The amplitude of a particle executing $SHM$ is $4 \,cm$. At the mean position the speed of the particle is $16\, cm/sec$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt 3 \,cm/s,$ will be .... $cm$
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d) At mean position velocity is maximum
i.e., $v = \omega \sqrt {{a^2} - {y^2}} $
$ \Rightarrow 8\sqrt 3 = 4\sqrt {{4^2} - {y^2}} $
$ \Rightarrow 192 = 16\,(16 - {y^2})$
$ \Rightarrow 12 = 16 - {y^2}$
$ \Rightarrow y = 2\,cm.$
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