A) \[a\omega \,\,\cos \,\,\theta \]
B) \[a\omega \]
C) \[a\omega \,\,\sin \,\,\theta \]
D) \[\text{none of these}\]
Correct Answer: B
Solution :
Key Idea: Rate of change of displacement is velocity \[i.e.,\]\[v=\frac{dx}{dt}\]. Given equation is \[x=a\,\sin \,\left( \omega t-\theta \right)\] Differentiating the equation with respect to \[t\] and using\[\frac{d}{d\,\theta }\sin \,\theta =\cos \,\,\theta ,\] \[v=\frac{dx}{dt}=a\omega \,\cos \,\,\theta \left( \omega t-\theta \right)\] Velocity is maximum when\[\cos \,\left( \omega t-\theta \right)=1\]Hence, \[{{v}_{\max }}=a\omega \] Note: when particle passes through its equilibrium position, then its velocity is maximum.
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