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

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A periodic but not simple harmonic motion

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

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

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

answer is C.

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Detailed Solution

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

Rightarrow x = frac{1}{2}left( {1 - cos 2omega t} right)

Rightarrow x = frac{1}{2} - frac{1}{2}cos 2omega t

So the particle is executing SHM with amplitude 1/2 and frequency

i.e.

. The equilibrium position of the particle is at x = 1/2.

Now time period of x = cos ωt is

frac{{2pi }}{omega } = 2left( {frac{pi }{omega }} right)

So option (3) is right.

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