Isochronism: Analysis of the Brachistochrone andthe Tautochrone as they relate to the Simple and

Cycloidal Pendulum

Covello, B., Stambaugh, N.

April 1, 2013

i

Contents

1 The Simple Pendulum 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Taylor Expression for the Second Order Non-Linear ODE . . . . . . 31.3 An Alternative Approach . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Elliptic Integrals of the First Kind . . . . . . . . . . . . . . . . . . . 8

2 The Brachistochrone 102.1 Euler-Lagrange Solution . . . . . . . . . . . . . . . . . . . . . . . . 112.2 The Cycloid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 The Tautochrone 133.1 Establishing a Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4 The Cycloidal Pendulum 164.1 The Evolute of a Cycloid is a Cycloid . . . . . . . . . . . . . . . . . 16

5 References 18

ii

1 The Simple Pendulum

1.1 Introduction

In 1599, Galileo studied cycloids, the path drawn by a restricted point on the edgeof a circle as it rolls along a flat plane. Soon after this it was discovered thatthis curve is the solution to the tautochrone problem (Hyugenes, 1659) and thebrachistochrone problem (Bernoulli, 1697). Hyugenes made his discovery whileattempting to find a pendulum whose period was truly independent of amplitude.The cycloidal pendulum bypasses the need for the small angle approximation com-monly utilized in introductory physics courses. Bernoulli recognized that his so-lution to the brachistochrone problem gave the same curve, but to this day it isconsidered a coincidence. This project aims to elucidate the intrinsic propertiesthe cycloid path presents to the formulation of the aforementioned problems. Thedifferences amongst the regular pendulum and the cycloidal pendulum will be an-alyzed using Taylor series approximations and Lagrangian mechanics.

Consider the classical pendulum with arm length l, having one end fixed ata point and the other end attached to a mass m swinging with angle θ from theverticle position. The acceleration of such a system may be described by thefollowing:

F = ma = −mg sin θ (1)

x = lθ

v = lθ̇

a = lθ̈ (2)

Where F is the restoring force, x is the arc length, v is the velocity, and a isacceleration. By substituting 2 into 1 we form:

mlθ̈ = −mg sin θ

θ̈ =−gl

sin θ (3)

1

If we let K represent kinetic energy and V represent potential energy, then theLagrangian L may be represented as:

L = T − V =1

2lθ̇2 −mgl(1− cos θ) (4)

Substituting this into the Euler Lagrange Equation:

d

dt

(∂L

∂θ̇

)− ∂L

∂θ= 0

d

dt

(∂ 1

2lθ̇2 −mgl(1− cos θ)

∂θ̇

)−∂(1

2lθ̇2 −mgl(1− cos θ))

∂θ= 0

d

dt(ml2θ̇) +mgl sin θ = 0

ml2θ̈ = −mgl sin θ

θ̈ =−gl

sin θ (5)

Two aspects of this equation should be noted. First, the Euler Lagrange formula-tion agrees with the normal Newtonian mechanic definition of force as expected.Second, this is a second order nonlinear ODE for which there are no concreteanalytical methods for solving exactly. In many elementary physics courses, it istaught that the frequency of the pendulum is completely independent of ampli-tude. This holds true given a restriction on θ to be small, and approximatingsin θ ≈ θ. This approximation is conveinent, as it reduces the equation to a secondorder linear ODE given by:

θ̈ =−glθ (6)

This equation represents the simple harmonic oscillator with a period given by:

T = 2π

√l

g

What are the consequences of this approximation, and how might one be able tomodify the non-linear solution so no small angle approximations need be made?We first turn our attention to attempting to solve our second ordered non-linearODE through a Taylor approximation series.

2

1.2 Taylor Expression for the Second Order Non-Linear ODE

Consider the differential equation −y′′(t) = sin(y(t)). Using the series expansionfor sine, and assuming a power series expansion for y(t) =

∑ant

n, we search forthe recursive pattern in the coefficients. Recall y′′(t) =

∑∞n=0(n+ 2)(n+ 1)ant

n.

−y′′(t) =y − 1

3!y3 +

1

5!y5 − 1

7!y7 + . . .

−1(2a2 + 6a3t+ 12a4t2 + 20a5t

3 + 30a6t4) =a0 + a1t+ a2t

2 + a3t3 . . .

− 1

3!

(a0 + a1t+ a2t

2 + a3t3 . . .

)3+

1

5!

(a0 + a1t+ a2t

2 + a3t3 . . .

)5− 1

7!

(a0 + a1t+ a2t

2 + a3t3 . . .

)7+ . . .

There is no obvious pattern within this first equation, indeed, it seems torepresent the chaotic nature so characterized by non-linear second order ODE.Yet we continue, collecting together the constant terms first. This is given by:

−2a2 = a0 −1

3!a30 +

1

5!a50 −

1

7!a70 + . . .

−2a2 = sin(a0)

As above, we now collect the linear terms n = 1. We are to imagine all themultiplicative possibilities that will generate a first degree linear term. Thus,we expect all constant to resemble the form an−10 a1. This restriction comes fromthe fact that any addition a1 terms would generate higher degrees of t. In orderto describe the possibilities of these terms we turn our attention to the field ofcombinatorics to find a generation of nC1 = n terms.

−6a3 = a1 −1

3!(3a20a1) +

1

5!(5a40a1)−

1

7!(7a60a1) + . . .

−6a3 = a1(1−1

2!(a20) +

1

4!(a40)−

1

6!(a60) + . . .)

−6a3 = a1 cos(a0)

For the n = 2 term, a t2 is generated in one of two ways: either a constantmultiplied by t2 or two linear terms multiplied together. This may be visualized

3

below:

−12a4 = a2 −1

3!(3a20a2 + 3a21a0) +

1

5!(5a40a2 + 10a21a

30)−

1

7!(7a40a2 + 21a21a

50) + . . .

−12a4 = a2 −1

2!a20a2 −

1

2!a21a0 +

1

4!a40a2 +

2

4!a21a

30 −

1

6!a40a2 −

3

6!a21a

50 + . . .

−12a4 = a2 −1

2!a20a2 +

1

4!a40a2 −

1

6!a60a2 −

1

2!a21a0 +

2

4!a21a

30 −

3

6!a21a

50 + . . .

−12a4 = a2

(1− 1

2!a20 +

1

4!a40 −

1

6!a60 . . .

)− a21

2

(2

2!a0 −

4

4!a30 +

6

6!a50 − . . .

)−12a4 = a2 cos(a0)− a21

(a0 −

1

3!a30 +

1

5!a50 − . . .

)−12a4 = a2 cos(a0)−

a212

sin(a0)

We continue our established routine with n = 3.

−20a5 = a3 −1

3!(3a20a3 + 6a0a1a2 + a31) +

1

5!(5a40a3 + 20a30a1a2 + 10a20a

31)

− 1

7!(7a60a3 + 42a50a1a2 + 35a40a

31) + . . .

−20a5 = a3 −1

2!a20a3 − a0a1a2 −

1

3!a31 +

1

4!a40a3 +

1

3!a30a1a2 +

1

4!2a20a

31

− 1

6!a60a3 −

1

5!a50a1a2 −

1

6!5a40a

31 + . . .

−20a5 = a3(1−1

2!a20 +

1

4!a40 −

1

6!a60 + . . .)− a1a2(a0 −

1

3!a30 +

1

5!a50 . . .)

− 1

6a31(1−

1

2a20 +

1

4!a40 + . . .)

−20a5 = a3 cos(a0)− a1a2 sin(a0)−1

6a31 cos(a0)

In order to continue to have enough terms for generalization, we must includeadditional terms given by y9.

−30a6 =a4 −1

3!(3a21a2 + 3a0a

22 + 6a0a1a3 + 3a20a4)

+1

5!(5a0a

41 + 30a20a

21a2 + 10a30a

22 + 20a30a1a3 + 5a40a4)

− 1

7!(35a30a

41 + 105a40a

21a2 + 21a50a

22 + 42a50a1a3 + 7a60a4)

+1

9!(126a50a

41 + 252a60a

21a2 + 36a70a

22 + 72a70a1a3 + 9a80a4)

4

Note that the sum of n in each term Xay0an is n = 4. For example:

4 → a43+1 → a1a32+2 → a22

2+1+1 → a21a21+1+1+1 → a41

The a0’s then fill in as needed to make the correct length partition based on thepower of y that we’re dealing with. These terms may then be separated basedupon correct partitions:

−30a6 =a4 −1

3!3a20a4 +

1

5!5a40a4 −

1

7!7a60a4 +

1

9!9a80a4

− 1

3!6a0a1a3 +

1

5!20a30a1a3 −

1

7!42a50a1a3 +

1

9!72a70a1a3

− 1

3!3a0a

22 +

1

5!10a30a

22 −

1

7!21a50a

22 +

1

9!36a70a

22

− 1

3!3a21a2 +

1

5!30a20a

21a2 −

1

7!105a40a

21a2 +

1

9!252a60a

21a2

+1

5!5a0a

41 −

1

7!35a30a

41 +

1

9!126a50a

41

Simplifying the factorials, and factoring out so that the first coefficient on eachline is 1 or a0,

−30a6 =a4

(1− 1

2!a20 +

1

4!a40 −

1

6!a60 +

1

8!a80 ± . . .

)− a1a3

(a0 −

1

3!a30 +

1

5!a50 −

1

7!a70)± . . .

)− a22

2

(a0 +

1

3!a30 −

1

5!a50 +

1

7!a70 ± . . .

)− a21a2

2

(1 +

1

2a20 −

1

4!a40 +

1

6!a60 ± . . .

)+a414!

(a0 −

1

3!a30 +

1

5!a50 ± . . .

)Finally, we can make all of our substitutions for sin(a0) and cos(a0).

−30a6 =a4 cos(a0)− a1a3 sin(a0)−a222

sin(a0)−a21a2

2cos(a0) +

a414!

sin(a0)

5

1. There is one term for each partition of n. For each partition, label it λ1 +λ2 + . . .+ λk.

2. The (non-numeric) coefficient out front is aλ1aλ2 . . . aλk .

3. The numeric coefficient is 1/m, where m is the number of indistinguishablerearrangements of the partition. So if a part appears r times, include a factorof 1/r!.

4. The sign is the sign of ik, so + + - - + + - - + + - - ...

5. If k is odd, then finish with cos(a0), otherwise, sin(a0).

Thus, we get:

−42a7 =a5 cos(a0)− a1a4 sin(a0)− a2a3 sin(a0)−1

2a21a3 cos(a0)−

1

2a22a1 cos(a0)

+1

6a31a2 sin(a0) +

1

120a51 cos(a0)

Summary without recursion:

a2 = −1

2sin(a0)

a3 =1

6a1 cos (a0)

a4 =1

24a21 sin (a0) +

1

24cos (a0) sin (a0)

a5 =1

120a31 cos (a0) +

1

120a1 cos (a0)

2 − 1

40a1 sin (a0)

2

a6 = − 1

720a41 sin (a0)−

11

720a21 cos (a0) sin (a0)−

1

720cos (a0)

2 sin (a0) +1

240sin (a0)

3

a7 = −11a31 cos (a0)

5040− a51 cos (a0)

5040− a1 cos (a0)

3

5040+

1

336a31 sin (a0)

2 +11

1680a1 sin (a0)

2

If we only look at the case where a1 = 0 (that is, it starts from rest ...), thenit will be an even function (so all aodd = 0) ...

a2 = −1

2sin(a0)

a4 =1

24cos (a0) sin (a0)

a6 = − 1

720cos (a0)

2 sin (a0) +1

240sin (a0)

3

6

Note that if sin(a0) = 0, then all terms are zero. This corresponds to releasingit at rest at the top of bottom, so it will never move. This is easily verified byexamining the following at sin(a0) = 0:

a2 = −1

2sin(a0) = 0

a4 =1

24cos (a0) sin (a0) = 0

a6 = − 1

720cos (a0)

2 sin (a0) +1

240sin (a0)

3 = 0

Indeed, it may be noted that all terms of an are characterized with sine terms.If sin(a0) = 1 and cos(a0) = 0, then we get the solution

y(t) =π

2− 1

2t2 +

1

240t6 − 1

1920t10 +

11

13977600t14

1.3 An Alternative Approach

Recently, a different approach has been suggested in the literature as follows:Assume y′′(t) = − sin(y(t)), y(0) = π

2and y′(t) = 0. Then by assumption, y′′(0) =

− sin(t) = 1.

y′′′(t) =d

dt(− sin(y(t)))

= −y′(t) cos(y(t))

y′′′(0) = − cosπ

2· 0 = 0

Continuing in this way,

y(4)(t) = y′(t)2 sin y(t)− cos y(t)y′′(t))

= y′(t)2 sin(y(t)) + cos(y(t)) sin(y(t))

= sin(y(t))y′(t)2 + cos(y(t)))

y(4)(0) = 0

y(t) = y(0) + y′(0)t+1

2!y′′(0)t2 +

1

4!y(4)(0)t4 + . . .

=π

2+ 0t+

1

2!(−1)t2 +

1

3!0t3 +

1

4!0t4 +

1

5!0t5 +

1

6!3t3 + . . .

7

=π

2− 1

2t2 +

1

240t6 + . . .

This approach is quickly getting out of control, while the previous one has theaesthetic advantages of quickly explaining why so many terms drop out and beinga better way to jump through the recursion.

We now turn our attention back to the simple pendulum without the smallangle approximation in search of other means to confirm this power series.

1.4 Elliptic Integrals of the First Kind

Recall that our second order non-linear ODE without the small angle approxima-tion is given by:

d2θ

dt2=−gl

sin θ

We allow gl

= ω2 with the system having initial conditions to be given by θ(0) = θ0and θ̇(0) = φ0 Multiplying each side by dθ

dt, we write:

d2θ

dt2= −ω2 sin θ

At this point we could find∫

d2θdt2

+ ω2 sin θ = 0, however, we simply use the Lawof Conservation of Energy from physics to simply derive our next equation.

Recall that the total energy E of a system must remain constant. Thus as thependulum swings for a given time t, conversion of potential and kinetic energytake place at the same rate. Recall our kinetic energy T = 1

2lθ̇2, and our potential

energy V = mgl(1− cos θ). We chose our zero potential level to be at the bottomof the pendlum. We now note that ∆T = −∆V . This yields the following:

1

2θ̇2 = ω2 cos θ0 − ω2 cos θ

.

dt =dθ

ω√

2(cos θ0 − cos θ)

8

We now wish to find the period given from 0 to θ0, noting that from 0 to θ0 ourtime t = T

4, where T represents the period. Additionally, we use the trigonometric

identity cos 2x = 1− sin2 x to yield:

T

4=

1

ω√

2

∫ θ0

0

dθ√sin2 θ0

2− sin2 θ

2

(7)

We now make the subsitution u =sin θ

2

sinθ02

, finding θ = 2 arcsin(u sin θ02

). Recall

ddx

arcsin(x) = dx√1−x2 . This yields:

dθ =2 sin θ0

2

1− u2 sin2 θ02

du

Substituting this into equation 7 yields:

T

4=

1

2ω

∫ 1

0

2 sin θ02du√

1− u2 sin2 ω2

√sin2 θ0

2−(

sin2 θ02u

)2

=1

ω

∫ 1

0

sin θ02du√

1− u2 sin2 θ02

√sin2 θ0

2(1− u2

=1

ω

∫ 1

0

du√1− u2 sin2 θ0

2

√1− u2

We now introduce the elliptic modulus k = sin θ02

. Thus, our resulting integralis given by:

T

4=

1

ω

∫ 1

0

du√1− u2k2

√1− u2

(8)

This is known as a complete elliptical integral of the first kind K. This integralmay be expressed as a special case of the Gauss hypergeometric function multipliedby π

2, where a = 1

2, b = 1

2, c = 1, x = k2. This may be represented as 1F2(

12, 12, 1, k2).

Thus,

K(k) =π

2 1F2(

1

2,1

2, 1, k2) =

π

2

∞∑n=0

[ (2n)!

22n(n!)2

]2k2n =

π

2

∞∑n=0

[P2n(0)

]k2n (9)

P2n is the Legendre Polynomial. This expression is equivalent to:

9

We may now calculate the exact period of the non-linear pendulum given by:

T =4

ωK(k) = 2π

√l

g

∞∑n=0

((2n− 1)!!

(2n)!!

)2

k2n (10)

A period dependence graph based on k along with the phase portrait for thependulum is modeled below:

2 The Brachistochrone

We now turn our attention to the brachistochrone. Arising from the greek word“shortest time”, the Brachistochrone problem aims to minimize a path from a fixedpoint a to another fixed point b under the influence of gravity and starting fromrest. In other words:

We are given two fixed points in a vertical plane. A particle starts from rest atone of the points and travels to the other under its own weight. Find the path

that the particle must follow in order to reach its destination in the briefest time.

We begin with y(x0) = y0 and y(0) = 0, choosing a zero of potential energy givenby:

V (y) = −mgy

10

The kinetic energy is therefore

T (y) = −V (y) =1

2mv2.

And the speed of the particle is given by

v(y) =√

2gy.

The distance traveled is

ds =√

(dx)2 + (dy2) =

√√√√1 +

(dy

dx

)2

dx.

The change in time required to travel ds is

dt =ds

v=

√1 +

(dydx

)2√

2gydx,

and the total time of descent is given by

T =

∫ x0

0

√1 +

(dydx

)2√

2gydx.

Our Lagrangian may be written as

L =

√1 + y′2√y

2.1 Euler-Lagrange Solution

The Euler-Lagrange equation may be represented by:

d

dx

(∂L

∂y′

)=∂L

∂y=⇒ d

dx

(1√yy′

1√1 + y′2

)= −

√1 + y′2

2y√y

− y′2

2y√y(1 + y2)

+y′′√

y(1 + y2)− y′2y′′

√y(1 + y′2)

32

=

√1 + y2

2y√y

We now multiply by 2y√y(1 + y′2)3/2 to get:

−2yy′′ = 1 + y′2

11

∫2yy′′

1 + y2=

∫y′

y=⇒ ln(1 + y′2) = −lny + A

=⇒ 1 + y′2 =B

y

where we let B = eA. We now integrate to solve for y’:√ydy

√B − y

= dx

We now let y = B sin2 φ, dy = 2B sinφ cosφdφ. The equation above simplifies to:

2B sin2 φdφ = dx (11)

Recalling that sin2 φ = (1 − cos 2φ)/2 we integrate 11 to get B(2φ − sin 2φ + C).Thus 2y = B(1− cos 2φ). Let θ = 2φ. Thus, we have:

x = a(θ − sin θ) + d, y = a(1− cos θ) (12)

Having initial conditions (x, y) = (0, 0), and θ0 = 0 leads to the final equations forthe brachistochrone solution:

x = a(θ − sin θ), y = a(1− cos θ) (13)

Thus, we arrive at the parameterization for a cycloid.

2.2 The Cycloid

In finding the solution to the brachistochrone, we have stumbled upon the cycloid.A cycloid may be defined as the path of a point on the rim of a circle as it rollsalong a straight line.

x = a(θ − sin θ), y = a(1− cos θ) (14)

Where a is the radius of the generating circle. The cycloid satisfies the equation(dx

dy

)2

=2a

y− 1

It is important to note that in our brachistochrone problem, the cycloid is inverted,as our path is falling under the force of gravity.

12

3 The Tautochrone

Having found that the solution for the brachistochrone is a cycloid, the naturalquestion arises: How long does it take for an object to travel the path of an invertedcycloid? In other words, what is its period?.We first seek the time it takes to for the object to travel from the top of the cycloidto the bottom of the cycloid. Recall that a transfer of kinetic and potential energywill take place such that:

1

2mv2 = mgy =⇒ v =

√2gy =

ds

dt=⇒ dt =

√(dx)2 + (dy)2√

2gy(15)

Taking the derivates of 14 yield:

x′ = a(1− cos θ), y′ = a sin θ =⇒ x′2 + y′2 = 2a2(1− cos θ) (16)

By substituting 16 into 15 we find:

dt =a√

2(1− cos θ)dθ√2ga(1− cos θ)

dθ

Thus the time of travel from the top of the cycloid to the bottom may be givenby:

t =

∫ π

0

a√

2(1− cos θ)dθ√2ga(1− cos θ)

dθ =

√a

gπ (17)

Let us recalculate the time, this time beginning at some initial angle y0 and θ0:

t =

∫ 0

y0

√(dx)2 + (dy)2√

2g(y0 − y)=

√a

g

∫ π

θ0

√1− cos θ

cos θ0 − cos θdθ (18)

Using the identity cos θ = 2 cos2 θ2− 1, we find:

t = π

√a

g

∫ π

θ0

sin θ2dθ√

cos2 θ2− cos2 θ

2

(19)

13

We perform a u substitution in which u =cos θ

2

cos θ2

, du = − sin θ2dθ

2 cos θ2

, so that:

t = −2

√a

g

∫ 0

1

du√1− u2

= π

√a

g

Thus, the total time period will be 4t

T = 4π

√a

g(20)

We shown two important facts.

1) The period is completely independent of amplitude when the objecttravels on the path of a cycloid.

2) If an object begins on the path of a cycloid, the time it takes for it

to travel to the lowest gravitational point will be π√

agregardless of

the beginning initial height.

Point number 2 is an important point, for it answers the following question asdescribed by the tautochrone problem: What is the path in which the time takenby an object sliding without friction in a uniform gravitational field to its loweststarting point is independent of its starting point? We have shown that the solu-tion to the tautochrone is a cyloid.

3.1 Establishing a Bridge

In 1658 Hyugen’s proved that the path of isochronous oscillation is the tautochrone,and that the path is a cycloid. 40 years later, Bernoulli discovered the solutionto the brachistochrone. Bernoulli noted that the curves were related. Hyugen’stautochrone is a complete cycle of the inverted cylcoid, and the brachistochrone isa segment of an inverted cycloid an initial point at an apex.

There are several differences to be noted between the tautochrone and thebrachistochrone. First, in a tautochrone, the total height difference is dependentupon the total horizontal distance. In the brachistochrone, the total horizon-tal length and maximum difference of height must be independent, as there is abrachistochrone solution between any two points.

In 2010, Munoz et al noted the following: “The time it takes for a point likeparticle to descend along an inverted cycloid to a minimum of the path is indepen-dent of ∆y, given by: Tα(∆y)0. This is in stark contrast to an inclined plane in

14

which Tα(∆y)12 .” They began to construct a paths that complied with Tα(∆y)β.

They shought to introduce a family of curve that complied with this relation forvalues −1

2< β ≤ 1

2. In their paper they found that a curve σ is identified by four

different parameters, β, xL, yL, h such that:

xσ(y) = xL + h

∫ y−yLh

0

√η2β−1 − 1dη (21)

Through this family of curves they proved the following:Lemma 1 There does not exist a continuously differentiable bounded curve inthe plane, in which a classical non-relativistic point like particle could move un-der the influence of a spatially homogenous time independent gravitational fieldg(x, t) = g, starting with zero intial velocity at position xi, complying with a con-dition ∆tα(∆y)β for values of β > 1

2, or β ≤ −1

2where ∆y := −g · (xi − xL) for a

fixed position xL, δt being the lapse of time required for the particle to move fromxi to xL.

For −12< β < 1

2, the solutions exist, and are given in 21, and are unique up to

an arbitrary combination of reflections, translations, and scale transformationsThe paper also describes a curve with a fixed β and a fixed total height dif-

ference H > 0, proving that there is a minimum total time of descent which theycall σβH . This is given by the parametric equation:

xσH(y) = H

∫ yH

0

√η2β−1 − 1dη , y ∈ (0, H] (22)

They persist to prove the following lemma:Lemma 2 Each element in a family of paths as defined in 22 has a fixed horizontallength to total height difference ratio, independent of H. This ratio is monotonicallydecreasing function of β. A particle starting from rest at height H would descendalong a given path σβH in a time which is also a monotonically decreasing functionof β for a fixed H. Thus, every σβH with 0 < β ≤ 1

2is shorter and swifter than

the half tautochrone and every σβH with −12< β < 0 is longer and slower than

the tautochrone. The path of quickest descent for a given height is a segment of avertical straight line. As β approaches −1

2with a fixed height, the total arch length

of the path grows without limit, as does the total time of descent.

The researchers said the following of their research:

There are some interesting consequences and questions that we have left aside:

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• These curves are independent of the mass of the particle: m is absent in everyexpression, starting from because gravitational mass and inertial mass werecanceled out even before writing this equation. This is just a consequence ofthe equivalence principle.

• The eerily ubiquitous constant Γ12

=√π, present in all our calculations, just

begs the question: are there possible generalizations of our curves in theframe of strong gravitational fields?

• In one-dimensional physics, the harmonic oscillator potential is the isochronouspotential, i.e., this is the only potential for which the frequency is strictlyindependent of the amplitude. Can this potential be derived from the Euler-Lagrange equations of the tautochrone curve? Can this potential be gener-alized for some or all of our curves?

These questions provide a general direction from which to take future researchto truly ascertain if there is a connection between the tautochrone and brachis-tochrone. For now, we turn our attention to the newly established isochronouspathway.

4 The Cycloidal Pendulum

Having found a path in which the period is truly independent of the amplitude,we have now found a way to make a pendulum truly isochronous, completely by-passing the need to make a small angle approximation.

4.1 The Evolute of a Cycloid is a Cycloid

How would one go about constructing a pendulum whose mass follows the path ofa cycloid? To answer this question we resort to evolutes, which are defined as theenvelope of the normals to a given curve. the parametric equation of an evolute:

α = x− (1 + y′2)y′

y′′, β = y +

1 + y′2

y′′(23)

By substituting 15 into 23 we get:

α = −a(sinθ + θ), β = a(3− cosθ)

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Substituting θ = π + θ′ we find:

x′ = a(sin θ′ − θ′), y′ = a(1 + cosθ′)

Where x′ = α + aπ and y′ = β − 2a This shows that the evolute of a cycloid isanother cycloid. We now come upon the structure of the cycloid pendulum:

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5 References

(1) Belendez, A, Exact solution for the nonlinear pendulum, Revista Brasileira deEnsino de Fisica. 4 (2007), pp. 645-648(2) T.H. Fay, The pendulum equation, Int. J. Math. Educ. Sci. Techn. 33 (2002),pp. 505519.(3) J. Benacka, Three spreadsheet models of a simple pendulum, SiE 3 (2008), pp.5969.(4) J. Benacka, A better cosine approximate solution to pendulum equation. Int.J. Math. Educ. Sci. Techn. 40 (2008), pp. 307308(5) M.I. Molina, Simple linearization of the simple pendulum for any amplitude,Phys. Teach. 35 (1997), pp. 489490.(6) R.B. Kidd and S.L. Fogg, A simple formula for the large-angle pendulumperiod, Phys. Teach. 40 (2002), pp. 8183.(7) G.E. Hite, Approximations for the period of a simple pendulum, Phys. Teach.43 (2005), pp. 290292.(8) F.M.S. Lima and P. Arun, An accurate formula for the period of a simplependulum oscillating beyond the small-angle regime, Am. J. Phys. 74 (2006), pp.892895.(9) Flores, E.V. and Osler, T.J., 1999, The tautochrone under arbitrary potentialsusing fractional derivatives, American Journal of Physics, 67, 718722.(10) Osler, T.J. and Flores, E.V., 2001, The rotating tautochrone, Journal ofApplied Mechanics, 68, 353356.

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