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(C) The standard equation for linear $S.H.M$ is given by $x = a \sin(\omega t + \phi)$.Comparing the given equation $x = 5 \sin(4t - \frac{\pi}{6})$ w

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$16$

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(C) The standard equation for linear $S.H.M$ is given by $x = a \sin(\omega t + \phi)$.Comparing the given equation $x = 5 \sin(4t - \frac{\pi}{6})$ with the standard form,we get the amplitude $a = 5 \ cm$ and angular frequency $\omega = 4 \ rad/s$.The velocity $v$ of a particle in $S.H.M$ at a displacement $x$ is given by the formula:$v = \omega \sqrt{a^2 - x^2}$Substituting the given values $a = 5 \ cm$,$x = 3 \ cm$,and $\omega = 4 \ rad/s$:$v = 4 \sqrt{5^2 - 3^2}$$v = 4 \sqrt{25 - 9}$$v = 4 \sqrt{16}$$v = 4 	imes 4 = 16 \ cm/s$.Therefore,the velocity of the particle is $16 \ cm/s$.

The equation of motion of a particle performing linear $S.H.M$ is $x=5 \sin \left[4 t-\frac{\pi}{6}\right]$,where $x$ is its displacement in $cm$. The velocity of the particle when its displacement is $3 \ cm$ is: (in $cm/s$)

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