A) \[t=0\]
B) \[t=\frac{\pi }{4\omega }\]
C) \[t=\frac{\pi }{2\omega }\]
D) \[t=\frac{\pi }{\omega }\]
Correct Answer: D
Solution :
\[\vec{A}=\cos \omega t\,\hat{i}+\sin \omega t\,\hat{j}\] \[\vec{B}=\cos \frac{\omega t}{2}\hat{i}+\sin \frac{\omega t}{2}\hat{j}\] For\[\vec{A}\]and \[\vec{B}\]orthogonal \[\vec{A}\vec{B}=0\] \[(\cos \omega t\,\hat{i}+\sin \omega t\,\hat{j}).\left( \cos \frac{\omega t}{2}\hat{i}+\sin \frac{\omega t}{2}\hat{j} \right)=0\] \[\cos \omega t.\cos \frac{\omega t}{2}+\sin \omega t.\sin \frac{\omega t}{2}=0\] \[\cos \left( \omega t-\frac{\omega t}{2} \right)=0\] \[\Rightarrow \] \[\cos \frac{\omega t}{2}=0\] \[\frac{\omega t}{2}=\frac{\pi }{2}\Rightarrow \omega t-\pi \Rightarrow t=\frac{\pi }{\omega }\]You need to login to perform this action.
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