A particle excutes $SHM$ on a straight line path. The amplitude of oscillation is $2\,cm$. When the displacement of the particle from the mean position is $1\,cm$, the numerical value of magnitude of acceleration is equal to the numerical value of magnitude of velocity. The frequency of $SHM$ is (in $second^{-1}$)
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$\mathrm{V}=\omega \sqrt{\mathrm{a}^{2}-(\mathrm{a} / 2)^{2}}=\frac{\sqrt{3}}{2} \mathrm{a} \omega$
$\operatorname{acc}^{\mathrm{n}} \mathrm{f}=\frac{\omega^{2} \mathrm{a}}{2}$
$\therefore \mathrm{V}=\mathrm{f}$
$\frac{\omega^{2} a}{2}=\frac{\sqrt{3}}{2} a \omega$
$2 \pi n=\sqrt{3}$
$\mathrm{n}=\frac{\sqrt{3}}{2 \pi}$
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