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A particle executing a simple harmonic motion has a period of $6 \mathrm{~s}$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is
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The correct answer is:
$\frac{1}{2} \mathrm{~s}$
Equation for simple harmonic motion
$\begin{aligned} y &=a \sin \left(\frac{2 \pi}{T}\right) t \\ \frac{a}{2} &=a \sin \left(\frac{2 \pi}{T}\right) t \\ \frac{2 \pi}{T} t &=\frac{\pi}{6} \\ t &=\frac{T}{12}=\frac{6}{12}=\frac{1}{2} \mathrm{~s} \end{aligned}$
$\begin{aligned} y &=a \sin \left(\frac{2 \pi}{T}\right) t \\ \frac{a}{2} &=a \sin \left(\frac{2 \pi}{T}\right) t \\ \frac{2 \pi}{T} t &=\frac{\pi}{6} \\ t &=\frac{T}{12}=\frac{6}{12}=\frac{1}{2} \mathrm{~s} \end{aligned}$
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