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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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Verified Answer
The correct answer is:
$T / 12$
$y=a \sin \omega t=\frac{a \sin 2 \pi}{T} t$
At the mean position $y=\frac{a}{2}$
$$
\begin{array}{l}
\frac{a}{2}=\frac{a \sin 2 \pi t}{T} \\
t=\frac{T}{12}
\end{array}
$$
At the mean position $y=\frac{a}{2}$
$$
\begin{array}{l}
\frac{a}{2}=\frac{a \sin 2 \pi t}{T} \\
t=\frac{T}{12}
\end{array}
$$
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