Page:Elementary Text-book of Physics (Anthony, 1897).djvu/35

§ 21] a clock, or counter-clockwise. Motion from left to right in the diameter is also considered positive. Displacement to the right of the centre is positive, and to the left negative.

If a point start from $$X$$ (Fig. 10), the position of greatest positive elongation, with a simple harmonic motion, its distance $$s$$ from $$O$$ or its displacement at the end of the time $$t$$, during which the point in the circle has moved through the arc $$BX$$, is $$OC = OB \cos{\phi}$$. Now, $$OB$$ is equal to $$OX$$, the amplitude, represented by $$a$$. If $$\omega$$ represent the angular velocity of the moving point, we have $$\phi = \omega t$$. Hence we have To find the velocity at the point $$C$$, we must resolve the velocity of the point moving in the circle into its components parallel to the axes. The component at the point $$C$$ along $$OX$$ is $$V \sin{\phi}$$; or, since $$V = \omega a$$, remembering that motion from right to left is considered negative.

The acceleration at the point $$C$$ is the component along $$OX$$ of the acceleration of the point moving in the circle. The acceleration of $$B$$ is $$- \frac{v^2}{a}$$, the minus sign being given because this acceleration is directed opposite to the positive direction of the radius. The component at $$C$$ along $$OX$$ is This formula shows that the acceleration in a simple harmonic