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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
A
Simple harmonic motion of frequency $\frac{\omega}{2\pi}$
B
Simple harmonic motion of frequency $\frac{\omega}{\pi}$
C
Simple harmonic motion of frequency $\frac{3\omega}{2\pi}$
D
Non simple harmonic motion
Detailed Solution
$x = a\sin^2\omega t$
Velocity: $\frac{dx}{dt} = 2a\omega\sin\omega t\cos\omega t = a\omega\sin2\omega t$
Acceleration: $\frac{d^2x}{dt^2} = 2a\omega^2\cos2\omega t$
For SHM about the origin we need $\frac{d^2x}{dt^2} = -\omega'^2x$, i.e., acceleration proportional to $-x$.
Here $2a\omega^2\cos2\omega t$ is not proportional to $-a\sin^2\omega t$, so the condition $\frac{d^2x}{dt^2} \propto -x$ is not satisfied for the displacement x as given.
Hence the answer accepted is non simple harmonic motion.
Note: writing $x = \frac{a}{2} - \frac{a}{2}\cos2\omega t$ shows the motion is simple harmonic about the shifted mean position $x = \frac{a}{2}$ with frequency $\frac{\omega}{\pi}$; the key of the source PDF treats it as non-SHM because x is measured from the origin.
Velocity: $\frac{dx}{dt} = 2a\omega\sin\omega t\cos\omega t = a\omega\sin2\omega t$
Acceleration: $\frac{d^2x}{dt^2} = 2a\omega^2\cos2\omega t$
For SHM about the origin we need $\frac{d^2x}{dt^2} = -\omega'^2x$, i.e., acceleration proportional to $-x$.
Here $2a\omega^2\cos2\omega t$ is not proportional to $-a\sin^2\omega t$, so the condition $\frac{d^2x}{dt^2} \propto -x$ is not satisfied for the displacement x as given.
Hence the answer accepted is non simple harmonic motion.
Note: writing $x = \frac{a}{2} - \frac{a}{2}\cos2\omega t$ shows the motion is simple harmonic about the shifted mean position $x = \frac{a}{2}$ with frequency $\frac{\omega}{\pi}$; the key of the source PDF treats it as non-SHM because x is measured from the origin.