and because BD=semicircle BnS, it is plain that Dr or mP is equal to the arc nS, or to the arc mC; and therefore the ordinate NP Nm+mP=Nm+ arc Cm; hence P is a point in the cycloid by art. 328.

333. Cor. If OA be another semi-cycloid touching OFB in O, and placed in a contrary situation, as in the figure; it is plain that if a thread and weight P be suspended at O, and made to vibrate between the curves AO and OFB, the weight P will move in the arc of a cycloid ACB.

PROPOSITION LXXXIV.

334. Perpendicular to CO (fig. 135.) draw lq, and from the centre C, with radius equal to the arc CP, describe the semicircle Izq; draw pw parallel to PN, and indefinitely near to it, and make Co=are Cp; erect os perpendicular to lq, and draw the other lines as in the figure: then if the pendulum begin to vibrate from P, the velocity in the cycloid at any other point p, will be as the ordinate os; and the force by which the pendulum is accelerated in p, is as the arc PC which remains to be described.

For, by art. 314, the velocity at p is the same as would have been acquired by a body in falling perpendicularly from N to w, and the velocity at C, the same as would have been acquired by the same body in falling from N to C; but by art. 315, these velocities are as √(Nw) to (NC) and

(by 18, 4. Em. Geom.) NC: wC :: Cm2: CE2, and by division, NC: Nw :: Cm2 : Cm2-CE2; but Cq=CP=2Cm, and Co=Cp=2CE, therefore, NC: Nw : : Cq2 ; Cq2~Co2=lo × oq or ✔(NC): ✔(Nw) : : Cq : √(Cq2 —Co2)=√(lo × oq) that is, the velocity of the pendulum at C=√(NC): its velocity in p✔(Nw) : : Cq : √(Cq2—Co3 : : Cq ; os.

Again, let DC represent the force of gravity, or the weight of the body, this force may be resolved (art. 69.) into the two Dm, mC, the former of which, Dm, being parallel to FP, serves only to stretch the string, while the other, mC, is entirely employed in accelerating the pendulum at P, being parallel to the tangent at P; but CP=2CM, by art. 330; therefore, the force which accelerates P is proportional to CP.

335. Cor. Hence, it is evident that the part of the whole weight which the string sustains at any point P, is to the weight of the pendulum, as Dm to DC.

PROPOSITION LXXXV. ·

336. The time in which a pendulum vibrates in the arc of a cycloid LCP, (fig. 135.) is equal to the time in which a body would describe the arc of the semicircle Izq, with the velocity acquired in C continued uniformly.

Let nv be an ordinate indefinitely near to os, and draw st parallel to lq; then the triangles vst and Cso are similar, and vs: st=no :: Cs: os; but, by art. 334, the velocities are as Cs os, they are, therefore, as vs to no; consequently the spaces vs, no, will be described in the same time; because when the spaces are as the velocities, the times are equal.

In the same manner it may be proved that the other corresponding parts of the lines Cq, qz, will be described in the same time; and therefore the whole time of vibration through PCL, is equal to the time of describing lqz, with the velocity in C continued uniformly.

PROPOSITION LXXXVI.

337. The time of oscillation in a cycloid, is to the time of descent down its axis, as the circumference of a circle to its diameter, or as 3,14159: 1.

For the time in which the semi-circumference q is described by a body, is to the time in which the radius Cq would be described with the same velocity, as the circumference of a circle to its diameter.

But the time in which the semi-circumference, zq is described, is equal to the time of vibration in the cycloid PCL, by art. 336.

Now the time in which a body falls down mC is equal to the time in which PC=2mC would be described by the velocity acquired at C, (art. 282.) and the time of descent through mC, is equal to the time of descent through CD (art. 302.) Therefore, the time in which PC would be described by a velocity equal to that acquired at C continued uniformly, is equal to the time of descent through the diameter DC.

Consequently the time of one vibration through PCL, is to the time a body would be in descending down the diameter CD, as the circumference of a circle to its diameter.

338. Cor. 1. At all places in the same latitude, the oscilla tions of the same pendulum in a cycloid, are performed in equal times, whether those oscillations be made through great or small arcs.

For the times of oscillation are all in a given ratio to the time in which, a body falls through the axis CD, and are, therefore, all equal.

339. Cor. 2. If, with the radius OC, an are of a circle bCa be described, it is obvious that this are coincides with the cycloidal arc for some small distance on each side of the point C; therefore, the time of an oscillation in a small circular arc is nearly the same as in the cycloid.

340. Cor. 3. The times of vibration in different cycloids, or in small circular arcs of unequal radii, are as the square roots of the lengths of the pendulums, when the force of gravity is constant.

For in the same pendulum the time is as the time of descent through DC, that is as (DC), or as ✔(O€).

341. Cor. 4. The number of vibrations which the same pendulum makes in a given time, at the same place, is inversely as the square root of the length of the pendulum.

For let n be the number of vibrations, and t the time of each vibration; then the whole time is nt, which is given;

1

that is, nt 1, and n; but ta(L) the length of the

pendulum, by the last art. therefore nor

1

342. Cor. 5. Because nx, therefore n✓L1, and

La; or the lengths of pendulums are inversely as the square

of the number of vibrations made in a given time.

343. Cor. 6. When the force of gravity raries, the time of one vibration is as the square root of the length of the pendulum directly, and the square root of the force of gravity inversely.

For, by art. 275, the time t∞

where space described,

force of gravity; but in this case, s, therefore

and

1L

tx

or

声

because is a given quantity.

344. Cor. 7. When the length of the pendulum, or Lis

1

given, then to, and ƒœ; or the force of gravity is as fo the square of the time inversely,

345. Cor. 8. The force of gravity at the equator, is to the force of gravity in any given latitude, as the length of a pendulum vibrating seconds at the equator, to the length of a pendulum vibrating seconds in the given latitude.

For when the time t is given,✔ is to 1, and fxL.

346. Cor. 9. The space through which a body falls by the force of gravity in the time of one vibration of a pendulum in a cycloid, or small circular arc, is to half the length of the pendulum, as the square of the circumference of a circle, to the square of its diameter, or as 3,141592 : 1.

For, put the time of one vibration T, and the spaces a body would fall through in that time=S; the time of descent down half the length of the pendulum-t, and half the length of the pendulums, D the diameter, and C the circumference of a circle; then, by art. 337.

T:t::C:D,

and T2: t:: C2: D2.

But by art. 274, the spaces are as the squares of the times, that is, Ss: T2: t';

347.

therefore, ex equo S: s:: C2: D2.

Scholium. From the preceding articles we easily deduce some practical results.

1. To find the length of a pendulum that shall make any number n of vibrations in a given time, as one minute.

1

By art. 342, L; that is, the length is inversely as the

n

square of the number of vibrations; but to apply this to practice, we must have, front experiment, the length of a pendulum which vibrates once in a second, or 60 times in a minute, in a given latitude; now in the latitude of London, a pendulum vibrating seconds has been found to be 391 inches, therefore, by the above proposition, n2; 60: 391: L= 391 × 602 the length sought.

Ex. 1. Let it be required to ascertain the length of a pendulum that shall vibrate 30 times in a minute, or once in 2 seconds.