3i5

MISCELLANEOUS

TRACT 38.

[\/+ <*“) — \/[l 1 + ar)] = —* ( AD — ae). And

hence v = v'[~ r x (ad — ae)] the general expression forthe velocit}' at E. And when e arrives at c, it gives thegreatest velocity there = * (ad — ac)]. Which,

when w = 28800, w ~ 1, 2l — 3 feet, and cd = 6 inchesora foot, is ^(8 x 28800 x 16,'^ X ^'^~ 3 ) = 548feetper second. Which came out 555 ^ in the last problem, bo-using always Ac for ae in the value of/. But when the ex-tent of the vibrations is very small, as of an inch, as itcommonly is, this greatest velocity here will be ■v/S X 2S800x lfij'-j x = 9y nearly, which in the last problem

was 9/y- nearly.

To find the time, it is t or — = *,/— x —:-■w-wp

V ' Hwg ^/[c - V(‘ + * )J

making c — ad — ‘/(l i +a z ). To find the fluent the easier,multiply the numer. and denom. both by /[c + v'(H jr ")l >

so shall t — */— - x —--—— X V\c + V{l z + *’-)].

Swg — x 1 ) L v

Expand now the quantity \/[c + V (l 1 + a' 2 )] in a series,and put d = c + l, so shall t = s / x ^-^ (1 +;

'idl

40«/3 +S ,Pi + l<2dP + 513

---- 7 .- X s &c). Now

2048 aH‘ ‘

X

is = the arc to sine TT

9.(1 + l . 4rP + 5 rll + P 63 HFp X UHcliP

the fluent of the first term „and radius 1, which arc call a; and let p, a, be the fluents ofany other two successive terms, without the coefficients, thedistance of a from the first term a being n; then it is evi-dent that q = x l v = .t 2, 'a, and p = t 2 ”' 2 a. Assume theref.cl — bv — ex 1 ”' 1 Via 1 — x 2 ); then is q or x z v = bp—(2n — l)

e x i„i L - (2 n — 1 )ea' 2 .r™--i

ex w - 2 xV{d

(9n — 1 )(x :ir! x

X 2 ) 4 -

-v/K - * 2 )

= b p —

+

+

br — (2n — l)ea 2 p + ( c 2n— l)t\r*p-f-

ca'p = bp — (2 n — l)ea 2 F + 2nex 2 ?. Then, comparing thecoefficients of the like terms, we find 1 =: 2 en, and b =(J2n— l)ea-; from which are obtained e=—, and b —

Consequently a = ■ ^ ~ 1 •• — -■ — ~ —- , the general