A) \[t=\frac{\pi }{4\omega }\]
B) \[t=\frac{\pi }{2\omega }\]
C) \[t=\frac{\pi }{\omega }\]
D) \[t=0\]
Correct Answer: C
Solution :
For perpendicular vector, we have A.B = 0 \[[cos\omega t\hat{i}+sin\omega t\hat{j}].\left[ \cos \frac{\omega t}{2}\hat{i}+\frac{\sin \omega t}{2}\hat{j} \right]=0\] \[\Rightarrow \,\,\,\,\,\,\,\cos \omega t\cos \frac{\omega t}{2}+\sin \omega t.\sin \frac{\omega t}{2}=0\] \[[\because \,cos(A-B)=cosAcosB+sinAsinB]\] Þ \[\cos \left( \omega t-\frac{\omega t}{2} \right)=0\] \[\cos \frac{\omega t}{2}=0\Rightarrow \frac{\omega t}{2}=\frac{\pi }{2}\Rightarrow t-\frac{\pi }{\omega }\] Thus, time taken by vectors which are orthogonal, to each other is \[\frac{\pi }{\omega }\].You need to login to perform this action.
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