A particle executes linear simple harmonic motion with an amplitude of $2\, cm$. When the particle is at $1\, cm$ from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
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(c) Velocity $v = \omega \sqrt {{A^2} - {x^2}} $ and acceleration $ = {\omega ^2}x$
Now given, ${\omega ^2}x = \omega \sqrt {{A^2} - {x^2}} $
==> ${\omega ^2}.1 = \omega \sqrt {{2^2} - {1^2}} $
==> $\omega = \sqrt 3 $
$T = \frac{{2\pi }}{\omega } = \frac{{2\pi }}{{\sqrt 3 }}$
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