A particle is executing the motion $x = a\cos (\omega t - \theta)$. The maximum velocity of the particle is

The equation of motion given:

$x=a \cos (\cot -\theta) \text {. }$

The velocity, $v$ can be obtained by differentiating:-

$\begin{array}{c} v=\frac{d x}{d t}=-a \omega(\sin (\omega t-\theta)) . \\ v=-a \omega \sin (\omega t-\theta) . \end{array}$

Now, the maximum value of $\sin x=1$. we get:

$v=-a \omega\ \text { minimum velocity. }$

But at $\sin (\omega t-\theta)=-1$, we have,

$v=a w \text { max. velocity. }$

Hence, the correct option is : B.