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 second is
${{\sqrt 5 } \over {2\pi }}$
${{4\pi } \over {\sqrt 5 }}$
${{2\pi } \over {\sqrt 3 }}$
${{\sqrt 5 } \over \pi }$
Solution
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