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

Question

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?

Answer 1

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.

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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.

I guess it depends on one's definition. But generally in physics we seek definitions that are co-ordinate independent. One such definition would be that SHM is rectillinear motion wherefor the force on a particle is proportional to and directed against the particle's displacement from the particle's equilibrium (zero force) point.

Answer 2

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.