Amplitude of oscillation of a particle that executes S.H.M. is 2 cm. Its displacement from its mean position in a time equal to 1/6th of its time period is

\sqrt{2}

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\sqrt{3}

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1 / \sqrt{2}

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1 / \sqrt{3}

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Solution

The correct option is A $\sqrt{3}$ cm
$A = 2 c m$
$x = A$ $s i n$ $w t$
$= A s i n (\frac{2 π}{T} \times \frac{T}{6}) = \frac{A \times \sqrt{3}}{2}$
$x = \sqrt{3} c m$

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