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

Fig. 10 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

$s=a\cos {\omega t}.$

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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 $V=\omega a$,

$v=-\omega a\sin {\omega t},$

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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

$f=-{\frac {v^{2}}{a}}\cos {\omega t}{\text{ or }}f=-\omega ^{2}a\cos {\omega t}=\omega ^{2}s.$

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This formula shows that the acceleration in a simple harmonic