Pendulum.—Suppose that we have a body P (fig. 4) at rest, and that it is material, that is to say, has “mass.” And for simplicity let us consider it a ball of some heavy matter. Let it be free to move horizontally, but attached to a fixed point A by means of a spring. As it can only move horizontally and not fall, the earth’s gravity will be unable to impart any motion to it. Now it is a law first discovered by Robert Hooke (1635-1703) that if any elastic spring be pulled by a force, then, within its elastic limits, the amount by which it will be extended is proportional to the force. Hence then, if a body is pulled out against a spring, the restitutional force is proportional to the displacement. If the body be released it will tend to move back to its initial position with an acceleration proportioned to its mass and to its distance from rest. A body thus circumstanced moves with harmonic motion, vibrating like a stretched piano string, and the peculiarity of its motion is that it is isochronous. That is to say, the time of returning to its initial position is the same, whether it makes a large movement at a high velocity under a strong restitutional force, or a small movement at a lower velocity under a smaller restitutional force (see [Mechanics]). In consequence of this fact the balance wheel of a watch is isochronous or nearly so, notwithstanding variations in the amplitude of its vibrations. It is like a piano string which sounds the same note, although the sound dies away as the amplitude of its vibrations diminishes.
| Fig. 5. |
A pendulum is isochronous for similar reasons. If the bob be drawn aside from D to C (fig. 5), then the restitutional force tending to bring it back to rest is approximately the force which gravitation would exert along the tangent CA, i.e.
| g cos ACW = g | BC | = g | displacement BC | . |
| OC | length of pendulum |
Since g is constant, and the length of the pendulum does not vary, it follows that when a pendulum is drawn aside through a small arc the force tending to bring it back to rest is proportional to the displacement (approximately). Thus the pendulum bob under the influence of gravity, if the arc of swing is small, acts as though instead of being acted on by gravity it was acted on by a spring tending to drag it towards D, and therefore is isochronous. The qualification “If the arc of swing is small” is introduced because, as was discovered by Christiaan Huygens, the arc of vibration of a truly isochronous pendulum should not be a circle with centre O, but a cycloid DM, generated by the rolling of a circle with diameter DQ = ½OD, upon a straight line QM. However, for a short distance near the bottom, the circle so nearly coincides with the cycloid that a pendulum swinging in the usual circular path is, for small arcs, isochronous for practical purposes.
| Fig. 6. |
The formula representing the time of oscillation of a pendulum, in a circular arc, is thus found:—Let OB (fig. 6) be the pendulum, B be the position from which the bob is let go, and P be its position at some period during its swing. Put FC = h, and MC = x, and OB = l. Now when a body is allowed to move under the force of gravity in any path from a height h, the velocity it attains is the same as a body would attain falling freely vertically through the distance h. Whence if v be the velocity of the bob at P, v = √2gFM = √2g(h - x). Let Pp = ds, and the vertical distance of p below P = dx, then Pp = velocity at P × dt; that is, dt = ds/v. Also
| ds | = | 1 | = | 1 | , |
| dx | MP | √x(2l - x) |
whence
| dt = | ds | = | ldx | . | 1 | = | 1 | √ | l | . | dx | . | 1 | . |
| v | √x(2l - x) | √2g(h - x) | 2 | g | √x(p - x) | √1 - (x/2l) |