A particle executes simple harmonic motion according to the displacement equation $\mathrm{y} = 10 \cos (2\mathrm{\pi t}+\frac{\mathrm{\pi }}{6})\mathrm{cm};$, where t is in seconds. The velocity of the particle at $\mathrm{t} = \frac{1}{6}$ seconds will be:

Velocity v = $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\mathrm{d}}{\mathrm{dt}}[10 \cos (2\mathrm{\pi t}+\frac{\mathrm{\pi }}{6}\left)\right]$

                   $= -20\mathrm{\pi } \sin (2\mathrm{\pi t}+\frac{\mathrm{\pi }}{6}) {\mathrm{cms}}^{-1}$

$\mathrm{v}\left(\frac{1}{6}\right) = -20(3.14)\sin (\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}){\mathrm{cms}}^{-1}$

    = 62.8 $\mathrm{cm} {\mathrm{s}}^{-1}$(numerically)        [Using sin $\frac{\mathrm{\pi }}{2} = 1$]   

     = 0.628 ${\mathrm{ms}}^{-1}$