A particle executing simple harmonic motion has an amplitude of $6\, cm$. Its acceleration at a distance of $2 \,cm$ from the mean position is $8\,cm/{s^2}$. The maximum speed of the particle is ... $ cm/s$
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$A = {\omega ^2}y$
$\omega = \sqrt {A/y} = \sqrt {\frac{8}{2}} = 2\,rad/sec$
Now ${v_{\max }} = a\omega = 6 \times 2 = 12\,cm/sec$
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