A particle executes $SHM$ 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 $SHM$ (in $second^{-1}$) is :
Diffcult
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It is given that
$|v|=|a|$
$\Rightarrow \omega \sqrt{A^{2}-x^{2}}=\omega^{2} x$
$\Rightarrow \omega=\frac{\sqrt{A^{2}-x^{2}}}{x}$
$\Rightarrow 2 \pi f=\frac{\sqrt{A^{2}-x^{2}}}{x}$
$\Rightarrow f=\frac{1}{2 \pi}\left(\frac{\sqrt{A^{2}-x^{2}}}{x}\right)=\frac{1}{2 \pi}\left(\frac{\sqrt{2^{2}-1^{2}}}{1}\right)=\frac{\sqrt{3}}{2 \pi}$
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