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A particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude
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The correct answer is:
$T / 12$
$y=A \sin \omega t=\frac{A \sin 2 \pi}{T} t \Rightarrow \frac{A}{2}=A \sin \frac{2 \pi t}{T} \Rightarrow t=\frac{T}{12}$
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