## College Physics (4th Edition)

Let $t=0$ when the knob is at the bottom of the circle and let $\theta = 0$ be the angle at the bottom of the circle. Let $R$ be the radius of the circle that the knob follows. Let $x$ be horizontal component of the knob's position around the circle. Let $x = 0$ be the center of the circle. We can find an expression for $x$ in terms of time $t$: $\frac{x}{R} = sin~\theta$ $\frac{x}{R} = sin~(\omega~t)$ $x = R~sin~(\omega~t)$ Since the blade's back and forth motion is the same as the knob's horizontal component of motion, the blade's back and forth motion can be described with the equation $x = R~sin~(\omega~t)$, which is an equation for SHM. Therefore, the motion of the saw blade is SHM.