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A particle is executing simple harmonic motion with a period of T seconds and amplitude a in metre. The shortest time it takes to reach a point $\frac{a}{\sqrt{2}} m$ from its mean position in seconds is

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$\frac{T}{16}$

$\frac{T}{8}$

$\frac{T}{4}$

answer is C.

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Detailed Solution

$y = a \sin \frac{2 π}{T} t$

$\Rightarrow \frac{a}{\sqrt{2}} = a \sin \frac{2 π}{T} \cdot t$

$\Rightarrow \sin \frac{2 π}{T} t = \frac{1}{\sqrt{2}} = \sin \frac{π}{4}$

$\Rightarrow \frac{2 π}{T} t = \frac{π}{4}$

$∴ t = \frac{T}{8}$

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