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Questions  >>  JEEMAIN and NEET  >>  NEET PAST PAPERS  >>  2017
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A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration.- Then its time period in seconds is

$\begin{array}{1 1} (1) \large\frac{\sqrt 5}{\pi} \\ (2) \large\frac{\sqrt 5}{2 \pi} \\ (3)\large\frac{4 \pi}{\sqrt 5} \\ (4) \large\frac{2 \pi}{\sqrt 3} \end{array} $

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