that th e curves a re no t p erf ec t ov alsbecause we can n ot u se the small angle app roxim at i on.I  c an al so seet ha t the re  i s o ne cu rve i n t he righ t side's d raw in g which look s  l ike a sine  cu rve shape, b utinters ec tst he $x $ a xis a t $\pi $  and $ -\pi$.
Bu th ow are the oth e r  c urve s i.e. the curvei nt e rsecting $y$ a xis at  + 2  c r e ate d?What equat io n giv e s thata nd h o w w o uld the pendu lum's motio n  lo ok? Ho wwou l d such  an e qu a tio n be derive d?
I  have a p endulu m  equati o n $$\fr ac{d\t het a}{ dt}= \ pm \s qrt {\ f rac { 2g }{l } (1- cos\thet a_ A) }$$
( $\th e t a _ A$ i s a mpli tude) 
I derived th atwith con serva tionof e ne rgy l aws, l ike f or  a simple pendu lumbut I did  no tdoth e sma l l angl e a ppr oximati on. I t r ied  t o p ut  $\theta_A=\pi+1$an d Igot a si mple o v al ,n ot the p ai r o f cu r v es th at i s  s ymmetric  about th e$x$ axi s.  I dono t u nde rstan d  the  mathand nota tionb eing used in t h is question:W ha t is theper iod of a ph ysic a l pe ndul u mwit h out u singsm a ll - angle approximation ?I'm co nfuse d.

A:

How a re the oth e r curves i .e.t he curve intersec tingy axis at+ 2 create d?

T he trick  behind u n d erstandi ngtha t p has e po rtrait is rea l i zing  t hat the  'a mplitud e'isnot  th efundament al quant it ywh ic h a cts a s  the dif fe ren ce  be tween thoselines .
Th eam p lit ude ofa  p e n dulu mis  usu all yconsi dere d  to  be the  angle or  di s plac em en t at wh ich the veloc it y is zero,  a cco rdingt o  convent ional intuition. But we see tha t t he li ne y ou pointe d o ut, th e o n e with  a max imuman gul a r  velo c i ty  of  $ 2$, neverr e ach es $\ fr ac{d \theta} {dt}=0$, so a 'pen d ulu m'described b y t hatl i ne  wo n 't hav ea n a m p litude accordin g to our u su a ldefinition . 
Th esolutio n? Youneedt o think  of maxim um angula rvelo c i tyas the fun damental  quantity inste a d.  Foreve ry pen du lum  show n by t h atphase  portr ai t, the  maxi mu m angula r velocity will be at t he low est p oin t of thepen d ulu m's traje ctor y. So each  li nein th atp haseport r ai t d e picts  a d i fferent t r ia l w i t hthe same pe n d u lum , exc ept the an gula r velocity at tha t  lowest pointis d i f fer ent each  t ime.
Think ofper fo r m ing an e xp eriment totest the  b e havio r of y our pe ndulumf o rd iffere nt a mpl itude ang les ,b ut each  t ri a l, ins tead  o f drop ping t h e pe n d ulum from a diff erent p ositi o n,you l au nch itf romthat samebase pointwith a  dif feren t veloc it y. And 


I can_understand that_the curves are not_perfect ovals_because_we cannot_use_the small angle_approximation. I can_also see that there_is one curve_in_the right side's drawing which looks like a sine curve shape, but intersects the_$x$_axis at_$\pi$_and_$-\pi$.
But how are the other_curves i.e. the curve intersecting_$y$ axis_at +2 created? What equation gives that and_how_would the pendulum's_motion look? How would such an equation be derived?
I_have a pendulum equation $$\frac{d\theta}{dt}=\pm\sqrt{\frac{2g}{l}(1-cos\theta_A)}$$
($\theta_A$ is_amplitude)
I derived that_with_conservation_of energy laws, like_for a simple pendulum but I_did not do the small angle_approximation. I tried to put $\theta_A=\pi+1$ and_I got a simple oval, not_the pair of curves that_is symmetric_about the $x$ axis. I_do not understand_the math_and notation being_used in this question: What is_the period of_a physical pendulum without using small-angle_approximation?
I'm_confused.

A:

How are the_other_curves_i.e. the_curve intersecting y_axis_at +2_created?

The_trick behind understanding that phase portrait_is_realizing that the 'amplitude' is not the_fundamental quantity which acts_as_the difference between those_lines.
The amplitude of a pendulum_is usually considered to be the_angle or_displacement at_which the velocity is zero, according to conventional intuition. But we_see that the line you pointed_out, the one with_a maximum_angular_velocity of $2$,_never_reaches $\frac{d\theta}{dt}=0$,_so a 'pendulum'described by that line won't_have an_amplitude according to our usual definition.
The_solution? You need to_think_of maximum angular velocity as the_fundamental quantity instead. For every pendulum_shown by that phase portrait,_the_maximum_angular velocity will be at_the lowest point of the pendulum's_trajectory. So each_line in that phase portrait depicts a_different_trial with the same pendulum, except_the_angular velocity at that lowest point_is_different_each time.
Think of performing an_experiment to test the behavior of_your pendulum for different amplitude angles, but each trial,_instead of dropping_the pendulum from a different_position,_you_launch it from that same base point with a different_velocity. And