SiiB5j25iit3

TR^CT 38.

PROBLEMS.

347

Now if we suppose w = 1-grain, w = 5lb troy, or 28800grains, and 2/ = ab = 3 feet; the velocity at c becomes

8 x [ft-i x 28800

aV - - - = llllfa. So that

if a —Vo' inc. the greatest veloc. is ft. per sec.

it a — 1 inc. the greatest veloc. is 92||r ft. per sec.

if a = 6 inc. the greatest veloc. is 555 T 7 w ft* P er sec.

To find the time t, it is t or —* = 4V

wg

Vi* 1 — *?)"

Hence the correct fluent is t — — x arc to cosine — and

v wg a

radius 1 , for the time in de And when x — 0 , the wholetime in DC, or of half a vibration, is *7854/-^;; and cortseq.

the time of a whoie vibration through d d is 1’5708 /— .

Using the foregoing numbers, namely w — 1, w=28800,and 2/ = 3 feet; this expression for the time gives= 3531., the number of vibrations per second. But if w=2,there would be 250 vibrations per second ; and if w = 100,there would be 35? ® vibrations per second.

PROBLEM XXII.

To determine the same as in the last Problem, when theDistance cd bears some sensible proportion to the Length ab ;the Tension of the Thread however being still supposed aconstant Quantity.

Using here the same notation as in the last problem, andtaking the true variable length ae for ac, it is ae or eb:ce ::2w : — — ~' w * ; the whole motive force from the twm

AE VO* ■* **)

equal tensions w in ae and eb ; and theref. — x —-- =s

n _ _ w V0 a + i 2 )

/is the accelerative force at e. Theref. the fluxional equation

„ 2 - 4w?

is vv or 2gjs = — x

; and the fluents v* = ^ X

8wg

X —

— + .t 1 ). But when x — a, these are 0 =

V (f + a 1 ) ; therefore the correct fluents are v l — —- x

ill

} -a I

a.

&