Proposition 52

Table of Contents

PROPOSITION 52 Problem

According to hypothesis of uniform gravity acting uniformly downwards, to find the continuous curve MAN, upon which all the semi-oscillations can be completed in equal times.

Solution

Therefore let MAN be the continuous curve (Fig. 56) and on that AP = x, AM = t, and AN = s.

A new indeterminate [i. e. variable] z is assumed, and both x and t are thus given in terms of z, so that on putting z positive the part AM of the curve is produced, while on putting z negative the negative part AN is produced. Now since for the other part, x maintains the same value, x must be such a function of z, which remains the same, if z is taken either to be positive or negative, or x must be an even function of z.

Then t must be a function of this kind of z, in order that it produces s, if – z is put in place of z. But since the arc s falls on the other side of the axis, the value of this is negative with respect to the curve AM; whereby, if in the value t, – z is put in place of z, it must produce – s. Now let R be an odd function of z and S an even function of z, and put t = R + S; this becomes – s = – R + S ; hence t + s = 2R. Let the length of the isochronous pendulum be equal to a; since this is 2a = f + h , it follows that t + s = 2 2ax and hence 2 R = 2ax and x = R

Moreover, since x must be an even function of z, from this … expression, that by itself can be obtained; for since R is an odd function, the square of this is an even function.

Therefore let R = z; then z = 2ax and S must be an even function of 2ax or of x . From which done this equation is obtained : s = 2ax − S for all the continuous tautochrone curves. [p. 230] Let dS = Tdx ; then T is some odd function of 2ax x . Wherefore it becomes ds = adx −Tdx and 2 ax on putting PN = y. From which equation an infinite number of tautochronous curves are found. Q.E.I

Corollary 1

Therefore the curve AN found in this manner is a tautochrone with the continuous part AM of itself. Now by the preceding problem an infinite number of other curves AM are given, which joined with the curve AN produce isochronous oscillations.

Corollary 2

By the preceding proposition, all the curves AM, of which this is the equation : produces isochronous oscillations with the curve AN . But the length of the isochronous pendulum of these oscillations is equal to ( a + c )2 . 4

Corollary 3

Hence among these infinite curves AM with AN producing isochronous oscillations is that continuous with AN, in which c = a. And the length of the isochronous pendulum becomes equal to a, as we have assumed.

Corollary 4

If we put c = 0, then also this curve AM, of which the equation is dt = Tdx or 2ax t = S, is tautochronous with the curve AN . And in this case the length of the pendulum is a . [p. 231] Hence as often as T = 2bx , so also tautochronism is produced with the 4 right line AN, if thus it is inclined at an angle, such that the secant of the angle MAP is equal to ba .

Corollary 5

Since the curve AN must be normal to the axis AP at A, it is required that T vanishes on putting x = 0. Also likewise it follows from this, since a – T must be a positive quantity, even at the starting point A. For if T should become infinite on putting x = 0, agreeing with infinite modes, yet thus, as S vanishes on putting x = 0, the curve AN falls on the other part of the axis AP and the curve has a cusp at A and the body, after descending on MA, ascends on reflection on AN, which would be contrary to the nature of the oscillations.

Corollary 6

Therefore if T vanishes on putting x = 0, the radius of osculation at A, which is sds , dx as s = y in this place is equal to a and thus the oscillations agree with the smallest oscillations of the pendulum of length a, as we assumed.

Corollary 7

The part of the curve AN has a vertical tangent at D and a cusp there ; since the point is found from this equation a − T = 2ax on taking AC equal to the value of x from this equation. The other part too AM has a cusp, if somewhere it becomes : a + T = 2ax .

Corollary 8

If it happens that S = 0 and T = 0, then s = 2ax . Whereby the curve is a cycloid and the part AN is equal and similar to the curve AM. Hence it is a continuous cycloid curve, upon which all the oscillations are completed in the same time.

Example

Let T = 2bx , in which case the curve AN is also a tautochrone with the right line AC making an angle, the cosine of which is Moreover, there is obtained dy = a ; then the equation arises : b dx ( a 2 − 2a 2bx + 2bx − 2ax ) 2 ax .

Which equation also agrees with that, which we found for the curve in the preceding proposition, which constitutes a tautochrone with a straight line (452), if L is written for a and n for a . Whereby if b = a, an algebraic curve NAM also is found, which is a b tautochrone, the equation of which is : dy = dx a − 22 x2ax and the integral of this is :

Which is that curve, which constitutes a tautochrone with the vertical right line, as we found above (452). Now the length of the isochronous pendulum is equal to a, if the body is oscillating on this curve. But if it is moving on the right line AC and on part of the curve AN, the length of the isochronous pendulum is a4 . And if D is the cusp of the curve, … then AC = 8 , now the other root AM rises to infinity. Besides this algebraic tautochrone curve others are easily found.