544

MISCELLANEOUS

TRACT .38.

And when x = a , the same then becomes 2 = 4 /—. x log.

1 + v'2 = v/j X log. 1 + k/2 — \/* l°g* 1 + V2 =’688511, the whole number of vibrations niade by the de-scending weight.

But when the lower sign, or — , takes place, the fluent is

1*227091, the whole number of vibrations made by the lesseror ascending weight.

Schol. It is evident that the whole number of vibrations,in each case, is the same, whatever the length of the threadis. And that the greater number is to the less, as 1’5708 tothe hyp. log. of 1 + V2.

Farther, the number of vibrations performed in the sametime t, bj 7 an invariable pendulum, constantly of the samelength a, is = ’781190. For, the time of descending

the.space a, or the fluent of t = when x = a, is t =

Vjj. And, by the nature of pendulums, : */b : : 1

vibr. : V -- the number of vibrations performed in 1 second ;

hence l" : t : : V— • tV— — v/ — 7 ., the constant number ofa a v £/’

vibrations. '

So that the three numbers of vibrations, namely, of theascending, constant, and descending pendulums, are propor-tional to the numbers 1’5708, 1, and hyp. log. 1 + \/2, oras 1 5708, 1, and ’88137.; whatever be the length of thethread.

To determine the Circumstances of the Ascent and Descentof two unequal Weights, suspended at the two Ends of a Threadpassing over a Pulley: the Weight of the Thread and of thePulley being considered in the Solution.

3 X SPi

3*1416 X

gives IpV

4 x 193

*/— = arc to rad. 1 and vers. —. Which, when x — a .

4 irf a

193

PROBLEM XX.