Where does the negative sign disappear?

The defining equation for simple harmonic motion is such

$$a=-ω^2x$$

When we find the centripetal acceleration of an object in orbit we use the formula

$$a=ω^2r$$

As a consequence of the accleration being $$\frac{v^2}{r}$$

The two formulas for acceleration look extremely similiar, and displacement $$x$$ is essentially the $$r$$. So then what has happened to the negative sign?

• Though the "missing negative" mystery has been answered below, it's worth pointing out that similarities like this occur all over the place and don't necessarily mean anything (though they also often do). A meaningful follow-up question for you to consider is "If centripetal force really does have a minus sign in the equation, why is an object in orbit following simple harmonic motion?" Of course, it's also possible that you already understood that and it led you to the point we're at here. :-) Mar 10 at 20:12

The second fomula probably only intends to give the magnitude of the acceleration, since the description "centripetal acceleration" would already serve to give you the direction. Without that context you might say it is simply a sloppy formula, a better version would then be: $${\bf a} = -\omega^2 {\bf r}, \qquad \mbox{or} \quad \vec a = -\omega^2 \ \vec r$$ where vector quantities are used.

The negative sign "disappears" because in $$a = \omega^2r$$ we are only considering the magnitude of the centripetal acceleration and ignoring its direction.

Centripetal acceleration is directed towards the centre of the circle $$O$$, so in vector form we have

$$\vec a = - \omega^2 \vec r$$

where $$\vec r$$ is the vector from $$O$$ to the object. But in Cartesian co-ordinates we have $$\vec r = (x,y)$$, so the $$x$$ and $$y$$ components of acceleration are

$$a_x = - \omega^2 x \\ a_y = - \omega^2 y$$

In other words, the $$x$$ and $$y$$ co-ordinates of an object that is performing circular motion with a constant angular speed $$\omega$$ both perform simple harmonic motion.

Lets see where its come from

the position vector to the mass on a circular path with radius $$~r~$$ is:

$$\vec r= r\,\left[ \begin {array}{c} \cos \left( \omega\,t \right) \\ \sin \left( \omega\,t \right) \end {array} \right]$$ the velocity $$\vec v=\frac{d\vec r}{dt}=r\,\omega\,\left[ \begin {array}{c} -\sin \left( \omega\,t \right) \\ \cos \left( \omega\,t \right) \end {array} \right] \\ |\vec v|=v=\omega\,r$$ ans the acceleration

$$\vec a=\frac{d\vec v}{dt}=\omega^2\,r\, \left[ \begin {array}{c} -\cos \left( \omega\,t \right) \\ -\sin \left( \omega\,t \right) \end {array} \right] =-\omega^2\,\vec r$$

the magnitude of $$~|\vec a|=a=\omega^2\,r=\frac{v^2}{r}~$$