TRACT 38.

PROBLEMS.

337

l‘5708v/-~^ = L'5708^/-^ = l - 5708v / ^-, and the time of awhole vibration from B to c, or from c to b. is 3-1416 V --r—,

# 2g

where / = as =: ab is the length of'the pendulum, g-=16 T c-feet, or 193 inches, and S’1416 the circumference of a circlewhose diameter is 1.

Since the time of a body’s falling by gravity through {I,or half tlie length of the pendulum, is V'^y, which being inproportion to 3*1416y'—, as 1 to3‘14l6; therefore the di-ameter of a circle is to its circumference, as the time of fall-ing through half the, length of a pendulum, to the time ofone vibration.

If the time of the whole vibration be 1 second, this equa-tion arises, l" = 3-1416./—, and hence l — " e — ——-

3 - 1416 ’

4-9348

and g = 3-1416 2 x = 4-9348/. So that if one of these, gor Z, be given by experiment, these equations will give theother. When g, for instance, is supposed to be 16 xx feet,or 193 inches, then is Z = 4 7 y|y 3 - = 39-11 the length of apendulum to vibrate seconds. Or if l — 39-f-, the length ofthe seconds pendulum for the latitude of London, then isg — 4-9348/ = 193-07 inches = 16 t ' t Vb- feet, or nearly 16-^feet, for the space descended by gravity in the first secondof time in the latitude of London, also agreeing with experi-ment.

Hence the times of vibration of pendulums, are as thesquare roots of their lengths ; and the number of vibrationsmade in a given time, reciprocally as the square roots of thelengths. And hence also, the length of a pendulum vibrating

n times in a minute, or 60", is Z = 39|- x =

When a pendulum vibrates in a circular arc, as the lengthof the string is constantly the same, the time of vibration willbe longer than in a cycloid ; but the two times will approachnearer together as the circular arc is smaller; so that whenit is very small, the times of vibration will be nearly equal.And hence 39-|- inches is the length of a pendulum vibratingseconds in the very small arc of a circle.vol. hi. z

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