A particle executes simple harmonic motion with an amplitude of $4 \,cm$. At the mean position the velocity of the particle is $10\, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \,cm/s$ is
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(c) ${v_{\max }} = a\omega $
==> $\omega = \frac{{{v_{\max }}}}{a} = \frac{{10}}{4}$
Now, $v = \omega \sqrt {{a^2} - {y^2}} $
==> ${v^2} = {\omega ^2}({a^2} - {y^2})$
==> ${y^2} = {a^2} - \frac{{{v^2}}}{{{\omega ^2}}}$
$y = \sqrt{{a^2} - \frac{{{v^2}}}{{{\omega ^2}}}}$
$= \sqrt{{4^2} - \frac{5^2}{{({\frac{10}{4}})^2}}}$
$ = 2\sqrt 3 \,cm$
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