JEE MAIN - Physics (2005 - No. 40)
The function $${\sin ^2}\left( {\omega t} \right)$$ represents
a periodic, but not $$SHM$$ with a period $${\pi \over \omega }$$
a periodic, but not $$SHM$$ with a period $${{2\pi } \over \omega }$$
a $$SHM$$ with a period $${\pi \over \omega }$$
a $$SHM$$ with a period $${{2\pi } \over \omega }$$
Explanation
y = sin2$$\omega $$t
= $${{1 - \cos 2\omega t} \over 2}$$
$$ = {1 \over 2} - {1 \over 2}\cos \,2\omega t$$
$$ \therefore $$ Angular speed = 2$$\omega $$
$$ \therefore $$ Period (T) = $${{2\pi } \over {angular\,speed}}$$ = $${{2\pi } \over {2\omega }}$$ = $${\pi \over \omega }$$
So it is a periodic function.
As y = sin2$$\omega $$t
$${{dy} \over {dt}}$$ = 2$$\omega $$sin$$\omega $$t cos$$\omega $$t = $$\omega $$ sin2$$\omega $$t
$${{{d^2}y} \over {d{t^2}}}$$ = $$2{\omega ^2}$$ cos2$$\omega $$t which is not proportional to -y.
Hence it is is not SHM.
= $${{1 - \cos 2\omega t} \over 2}$$
$$ = {1 \over 2} - {1 \over 2}\cos \,2\omega t$$
$$ \therefore $$ Angular speed = 2$$\omega $$
$$ \therefore $$ Period (T) = $${{2\pi } \over {angular\,speed}}$$ = $${{2\pi } \over {2\omega }}$$ = $${\pi \over \omega }$$
So it is a periodic function.
As y = sin2$$\omega $$t
$${{dy} \over {dt}}$$ = 2$$\omega $$sin$$\omega $$t cos$$\omega $$t = $$\omega $$ sin2$$\omega $$t
$${{{d^2}y} \over {d{t^2}}}$$ = $$2{\omega ^2}$$ cos2$$\omega $$t which is not proportional to -y.
Hence it is is not SHM.
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