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3.5 Pendulum period 68
3.5 Pendulum period
Is it coincidence that g, in units of meters per second squared, is 9.81, very close to π^2 ≈ 9.87? Their proximity suggests a connection. Indeed, they are connected through the original definition of the meter. It was proposed by the the Dutch scientist and engineer Christian Huygens (science and engineering were not separated in the 17th century) – called ‘the most in- genious watchmaker of all time’ by the great physicist Arnold Sommerfeld [16, 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 to swing 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 = 2π
p l/g, as shown below, so
g =
4π^2 l T 2
Using the T = 2 s standard for the meter,
g =
4π^2 x1 m 4 s^2
= π^2 m s−^2.
So, if Huygens’s standard were used today, then g would be π^2 by defin- ition. Instead, it is close to that value. The story behind the difference is rich in physics, mechanical and materials engineering, mathematics, and history; see [17, 18, 19] for several views of a vast and fascinating subject.
Problem 3.11 How is the time measured? Huygens’s standard for the meter requires a way to measure time, and no quartz clocks were available. How could one, in the 17th century, ensure that the pendulum’s period is indeed 2 s?
Here our subject is to find how the period of a pendulum depends on its amplitude. The analysis uses all our techniques so far – dimensions ( Chapter 1 ), easy cases ( Chapter 2 ), and discretization (this chapter) – to learn as much as possible without solving differential equations.
Chapter 3. Discretization 69
m
l θ
Here is the differential equation for the motion of an ideal pen- dulum (one with no friction, a massless string, and a miniscule bob):
d^2 θ dt^2
g l
sin θ = 0,
where θ is the angle with respect to the vertical, g is the gravita- tional acceleration, and l is the mass of the bob.
Instead of deriving this equation from physical principles (see [20] for a derivation), take it as a given but check that it makes sense.
Are its dimensions correct?
It has only two terms, and they must have identical dimensions. For the first term, d^2 θ/dt^2 , the dimensions are the dimensions of θ divided by T^2 from the dt^2. (With apologies for the double usage, this T refers to the time dimension rather than to the period.) Since angles are dimensionless (see Problem 3.12 ), d^2 θ dt^2
= T−^2.
For the second term, the dimensions are h (^) g l
sin θ
i
h (^) g l
i × [sin θ].
Since sin θ is dimensionless, the dimensions are just those of g/l, which are T−^2. So the two terms have identical dimensions.
Problem 3.12 Angles Why are angles dimensionless?
Problem 3.13 Where did the mass go? Use dimensions to show that the differential equation cannot contain the mass of the bob (except as a common factor that divides out).
Because of the nonlinear factor sin θ, solving this differential equation is difficult. One can compute a power-series solution, and call the resulting infinite series a new function. That procedure, when applied to another differential equation, is the origin of the Bessel functions. However, the so-called elementary functions – those built from sin, cos, exp, ln, and powers – do not contain a solution to the pendulum equation.
Chapter 3. Discretization 71
Does this period have correct dimensions?
Pause to sanity check this result by asking: ‘Is each portion of the formula reasonable, 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 of left fields means an idea comes almost out of nowhere, surprising all with its craziness.] The first sanity check is dimensions. They are correct in the approximate spring differential equation; but let’s also check the dimen- sions of the period T = 2π
p l/g that results from solving the equation. In the symbolic factor
p l/g, the lengths cancel and leave only T^2 inside the square root. So
p l/g is a time – as it should be.
What about easy cases?
Another sanity check is easy cases. For example, imagine a huge gravita- tional field strength g. Then gravity easily and rapidly swings the bob to and fro, making the period tiny. So g should live in the denominator of T – and it does.
Problem 3.15 Another easy case? Can you use easy cases to explain why l belongs in the numerator?
Didn’t the 2π come from solving differential equations, contrary to the earlier promise to avoid solving differential equations?
The dimensions and easy-cases tests confirm the
p l/g factor. But how to explain the remaining piece: the numerical factor of 2π that arose from the solution to the ideal-spring differential equation. However, we want to avoid solving differential equations. Can our techniques derive the 2π?
3.5.1 Small amplitudes and Huygens’ method
m
l θ
Dimensions and easy cases rarely explain a dimension- less constant. Therefore explaining the factor of 2π prob- ably requires a new idea. It too is due to Huygens. His idea [16, p. 79ff] is to analyze the motion of a conical pendulum: a pendulum moving in a horizontal circle. Although its motion is two dimensional, it is at constant speed, so it is easy to analyze without solving differen- tial equations.
3.5 Pendulum period 72
Even if the analysis of the conical pendulum is simple, how is it relevant to the motion of a one-dimensional pendulum?
Projecting the two-dimensional motion onto a screen produces one-dimensional pendulum motion, so the period of the two-dimensional motion is the same as the period of the one-dimensional motion! This statement is slightly false when θ 0 is large. But when θ 0 is small, which is the extreme analyzed here, the equivalence is exact.
To project onto one-dimensional motion with amplitude θ 0 , give the con- ical pendulum the constant angle θ = θ 0. The plan is to use the angle 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 l and θ 0?
To find the speed, find the inward force in two ways:
- To move in a circle of radius r at speed v, the bob requires an inward force
F =
mv^2 r
where m is the mass of the bob (it anyway divides out later).
mg
T
F
- The two forces on the bob are from gravity and from the string mg tension. Since the bob has zero vertical acceleration – it has no vertical motion at all – the vertical component of the tension force cancels gravity:
T cos θ 0 = mg.
Therefore, the horizontal component of tension is the net force on the mass, so that net force is
F = T sin θ 0 = T cos θ| {z 0 } mg
tan θ 0 = mg tan θ 0.
Equating these two equivalent 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 circle is r = l sin θ 0 , the bob’s speed is
v =
p gl tan θ 0 sin θ 0.
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T
p l/g
= f(θ 0 ),
where f is a dimensionless function. Since T/
p l/g goes to 2π as θ 0 (the ideal-spring limit), simplify slightly by pulling out the factor of 2π: T p l/g
= 2πh(θ 0 ),
where the dimensionless function h has the simple endpoint value h( 0 ) =
- The function h contains all the information about how the period of a pendulum depends on its amplitude. 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 ques- tion can be answered for any case, the answer identifies a likely trend for the whole amplitude range. Two easy cases are the extremes of the ampli- tude range. One extreme is already analyzed case θ 0 = 0 ; it reproduces the differential equation and behavior of an ideal spring. But that analy- sis does not predict the behavior of the pendulum when θ 0 is nonzero but still small. Since the low-amplitude extreme is not easy to analyze, try the large-amplitude extreme.
How does the period behave at large amplitudes? What is a large amplitude?
A large amplitude could be θ 0 = π/2. That case is, however, hard to ana- lyze. The exact value of h(π/2) is the following awful expression, as can be shown using conservation of energy ( Problem 3.18 ):
h(π/2) =
π
Z (^) π/
0
dθ √ cos θ
Is this expression less than, equal to, or more than 1 ?! Who knows. The integral looks unlikely to have a closed form, and numerical evaluation is difficult without a computer ( Problem 3.19 ).
Problem 3.18 General expression for h Use conservation of energy to show that the period of a pendulum with am- plitude θ 0 is
T (θ 0 ) = 2
√ 2
s l g
Z (^) θ 0
0
dθ √ cos θ − cos θ 0
.
Chapter 3. Discretization 75
In terms of h, the equivalent statement is that
h(θ 0 ) =
√ 2 π
Z (^) θ 0
0
dθ √ cos θ − cos θ 0
.
For horizontal release, θ 0 = π/2, whereupon
h(π/2) =
√ 2 π
Z (^) π/
0
√^ dθ cos θ
.
Problem 3.19 Numerical evaluation for horizontal release Why do the discretization recipes, such as the ones in Section 3.2 and Section 3.3 , fail for the integrals in Problem 3.18? Use or write a program to evaluate h(π/2) numerically.
θ 0
h(θ 0 )
π
11
Since π/2 was not a helpful extreme, be even more extreme:^3 Try θ 0 = π: releasing the pendulum bob from the highest possible point. That release loca- tion fails if the pendulum bob is connected to the support point by only a string – the pendulum would collapse downwards rather than oscillate. This be- havior is not described by the pendulum differential equation, which assumes that the pendulum bob is constrained to move in a circle of radius l. Fortu- nately, the experiment is easy to improve, because it is a thought experi- ment. So, replace the string with a material that can maintain the constraint: Let’s splurge on a rigid but massless steel rod. The improved pendulum does not collapse even when θ 0 = π. Balanced at θ 0 = π, the pendulum bob will hang upside down forever; in other words, T (π) = ∞. For smaller amplitudes, the period is finite, so the period most probably increases as amplitude increases toward π. Stated in dimensionless form, h(θ 0 ) most probably increases monotonically toward infinity.
(^3) One definition of insanity is repeating an action but expecting a different result.
Chapter 3. Discretization 77
0
1
0 θ 0
f(θ) = sin θθ
The complexity arises because the sin θ factor makes the equation nonlinear. If only that factor were θ, then the equation would be linear and tractable. And sin θ is almost θ: The functions θ and sin θ match as θ goes to 0. However, as θ grows – i.e. for larger amplitudes – θ and sin θ part company. To expli- cate the comparison, rewrite the differential equa- tion in this form:
d^2 θ dt^2
g l
θf(θ) = 0,
where the ratio f(θ) ≡ (sin θ)/θ encapsulates the difference between the pendulum and the ideal spring. When f(θ) is close to 1 , the pendulum acts like an ideal spring; when f(θ) falls significantly below 1 , the simple- harmonic approximation falls in accuracy. Having isolated the complexity into f(θ), the next step is to approximate f(θ) until the pendulum equation becomes easy to solve.
3.5.3 Adding discretization
The differential equation’s nonlinearity is now represented by a changing f(θ). When change and complexity appear in the same sentence, pull out the discretization tool. In other words, replace the slowly changing f(θ) with a simpler, constant value.
0
1
0 θ 0
f( 0 )
The simplest choice is to replace f(θ) with f( 0 ). Since f( 0 ) = 1 , the differential equation becomes
d^2 θ dt^2
g l
θ = 0.
It is once again the ideal-spring equation, which produces a period independent of amplitude. So the simplest discretization f(θ) −→ f( 0 ) is too crude to provide new information about how the period depends on am- plitude.
What about discretizing using the other extreme of θ?
The absolute pendulum angle |θ| lives in the range [0, θ 0 ]. Since the first endpoint θ = 0 was not a useful angle for discretizing, try the other end- point θ 0.
3.5 Pendulum period 78
0
1
0 θ 0
f(θ 0 )
In other words, replace the changing f(θ) not with f( 0 ) but with the slightly smaller constant f(θ 0 ). That change replaces f(θ) with a straight line, and turns the pendulum differential equation in- to
d^2 θ dt^2
g l
θf(θ 0 ) = 0.
Is this equation linear? What physical system does it describe?
This equation is linear! Even better, it is familiar: It describes an ideal spring on a planet with slightly weaker gravity than earth’s:
d^2 θ dt^2
geff z }| { gf(θ 0 ) l
θ = 0,
where the gravity on the planet is geff ≡ gf(θ 0 ). Since an ideal spring has period T = 2π
p l/g, this ideal spring has period
T = 2π
s l geff = 2π
s l gf(θ 0 )
To compare this result with the ideal-spring period, rewrite it in dimen- sionless form using dimensionless quantities. One quantity, the amplitude θ 0 , is already dimensionless. The period T is not dimensionless, but the dimensionless period h(θ 0 ) is defined as
h(θ 0 ) ≡
T
2π
p l/g
The 2π in the definition makes the small-amplitude limit come out simple: h( 0 ) = 1. With that definition for h(θ 0 ), the the discretization f(θ) −→ f(θ 0 ) predicts
h(θ 0 ) =
s l gf(θ 0 )
s l g = f(θ 0 )−1/2.
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3.6 Summary and problems 80
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exact coefficient of θ^20 is 1/16 rather than 1/12; see for example [21] for the following infinite series:
h(θ 0 ) = 1 +
θ^20 +
θ^40 + · · ·.
Problem 3.21 Slope revisited Use the preceding result for h(θ 0 ) to check your conclusion in Problem 3. about the slope of h(θ 0 ) at θ 0 = 0.
3.6 Summary and problems
Discretization turns calculus on its head. Whereas calculus analyzes a chang- ing process by dividing it into ever finer intervals, discretization simplifies a changing process by lumping it into one unchanging process. Discretiza- tion turns curves into straight lines, so difficult integrals turn into rectan- gles, and mildly nonlinear differential equations turn into linear differen- tial equations. Even though lumping sacrifices accuracy, it provides great simplicity.
Problem 3.22 FWHM for another decaying function Use the FWHM heuristic to estimate Z (^) inf
− inf
dx 1 + x^4 .
Then compare the estimate with the exact value of π/
√
- For a fun additional problem, derive that exact value. Problem 3.23 Hypothetical pendulum equation Suppose the pendulum equation had been d^2 θ dθ^2
tan θ = 0.
How would the period T depend on amplitude θ 0? In particular, as θ 0 in- creases, would T decrease, remain constant, or increase? If T would not re- main constant, what is the slope at zero amplitude, i.e. T 0 ( 0 )? For small but nonzero θ 0 , find an approximate expression for the dimension- less period h(θ 0 ) and use it to check your previous conclusions. Problem 3.24 Gaussian one-sigma tail The tail of the Gaussian distribution is important in statistical inference. The Gaussian probability density function with zero mean and unit variance is
Chapter 3. Discretization 81
p(x) = e−x^2 / √ 2π
.
In this problem you will estimate the area of the one-sigma tail: Z (^) ∞
1
e−x^2 / √ 2π
dx.
a. Sketch the above Gaussian and shade the one-sigma tail. b. Use the criterium of falling 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 value: Z (^) ∞
1
e−x (^2) / √ 2π
dx = 1 − erf(1/
√ 2 ) 2 ≈ 0.159,
where erf(z) is the so-called error function (defined especially because of the important of this kind of problem).
Problem 3.25 Distant Gaussian tails Problem 3.24 asked about the one-sigma tail of a Gaussian. Now use dis- cretization to estimate the area of the n-sigma tail of a Gaussian, for large n:
t(n) ≡
Z (^) ∞
n
e−x^2 / √ 2π
dx.
