A particle is executing simple harmonic motion with a period of $T$ seconds and amplitude a 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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(c) $y = a\sin \frac{{2\pi }}{T}t$
==> $\frac{a}{{\sqrt 2 }} = a\sin \frac{{2\pi }}{T} \cdot t$
==> $\sin \frac{{2\pi }}{T}t = \frac{1}{{\sqrt 2 }} = \sin \frac{\pi }{4}$
==> $\frac{{2\pi }}{T}t = \frac{\pi }{4}$
==> $t = \frac{T}{8}$
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