82 8282 8268 4.5 Pendulum period2009-05-04 03:39:17 UTC / rev 24ffb11e86b9+ef(t)=nene−(t−n)2/2n.tne−t√8n ln 2nn/enThe first factor is a constant, the peak height.The second factor is the familiar Gaussian.This one is centered at t = n and con-tains 1/2n in the exponent but otherwiseit’s just a Gaussian. It falls by a factor of 2when (t − n)2/2n = ln 2, which is whent±= n ±√2n ln 2.The FWHM is t+− t−, which is√8n ln 2. The approximate area under ef(t),which is n!, is thenn! ≈nen×√8n ln 2.This approximation reproduces the most important factors of Stirling’s ap-proximation: the nnin the numerator and the enin the denominator. Stir-ling’s approximation contains√2π instead of√8 ln 2 – a change of only6%.Problem 4.9 Coincidence?The FWHM approximation for the area under a Gaussian (Section 4.3) was alsoaccurate to 6%. Coincidence?Problem 4.10 More accurate constant factorWhere does the more accurate constant factor of√2π come from?4.5 Pendulum periodIs it coincidence that g, in units of meters per second squared, is 9.81, veryclose to π2≈ 9.87? Their proximity suggests a connection. Indeed, theyare connected through the original definition of the meter. It was proposedby the the Dutch scientist and engineer Christian Huygens (science andengineering were not separated in the 17th century) – called ‘the most in-genious watchmaker of all time’ by the great physicist Arnold Sommerfeld83 8383 834 Discretization 692009-05-04 03:39:17 UTC / rev 24ffb11e86b9+[21, p. 79]. Huygens’s portable definition of the meter required only a pen-dulum clock: Adjust the bob’s length l until the pendulum requires 1 s toswing from one side to the other; in other words, until its period is T = 2 s.A pendulum’s period (for small amplitudes) is T = 2πpl/g, as shownbelow, sog =4π2lT2.Using the T = 2 s standard for the meter,g =4π2x1 m4 s2= π2m s−2.So, if Huygens’s standard were used today, then g would be π2by defin-ition. Instead, it is close to that value. The story behind the difference isrich in physics, mechanical and materials engineering, mathematics, andhistory; see [22, 23, 24] for several views of a vast and fascinating subject.Problem 4.11 How is the time measured?Huygens’s standard for the meter requires a way to measure time, and no quartzclocks were available. How could one, in the 17th century, ensure that the pendu-lum’s period is indeed 2 s?Here our subject is to find how the period of a pendulum depends on itsamplitude. The analysis uses all our techniques so far – dimensions (Chap-ter 2), easy cases (Chapter 3), and discretization (this chapter) – to learn asmuch as possible without solving differential equations.mlθHere is the differential equation for the motion of an ideal pen-dulum (one with no friction, a massless string, and a minisculebob):d2θdt2+glsin θ = 0,where θ is the angle with respect to the vertical, g is the gravi-tational acceleration, and l is the mass of the bob.Instead of deriving this equation from physical principles (see [25] for aderivation), take it as a given but check that it makes sense.84 8484 8470 4.5 Pendulum period2009-05-04 03:39:17 UTC / rev 24ffb11e86b9+Are its dimensions correct?It has only two terms, and they must have identical dimensions. For thefirst term, d2θ/dt2, the dimensions are the dimensions of θ divided by T2from the dt2. (With apologies for the double usage, this T refers to the timedimension rather than to the period.) Since angles are dimensionless (seeProblem 4.12),d2θdt2= T−2.For the second term, the dimensions arehglsin θi=hgli×[sin θ].Since sin θ is dimensionless, the dimensions are just those of g/l, whichare T−2. So the two terms have identical dimensions.Problem 4.12 AnglesWhy are angles dimensionless?Problem 4.13 Where did the mass go?Use dimensions to show that the differential equation cannot contain the mass ofthe bob (except as a common factor that divides out).Because of the nonlinear factor sin θ, solving this differential equation isdifficult. One can compute a power-series solution, and call the resultinginfinite series a new function. That procedure, when applied to anotherdifferential equation, is the origin of the Bessel functions. However, theso-called elementary functions – those built from sin, cos, exp, ln, andpowers – do not contain a solution to the pendulum equation.85 8585 854 Discretization 712009-05-04 03:39:17 UTC / rev 24ffb11e86b9+cos θsin θθ1unit circleθSo, use easy cases to simplify the source of the prob-lem, namely the sin θ factor. One easy case is the ex-treme case θ → 0. To approximate sin θ in that limit,mark θ and sin θ on a quarter-section of the unit cir-cle. By definition, θ is the length of the arc. Also bydefinition, sin θ is the altitude of the enclosed righttriangle. When θ is small, the arc is almost exactlythe altitude. Therefore, for small θ:sin θ ≈ θ.It is a tremendously useful approximation.Problem 4.14 Slightly better approximationThe preceding approximation replaced the arc with a straight, vertical line. Amore accurate approximation replaces the arc with the chord (a straight but non-vertical line). What is the resulting approximation for sin θ, including the θ3term?In this small-θ extreme, the pendulum equation turns intod2θdt2+glθ = 0.It looks like the ideal-spring differential equation analyzed in Section 2.5:d2xdt2+kmx = 0,where m is the mass and k is the spring constant (the stiffness). Comparingthe two equations produces this correspondence:x → θ;km→gl.Since the oscillation period for the ideal spring isT = 2πrmk,86 8686 8672 4.5 Pendulum period2009-05-04 03:39:17 UTC / rev 24ffb11e86b9+the oscillation period for the pendulum, in the θ → 0 limit, isT = 2πslg.Does this period have correct dimensions?Pause to sanity check this result by asking: ‘Is each portion of the formulareasonable, or does it come out of left field.’ [For non-American readers,left field is one of the distant reaches of a baseball field. To come out ofleft fields means an idea comes almost out of nowhere, surprising all withits craziness.] The first sanity check is dimensions. They are correct in theapproximate spring differential equation; but let’s also check the dimen-sions of the period T = 2πpl/g that results from solving the equation. Inthe symbolic factorpl/g, the lengths cancel and leave only T2inside thesquare root. Sopl/g is a time – as it should