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The periodic time of a body executing simple harmonic motion is $3 \mathrm{sec}$. After how much interval from time $t=0$, its displacement will be half of its amplitude
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The correct answer is:
$\frac{1}{4} \mathrm{sec}$
$\begin{aligned} & y=a \sin \frac{2 \pi}{T} t \Rightarrow \frac{a}{2}=a \sin \frac{2 \pi t}{3} \Rightarrow \frac{1}{2}=\sin \frac{2 \pi t}{3} \\ & \Rightarrow \sin \frac{2 \pi t}{3}=\sin \frac{\pi}{6} \Rightarrow \frac{2 \pi t}{3}=\frac{\pi}{6} \Rightarrow t=\frac{1}{4} \mathrm{sec}\end{aligned}$
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