First slide

Simple hormonic motion

Question

The function $({\sin ^2}\omega t)$ represents

Easy

A

A periodic but not simple harmonic motion
B

A simple harmonic motion with a period $\frac{{2\pi }}{\omega }$
C

A simple harmonic motion with a period half that of $\cos \omega t$
D

A simple harmonic motion with a period twice that of $\cos \omega t$

Solution

$x = \sin^{2} ω t = \frac{1}{2} \times (2 \sin^{2} ω t)$

$\Rightarrow x = \frac{1}{2}\left( {1 - \cos 2\omega t} \right)$
$\Rightarrow x = \frac{1}{2} - \frac{1}{2}\cos 2\omega t$
So the particle is executing SHM with amplitude 1/2 and frequency
$\frac{{2\pi }}{{2\omega }}$
i.e.
$\frac{\pi }{\omega }$
. The equilibrium position of the particle is at x = 1/2.
Now time period of x = cos ωt is
$\frac{{2\pi }}{\omega } = 2\left( {\frac{\pi }{\omega }} \right)$
.
So option (3) is right.

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