TRACT 38.

PROBLEMS.

329

upward motion ceases, and the cylinder descends again to thesame depth as at first, after which it again returns ascendingas before; and so on, continually playing up and down tothe same highest and lowest points, like the vibrations of apendulum, the motion ceasing in both cases i:i a similar man-ner at the extreme points, then returning, it gradually acce-lerates till arriving at the middle point, where it is thegreatest, then gradually retarding all the way to the nextextremity of the vibration, thus making all the vibrations inequal times, to the same extent between the highest aridlowest points, except that, by the small tenacity and frictionStc, of the water against the sides of the cylinder, it will begradually and slowly retarded in its motion, and the extentof the vibrations decrease till at length the cylinder, like thependulum, come to rest in the middle point of its vibrations,where it naturally floats in its quiescent state, with the partna of its length above the water.

The quantity 7 of the greatest velocity will be found, bysubstituting na for x, in the general value of the velocity'8^ —-, when it becomes 8 nV— = 4V« very nearly,

2am 7 2m y T v J 7

the value of m being ’925, and consequently that of n = I —m = -075.

To find the time t answering to any space x. Here

form the fluent is t — \*/2.ma x A, where a denotes the

the radius, and a is the quadrantal arc = 1'5708 ; then theflu. becomes \V2ma x 1-5708”*17 */a X 1'5708 = ’267 t/afor the time of a semivibration; hence the time of each wholevibration is •534 Va = j 8 jVa, which time therefore dependson the length of the cylinder a. To make this time — 1second, a must be = (y) 1 very nearly^^ feet, or 42 inches.That is, the oaken cylinder of 42 inches length makes its

X

X

X

and by the 13th

t

2 nax —

0

circular arc to radius 1 and versed sine —. Now at the mid-

na

die of a vibration x is = na, and then the vers. — ~ n — — \

' ii rt tt, i