Question

If position of a particle is represented by a function ${\sin }^2\omega t$, then the particle executes

• A. a periodic but not simple harmonic motion with a period $\frac{2\pi }{\omega }$
• B. a periodic, but not simple harmonic motion with a period $\frac{\pi }{\omega }$
• C. a simple harmonic motion with a period $\frac{2\pi }{\omega }$
• D. a simple harmonic motion with a period $\frac{\pi }{\omega }$

Answer 1

The correct option is D a simple harmonic motion with a period $\frac{\pi }{\omega }$

Given $y={\sin }^2\omega t=\frac{1}{2}-\left(\frac{1}{2}\right)\cos 2\omega t$
$v=\frac{dy}{dt}=\omega \sin 2\omega t$
$a=\frac{dv}{dt}=-2{\omega }^2\cos 2\omega t$
$a=-4{\omega }^{2}y$
$\Rightarrow a\propto -y$

The particle executes $SHM$ and it is periodic with time period

$T=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$