This series gives the time whatever may be the extent of the oscil β lations; but if they be very small, may be omitted in most 2 cases; then g (11) As this equation does not contain the arcs, the time is independent of their amplitude, and only depends on the length of the thread and the intensity of gravitation; and as the intensity of gravitation is invariable for any one place on the earth, the time is constant at that place. It follows, that the small oscillations of a pendulum are performed in equal times, whatever their comparative extent may be. The series in which the time of an oscillation is given however, shows that it is not altogether independent of the amplitude of the arc. In very delicate observations the two first terms are retained; so that for as ẞ is the versed sine of the arc a, when the arc is very small, is the correction due to the magnitude of the arc described, and is the equation alluded to in article 9, which must be applied to make the times equal. This correction varies with the arc when the pendulum oscillates in air, therefore the resistance of the medium has an influence on the duration of the oscillation. 108. The intensity of gravitation at any place on the earth may be determined from the time and the corresponding length of the pendulum. If the earth were a sphere, and at rest, the intensity of gravitation would be the same in every point of its surface; because every point in its surface would then be equally distant from its centre. But as the earth is flattened at the poles, the intensity of gravitation increases from the equator to the poles; therefore the pendulum that would oscillate in a second at the equator, must be lengthened in moving towards the poles. If h be the space a body would describe by its gravitation during the time T, then 2h = gT, and because T2 = π2 h = '. r. ; therefore (13) If r be the length of a pendulum beating seconds in any latitude, this expression will give h, the height described by a heavy body during the first second of its fall. The length of the seconds pendulum at London is 39.1387 inches; consequently in that latitude gravitation causes a heavy body to fall through 16.0951 feet during the first second of its descent. Huygens had the merit of discovering that the rectilinear motion of heavy bodies might be determined by the oscillations of the pendulum. It is found by experiments first made by Sir Isaac Newton, that the length of a pendulum vibrating in a given time is the same, whatever the substance may be of which it is composed; hence gravitation acts equally on all bodies, producing the same velocity in the same time, when there is no resistance from the air. Isochronous Curve. 109. The oscillations of a pendulum in circular arcs being isochronous only when the arc is very small, it is now proposed to investigate the nature of the curve in which a particle must move, so as to oscillate in equal times, whatever the amplitude of the arcs may be. The forces acting on the pendulum at any point of the curve are the force of gravitation resolved in the direction of the arc, and the resistance of the air which retards the motion. The first is the second, which is proportional to the square of the velocity, is expressed by ds - n di in which n is any number, for the velocity is directly as the element of the space, and inversely as the element of The integral of which will give the isochrorous curve in air; but the most interesting results are obtained when the particle is assumed to move in vacuo; then n = 0, and which, multiplied by 2ds and integrated, gives an arbitrary constant quantity. ds2 - 2gz, c being dt2 Letz hat m, fig. 29, where the motion begins, the velocity being zero at that point, then will c = 2gh, and therefore Whatever the nature of the required curve may be, s is a function of z; and supposing this function developed according to the powers of, its differential will have the form, Substituting this value of ds in the preceding equation, it becomes The integral of this equation, taken from z = h to z=0, will give the time employed by the particle in descending to C, the lowest point of the curve. But according to the conditions of the problem, the time must be independent of h, the height whence the particle has descended; consequently to fulfil that condition, all the terms of the *E value of dt must be zero, except the first; therefore b must be zero, and i + 1 = 1, or i; thus ds= az dz; the integral of which is s = 2az1, the equation to a cycloid DzE, fig. 30, with a horizontal base, the only curve in vacuo having the property required. Hence the oscillations of a pendulum moving in a cycloid are rigorously isochronous in vacuo. If r2BC, by the properties of the cycloid r = 2a2, and if the preceding value of ds be put in It is unnecessary to add a constant quantity if z = h when t = 0. If T be the time that the particle takes to descend to the lowest point in the curve where z = 0, then Thus the time of descent through the cycloidal arc is equal to a semioscillation of the pendulum whose length is r, and whose oscillations are very small, because at the lowest point of the curve the cycloidal arc ds coincides with the indefinitely small arc of the osculating circle whose vertical diameter is 2r. 110. The cycloid in question is formed by supposing a circle ABC, fig. 30, to roll along a straight line ED. The curve EAD traced by a point A in its circumference is a cycloid. In the same manner the cycloidal arcs SD, SE, may be traced by a point in a circle, rolling on the other side of DE. These arcs are such, that if we imagine a thread fixed at S to be applied to SD, and then unrolled so that it may always be tangent to SD, its extremity D will trace the cycloid DzE; and the tangent zS is equal to the corresponding arc DS. It is evident also, that the line DE is equal to the circumference of the circle ABC. The curve SD is called the is preferred. The properties of the isochronous curve were discovered by Huygens, who first applied the pendulum to clocks. 111. The time of the very small oscillation of a circular pendulum is expressed by T=TA g r being the length of the pendulum, and consequently the radius of the circle AmB, fig. 31. Also t=22 is the time employed by a heavy g fig. 31. A m B body to fall by the force of gravitation through a height equal to ≈. Now the time employed by a heavy body to fall through a space equal to twice the length of the pendulum will be t = 4r fig. 32. E D that is, the time employed to move through the arc Am, which is half an oscillation, is to the time of falling through twice the length of the pendulum, as a fourth of the circumference of the circle AmB to its diameter. But the times of falling through all chords drawn to the lowest point A, fig. 32, of a circle are equal: for the accelerating force F in any chord AB, is to that of gravitation as AC: AB, or as AB to AD, since the triangles are similar. But the forces being as the spaces, the times are equal: for as F:g::AB: AD and Tt :: it follows that T=t. AB AD F g A C B 112. Hence the time of falling through the chord AB, is the same with that of falling through the diameter; and thus the time of falling through the arc AB is to the time of falling through the chord 2, that is, as one-fourth of the circumference to the dia AB as |