# Does the equation $x = a\sin^2\omega t$ represent a simple harmonic motion?

I've asked this question due a doubt in the following question.

The displacement of a particle along the $$x$$-axis is given by $$x = a\sin^2\omega t$$. The motion of the particle corresponds to

(1) simple harmonic motion of frequency $$\omega/\pi$$
(2) simple harmonic motion of frequency $$3\omega/2\pi$$
(3) non simple harmonic motion
(4) simple harmonic motion of frequency $$\omega/2\pi$$

(a) The answer given is option (3). This is is accordance to the following excerpt of the NCERT textbook.

$$\sin^2\omega t=\frac12-\frac12\cos2\omega t$$ The function is periodic having a period $$T=\pi/\omega$$. It also represents a harmonic motion with the point of equilibrium occuring at $$\frac12$$ instead of zero.

(b) But according to the book Concepts of Physics by HC Verma, and various others found on the internet, simple harmonic motion is defined as the one in which $$F = -kx$$ and sine function is one of its solutions.

If the equation is rearranged, as shown in the book, should it be considered as simple harmonic motion with equilibrium position at $$+1/2$$ unit and angular frequency of $$2\omega$$?

What is the exact difference between harmonic motion and simple harmonic motion?

• Does this answer your question? physics.stackexchange.com/questions/322609/… Sep 1, 2021 at 14:48
• Regardless of the answer, this is a bad question because it depends on the precise definition of simple harmonic motion, and no actual physicist cares about that. The motion is harmonic around $x=1/2$, who cares what you call it? Sep 1, 2021 at 14:52

(b) $$x=A \sin^2\omega t$$ is an interesting 'borderline case'. Either answer: not SHM or SHM of frequency $$2\omega$$ could be defended.
In favour of not SHM is that $$x=A \sin^2\omega t$$ does not satisfy the differential equation $$\ddot x=-\omega^2 x$$, but the slightly different equation $$\ddot x=a_0-\omega'^2 x$$ in which $$a_0$$ is a constant and $$\omega'=2\omega$$.
In favour of SHM of frequency $$2\omega$$ is that, as shown using your quoted trigonometric identity, $$x=A \sin^2\omega t$$ gives a displacement time graph that is a perfect sinusoid, albeit displaced vertically by a fixed amount.
My view is in favour of $$x=A \sin^2\omega t$$ being SHM of frequency $$2\omega$$. I would prefer textbooks to define SHM as motion in which the displacement–time graph is a sinusoid (however shifted with respect to either axis), rather than as motion obeying $$\ddot x=-\omega^2 x$$. But that's just my view.