3.5 Pendulum period :40:05 UTC / rev 4d4a39156f1e. g = 4π2 l T 2. g = 4π2 x1 m 4 s 2 = π 2 m s Pendulum period 68


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1 Penduum period Penduum period Is it coincidence that g, in units of meters per second squared, is 9.8, very cose to ? Their proximity suggests a connection. Indeed, they are connected through the origina definition of the meter. It was proposed by the the Dutch scientist and engineer Christian Huygens (science and engineering were not separated in the 7th century) caed the most ingenious watchmaker of a time by the great physicist Arnod Sommerfed [6, p. 79]. Huygens s portabe definition of the meter required ony a penduum cock: Adjust the bob s ength unti the penduum requires s to swing from one side to the other; in other words, unti its period is T = 2 s. A penduum s period (for sma ampitudes) is T = 2 /g, as shown beow, so g = 42 T 2. Using the T = 2 s standard for the meter, g = 42 x m 4 s 2 = 2 m s 2. So, if Huygens s standard were used today, then g woud be 2 by definition. Instead, it is cose to that vaue. The story behind the difference is rich in physics, mechanica and materias engineering, mathematics, and history; see [7, 8, 9] for severa views of a vast and fascinating subject. Probem 3. How is the time measured? Huygens s standard for the meter requires a way to measure time, and no quartz cocks were avaiabe. How coud one, in the 7th century, ensure that the penduum s period is indeed 2 s? Here our subject is to find how the period of a penduum depends on its ampitude. The anaysis uses a our techniques so far dimensions (Chapter ), easy cases (Chapter 2), and discretization (this chapter) to earn as much as possibe without soving differentia equations
2 69 69 Chapter 3. Discretization 69 Here is the differentia equation for the motion of an idea penduum (one with no friction, a massess string, and a miniscue bob): dt 2 + g sin θ = 0, where θ is the ange with respect to the vertica, g is the gravitationa acceeration, and is the mass of the bob. Instead of deriving this equation from physica principes (see [20] for a derivation), take it as a given but check that it makes sense. Are its dimensions correct? It has ony two terms, and they must have identica dimensions. For the first term, /dt 2, the dimensions are the dimensions of θ divided by T 2 from the dt 2. (With apoogies for the doube usage, this T refers to the time dimension rather than to the period.) Since anges are dimensioness (see Probem 3.2), [ d 2 ] θ dt 2 = T 2. For the second term, the dimensions are [ g ] [ g ] sin θ = [sin θ]. Since sin θ is dimensioness, the dimensions are just those of g/, which are T 2. So the two terms have identica dimensions. θ m Probem 3.2 Anges Why are anges dimensioness? Probem 3.3 Where did the mass go? Use dimensions to show that the differentia equation cannot contain the mass of the bob (except as a common factor that divides out). Because of the noninear factor sin θ, soving this differentia equation is difficut. One can compute a powerseries soution, and ca the resuting infinite series a new function. That procedure, when appied to another differentia equation, is the origin of the Besse functions. However, the socaed eementary functions those buit from sin, cos, exp, n, and powers do not contain a soution to the penduum equation
3 Penduum period 70 So, use easy cases to simpify the source of the probem, namey the sin θ factor. One easy case is the extreme case θ 0. To approximate sin θ in that imit, mark θ and sin θ on a quartersection of the unit circe. By definition, θ is the ength of the arc. Aso by definition, sin θ is the atitude of the encosed right triange. When θ is sma, the arc is amost exacty the atitude. Therefore, for sma θ: sin θ θ. It is a tremendousy usefu approximation. unit circe sin θ θ cos θ θ Probem 3.4 Sighty better approximation The preceding approximation repaced the arc with a straight, vertica ine. A more accurate approximation repaces the arc with the chord (a straight but nonvertica ine). What is the resuting approximation for sin θ, incuding the θ 3 term? In this smaθ extreme, the penduum equation turns into dt 2 + g θ = 0. It ooks ike the ideaspring differentia equation anayzed in Section.5: d 2 x dt 2 + k m x = 0, where m is the mass and k is the spring constant (the stiffness). Comparing the two equations produces this correspondence: x θ; k m g. Since the osciation period for the idea spring is m T = 2 k, the osciation period for the penduum, in the θ 0 imit, is T = 2 g
4 7 7 Chapter 3. Discretization 7 Does this period have correct dimensions? Pause to sanity check this resut by asking: Is each portion of the formua reasonabe, or does it come out of eft fied. [For nonamerican readers, eft fied is one of the distant reaches of a baseba fied. To come out of eft fieds means an idea comes amost out of nowhere, surprising a with its craziness.] The first sanity check is dimensions. They are correct in the approximate spring differentia equation; but et s aso check the dimensions of the period T = 2 /g that resuts from soving the equation. In the symboic factor /g, the engths cance and eave ony T 2 inside the square root. So /g is a time as it shoud be. What about easy cases? Another sanity check is easy cases. For exampe, imagine a huge gravitationa fied strength g. Then gravity easiy and rapidy swings the bob to and fro, making the period tiny. So g shoud ive in the denominator of T and it does. Probem 3.5 Another easy case? Can you use easy cases to expain why beongs in the numerator? Didn t the 2 come from soving differentia equations, contrary to the earier promise to avoid soving differentia equations? The dimensions and easycases tests confirm the /g factor. But how to expain the remaining piece: the numerica factor of 2 that arose from the soution to the ideaspring differentia equation. However, we want to avoid soving differentia equations. Can our techniques derive the 2? 3.5. Sma ampitudes and Huygens method Dimensions and easy cases rarey expain a dimensioness constant. Therefore expaining the factor of 2 probaby requires a new idea. It too is due to Huygens. His idea [6, p. 79ff] is to anayze the motion of a conica penduum: a penduum moving in a horizonta circe. Athough its motion is two dimensiona, it is at constant speed, so it is easy to anayze without soving differentia equations. θ m 7 7
5 Penduum period 72 Even if the anaysis of the conica penduum is simpe, how is it reevant to the motion of a onedimensiona penduum? Projecting the twodimensiona motion onto a screen produces onedimensiona penduum motion, so the period of the twodimensiona motion is the same as the period of the onedimensiona motion! This statement is sighty fase when θ 0 is arge. But when θ 0 is sma, which is the extreme anayzed here, the equivaence is exact. To project onto onedimensiona motion with ampitude θ 0, give the conica penduum the constant ange θ = θ 0. The pan is to use the ange to find the speed of the bob, then use the speed to find its period. What is the speed of the bob in terms of and θ 0? To find the speed, find the inward force in two ways:. To move in a circe of radius r at speed v, the bob requires an inward force F = mv2 r, where m is the mass of the bob (it anyway divides out ater). 2. The two forces on the bob are from gravity and from the string tension. Since the bob has zero vertica acceeration it has no vertica motion at a the vertica component of the tension force cances gravity: T cos θ 0 = mg. Therefore, the horizonta component of tension is the net force on the mass, so that net force is F = T sin θ 0 = T cos θ } {{ 0 tan θ } 0 = mg tan θ 0. mg T F mg mg Equating these two equivaent expressions for the inward force F gives mg tan θ 0 = mv 2 /r or v = gr tan θ 0. Since the radius of the circe is r = sin θ 0, the bob s speed is v = g tan θ 0 sin θ
6 73 73 Chapter 3. Discretization 73 Probem 3.6 Check dimensions Check that v = g tan θ 0 sin θ 0 has correct dimensions. The period is the circumference divided by speed: T = 2r v = 2 sin θ 0 cos θ 0 = 2. g tan θ0 sin θ 0 g As ong as θ 0 is sma, cos θ 0 is approximatey, so T 2 /g. This equation contains a negative resut: the absence of θ 0 ; therefore, period is independent of ampitude (for sma ampitudes). This equation aso contains a positive resut: the magic factor of 2, courtesy of Huygens and without soving differentia equations Large ampitudes The preceding resuts are vaid when the ampitude θ 0 is infinitesimay sma. When this restriction is removed, how does the period behave? Does the period increase, decrease, or remain constant as θ 0 is increased? First reformuate this question in dimensioness form by constructing dimensioness groups (Section 2.4.). The period T beongs to a dimensioness group T/ /g. Since the ampitude θ 0 is no onger restricted to be near zero, it can have an important effect on period, so θ 0 shoud aso join a dimensioness group. Since anges are dimensioness, θ 0 can make a dimensioness group by itsef. With these choices, the probem contains two dimensioness groups (Probem 3.7): T/ /g and θ 0. Probem 3.7 Dimensioness groups using the penduum variabes Check that the period T, ength, gravitationa strength g, and ampitude θ 0 produce two independent dimensioness groups. In constructing two usefu groups, why shoud the period T appear in ony one group? For the same purpose, why shoud θ 0 not appear in the same group as T? Two dimensioness groups produce this genera dimensioness form: or one group = f(other group), 73 73
7 Penduum period 74 T /g = f(θ 0 ), where f is a dimensioness function. Since T/ /g goes to 2 as θ 0 (the ideaspring imit), simpify sighty by puing out the factor of 2: T /g = 2h(θ 0 ), where the dimensioness function h has the simpe endpoint vaue h(0) =. The function h contains a the information about how the period of a penduum depends on its ampitude. In terms of h, the preceding question about the period becomes this question: Is the function h(θ 0 ) monotonic increasing, monotonic decreasing, or constant? This type of question suggests considering easy cases of θ 0 : If the question can be answered for any case, the answer identifies a ikey trend for the whoe ampitude range. Two easy cases are the extremes of the ampitude range. One extreme is aready anayzed case θ 0 = 0; it reproduces the differentia equation and behavior of an idea spring. But that anaysis does not predict the behavior of the penduum when θ 0 is nonzero but sti sma. Since the owampitude extreme is not easy to anayze, try the argeampitude extreme. How does the period behave at arge ampitudes? What is a arge ampitude? A arge ampitude coud be θ 0 = /2. That case is, however, hard to anayze. The exact vaue of h(/2) is the foowing awfu expression, as can be shown using conservation of energy (Probem 3.8): h(/2) = 2 /2 0 dθ cos θ. Is this expression ess than, equa to, or more than?! Who knows. The integra ooks unikey to have a cosed form, and numerica evauation is difficut without a computer (Probem 3.9). Probem 3.8 Genera expression for h Use conservation of energy to show that the period of a penduum with ampitude θ 0 is T(θ 0 ) = 2 θ0 dθ 2. g cos θ cos θ
8 75 75 Chapter 3. Discretization 75 In terms of h, the equivaent statement is that h(θ 0 ) = 2 θ0 0 dθ cos θ cos θ0. For horizonta reease, θ 0 = /2, whereupon h(/2) = 2 /2 0 dθ cos θ. Probem 3.9 Numerica evauation for horizonta reease Why do the discretization recipes, such as the ones in Section 3.2 and Section 3.3, fai for the integras in Probem 3.8? Use or write a program to evauate h(/2) numericay. Since /2 was not a hepfu extreme, be even more extreme: 3 Try θ 0 = : reeasing the penduum bob h(θ 0 ) from the highest possibe point. That reease ocation fais if the penduum bob is connected to the support point by ony a string the penduum woud coapse downwards rather than osciate. This behavior is not described by the penduum differentia equation, which assumes that the penduum bob is constrained to move in a circe of radius. Fortunatey, θ 0 the experiment is easy to improve, because it is a thought experi ment. So, repace the string with a materia that can maintain the constraint: Let s spurge on a rigid but massess stee rod. The improved penduum does not coapse even when θ 0 =. Baanced at θ 0 =, the penduum bob wi hang upside down forever; in other words, T() =. For smaer ampitudes, the period is finite, so the period most probaby increases as ampitude increases toward. Stated in dimensioness form, h(θ 0 ) most probaby increases monotonicay toward infinity. 3 One definition of insanity is repeating an action but expecting a different resut
9 Penduum period 76 Athough monotonic behavior is the simpest assumption, aternative assumptions are possibe. For exampe, h(θ 0 ) for sma θ 0, the dimensioness function h(θ 0 ) coud decrease from ; then fatten; then increase toward infinity as θ 0 approaches. Atough possibe, such behavior woud be surprising compared to the origina, penduum differentia equation. What woud such a nice, smooth differentia equation ike the penduum equation be doing producing such θ 0 a bady behaved, nonmonotonic soution? This compicated behavior is therefore unikey. As a rue of thumb, assume unti proven otherwise that nature does not pay nasty tricks. Probem 3.20 Sma but nonzero ampitude At θ 0 = 0, does h(θ 0 ) have zero or positive sope? In other words, which figure is the more ikey to be correct: h(θ 0 ) h(θ 0 ) θ 0 h (0) = 0 h (0) > 0 θ 0 As has been said in armscontro negotiations: Trust but verify. So, whie trusting the preceding rue of thumb, verify it by more accuratey anayzing the period at sma ampitudes. This anaysis seems ike it requires soving the origina penduum differentia equation, dt 2 + g sin θ = 0. To avoid this difficut task, et s isoate, encapsuate, and try to mitigate the equation s compexity
10 77 77 Chapter 3. Discretization 77 The compexity arises because the sin θ factor makes the equation noninear. If ony that factor were θ, then the equation woud be inear and tractabe. And sin θ is amost θ: The functions θ and sin θ match as θ goes to 0. However, as θ grows i.e. for arger ampitudes θ and sin θ part company. To expicate the comparison, rewrite the differentia equation in this form: 0 f(θ) = sin θ θ 0 θ 0 dt 2 + g θf(θ) = 0, where the ratio f(θ) (sin θ)/θ encapsuates the difference between the penduum and the idea spring. When f(θ) is cose to, the penduum acts ike an idea spring; when f(θ) fas significanty beow, the simpeharmonic approximation fas in accuracy. Having isoated the compexity into f(θ), the next step is to approximate f(θ) unti the penduum equation becomes easy to sove Adding discretization The differentia equation s noninearity is now represented by a changing f(θ). When change and compexity appear in the same sentence, pu out the discretization too. In other words, repace the sowy changing f(θ) with a simper, constant vaue. The simpest choice is to repace f(θ) with f(0). Since f(0) =, the differentia equation becomes dt 2 + g θ = 0. It is once again the ideaspring equation, which produces a period independent of ampitude. So 0 0 θ 0 the simpest discretization f(θ) f(0) is too crude to provide new information about how the period depends on ampitude. What about discretizing using the other extreme of θ? The absoute penduum ange θ ives in the range [0, θ 0 ]. Since the first endpoint θ = 0 was not a usefu ange for discretizing, try the other endpoint θ 0. f(0) 77 77
11 Penduum period 78 In other words, repace the changing f(θ) not with f(0) but with the sighty smaer constant f(θ 0 ). That change repaces f(θ) with a straight ine, and turns the penduum differentia equation into f(θ 0 ) dt 2 + g θf(θ 0) = θ 0 Is this equation inear? What physica system does it describe? This equation is inear! Even better, it is famiiar: It describes an idea spring on a panet with sighty weaker gravity than earth s: dt 2 + g eff { }} { gf(θ 0 ) θ = 0, where the gravity on the panet is g eff gf(θ 0 ). Since an idea spring has period T = 2 /g, this idea spring has period T = 2 = 2 g eff gf(θ 0 ). To compare this resut with the ideaspring period, rewrite it in dimensioness form using dimensioness quantities. One quantity, the ampitude θ 0, is aready dimensioness. The period T is not dimensioness, but the dimensioness period h(θ 0 ) is defined as h(θ 0 ) T 2 /g. The 2 in the definition makes the smaampitude imit come out simpe: h(0) =. With that definition for h(θ 0 ), the the discretization f(θ) f(θ 0 ) predicts h(θ 0 ) = / gf(θ 0 ) g = f(θ 0) /
12 79 79 Chapter 3. Discretization 79 Since f(θ 0 ) = (sin θ 0 )/θ 0, the dimensioness period becomes h(θ 0 ) ( ) sin /2 θ0 h(θ 0 ) =. θ 0 This prediction (gray curve) matches the exact dimensioness period (back curve) quite we at sma but nonzero ampitudes. The comparison is easiest to make in that imit of sma but nonzero ampitude θ 0. In that imit, the Tayor series for sine is θ 0 so sin θ θ θ3 6, sin θ 0 θ 0 θ2 6. Therefore h(θ 0 ) ( ) /2 θ Since θ 2 0 /6 is even smaer than θ 0, which is itsef sma, the right side further simpifies using the binomia approximation (for sma x): ( + x) /2 x 2. Then the dimensioness period becomes ( ) /2 h(θ 0 ) θ2 0 + θ Putting back the dimensiona quantities, the period is ( ) T 2 + θ2 0. g 2 Is this resut an underestimate or an overestimate? The discretization approximation used the owest possibe effective gravity g eff, namey its vaue at the endpoint θ = θ 0. Since weak gravity produces a ong period, the approximation overestimates the period. Indeed, the 79 79
13 Summary and probems 80 exact coefficient of θ 2 0 is /6 rather than /2; see for exampe [2] for the foowing infinite series: h(θ 0 ) = + 6 θ θ Probem 3.2 Sope revisited Use the preceding resut for h(θ 0 ) to check your concusion in Probem 3.20 about the sope of h(θ 0 ) at θ 0 = Summary and probems Discretization turns cacuus on its head. Whereas cacuus anayzes a changing process by dividing it into ever finer intervas, discretization simpifies a changing process by umping it into one unchanging process. Discretization turns curves into straight ines, so difficut integras turn into rectanges, and midy noninear differentia equations turn into inear differentia equations. Even though umping sacrifices accuracy, it provides great simpicity. Probem 3.22 FWHM for another decaying function Use the FWHM heuristic to estimate inf inf dx + x 4. Then compare the estimate with the exact vaue of / 2. For a fun additiona probem, derive that exact vaue. Probem 3.23 Hypothetica penduum equation Suppose the penduum equation had been dθ 2 + g tan θ = 0. How woud the period T depend on ampitude θ 0? In particuar, as θ 0 increases, woud T decrease, remain constant, or increase? If T woud not remain constant, what is the sope at zero ampitude, i.e. T (0)? For sma but nonzero θ 0, find an approximate expression for the dimensioness period h(θ 0 ) and use it to check your previous concusions. Probem 3.24 Gaussian onesigma tai The tai of the Gaussian distribution is important in statistica inference. The Gaussian probabiity density function with zero mean and unit variance is 80 80
14 8 8 Chapter 3. Discretization 8 p(x) = e x2 /2 2. In this probem you wi estimate the area of the onesigma tai: e x2 /2 2 dx. a. Sketch the above Gaussian and shade the onesigma tai. b. Use the criterium of faing by a factor of e (Section 3.2) to discretize and thereby estimate the area. c. Use the FWHM heuristic instead to estimate the area. d. Compare the two discretization estimates with the exact vaue: e x2 /2 dx = erf(/ 2) 0.59, 2 2 where erf(z) is the socaed error function (defined especiay because of the important of this kind of probem). Probem 3.25 Distant Gaussian tais Probem 3.24 asked about the onesigma tai of a Gaussian. Now use discretization to estimate the area of the nsigma tai of a Gaussian, for arge n: t(n) n e x2 /2 2 dx. 8 8
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