TRACT 38.

PROBLEMS.

519

equation between any two successive terms, and by means ofwhich the series may be continued as far as we please. Andhence, neglecting the coefficients, putting a = the first term,namely, the arc whose sine is —, and b, c, d, &c, the follow-er. £ 1! T n °'' — *V(°? “

ing terms, the senes is as follows, a -f- •—

2

+

3a a B—j'3 A /(a a —i 2 ) 5a 2 c — x 5 \/[a 2 — x 5 ) 0 XT v.

- -- -- --- &c. Now when x = 0,

4 6

this series = 0; and when x = a, the scries becomes -\p +g &c, where p = 3-1416, or the series is

+ — +

■iP( l + i* + y^ a * + 2-rl^ ft6 &c -)

So that, bj T taking in the coefficients, the general time ofpassing over any distance de will be

id

' 2.4"

,w(c + l) ,. . 1

X ip X (1 +

Swg

4r ll •

i a * ~ : ^7n • .r-V &C,

arc sm.

- I u“a — x A J{a?- — x?) 2 d + l 3(1=2— ^^/(ai — x*) 0 _ ,

~a 4 di ‘ 2 32r£J ’ 4 &C-

And hence, taking x — 0 , and doubling, the time of awhole vibration, or double the time of passing over cd will

be equal to X (1 -j- . i a 1 — —- 77 -^. —~^-a + +

-a 8 &c.)■1 '

/ w(c + 1)2 w g

4tf= + 2(tf+/= 1.3.5

-a 0 —

40iP + 8 dH + 12dP + 5/3 1 . 3 . 5 . 7 „

--. -« 8

2 . 4.6 . S

■, the

12Su'3l s '2.4.6" 204Srf'l'

Which, when a = 0, or c = Z, becomes only \p*/same as in the last problem, as it ought.

Taking here the same numbers as in the last problem,viz, l = 4-, a — i, w = 2, w = 28800, g = 16 ; then

iPV'~T~~T ~ '0040514, and the series is 1 -f- -003762 —•000175 +• -000003 &c = 1-006590; therefore -0040514 x1-006590 = - 0040a65 = is the time of one whole vi-bration, and consequently 245| vibrations are performed ina second ; which were 250 in the last problem.