# Can a periodic motion whose displacement is given by $x=\sin^2(\omega t)$, be considered as a SHM?

The definition of Simple Harmonic Motion is :

simple harmonic motion is a type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement

From the definition of SHM we can state that, in order to be considered as SHM, a periodic motion only has to meet the following criteria that is:

(where $a$ is acceleration and $x$ is the displacement from equilibrium position) $$a \propto -x$$ If we have a periodic motion where $$x= \sin^2(\omega t)$$ $$\implies v=\frac{dx}{dt}=2\omega \sin(\omega t)\cos(\omega t)=\omega \sin(2\omega t)$$ $$\implies a=\frac{d^2x}{dt^2}=2\omega^2\cos(2\omega t)=2\omega^2(1-2\sin^2(\omega t))$$

Since, $x=\sin^2(\omega t)$, we get, $$a=2\omega^2(1-2x)$$ then, can we say that $$a \propto -x$$ and thus declare that it is a SHM?

• Being proportional to $(1-2x)$ is not the same thing as being proportional to $-x$. Proportional to $-x$ means that the force goes to zero as $x$ goes to zero, for example. Mar 31, 2017 at 6:11
• @PhillS Then what kind of periodic motion is $x=sin^2(wt)$ referring to? Can you give me an example? I'm having a hard time visualising it. Mar 31, 2017 at 6:25

Since $sin^2 (\omega\,t) = \frac{1}{2}\left(1-\cos(2\,\omega\,t)\right)$, it is evident that a position time dependence of the form $x=sin^2 (\omega\,t)$ indeed describe simple harmonic motion centered on the point $x=\frac{1}{2}$ and with frequency $2\,\omega$. This is the meaning of the constant part in your force $1-2\,x$: it is an offset of the point where there is zero force.
But in SHM, by displacement don't we mean the displacement from the equilibrium point, and by equilibrium point don't we mean the point where the force is zero? But here the force is zero when the object is $\frac{1}{2}$ units away from equilibrium. This is confusing me.
• But in SHM, by displacement don't we mean the displacement from the equilibrium point, and by equilibrium point don't we mean the point where the force is zero? But here the force is zero when the object is $\frac{1}{2}$ units away from equilibrium. This is confusing me. Mar 31, 2017 at 6:44
Define $y=x-\frac{1}{2}$ so $\ddot{y}=\ddot{x}\propto 1-2x=-2y$. Since $y$ exhibits simple harmonic motion, $x$ exhibits offset simple harmonic motion.