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A particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:
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Verified Answer
The correct answer is:
$T / 12$
For equilibrium position
$\begin{aligned}
x(t) & =a \sin \omega t \\
\text {At } x(t) & =\frac{a}{2} \\
\therefore \sin \left(\frac{\pi}{6}\right) & =\sin \omega t\left\{\because \omega=\frac{2 \pi}{\mathrm{T}}\right\} \\
\text {or } \frac{\pi}{6} & =\frac{2 \pi t}{T} \\
\Rightarrow t & =\frac{T}{12}
\end{aligned}$
$\begin{aligned}
x(t) & =a \sin \omega t \\
\text {At } x(t) & =\frac{a}{2} \\
\therefore \sin \left(\frac{\pi}{6}\right) & =\sin \omega t\left\{\because \omega=\frac{2 \pi}{\mathrm{T}}\right\} \\
\text {or } \frac{\pi}{6} & =\frac{2 \pi t}{T} \\
\Rightarrow t & =\frac{T}{12}
\end{aligned}$
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