A) zero
B) \[\frac{2\pi }{3}\]
C) \[\pi \]
D) \[\frac{\pi }{6}\]
Correct Answer: B
Solution :
\[{{v}_{1}}=\frac{A}{2}=A\,\sin \,\omega t\] \[\Rightarrow \] \[\omega t={{30}^{o}}\] \[{{y}_{2}}=\frac{A}{2}=A\sin \left( \omega t+\frac{\pi }{2} \right)\] \[\Rightarrow \] \[\omega t+\frac{\pi }{2}={{150}^{o}}\] \[\therefore \]phase difference \[\phi ={{150}^{o}}-{{30}^{o}}\] \[=120=\frac{2\pi }{3}rad\]You need to login to perform this action.
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