A) zero
B) \[\pi /2\]
C) \[\pi /6\]
D) \[\pi /3\]
Correct Answer: B
Solution :
\[{{y}_{1}}=a\sin 2\pi \,vt\] and \[{{y}_{2}}=a\sin (2\pi vt+\text{o}|)\] \[y={{y}_{2}}-{{y}_{1}}=a[\sin (2\pi vt+o|)-\sin 2\pi vt]\] \[=2a\sin \frac{\text{o }\!\!|\!\!\text{ }}{2}\cos \left[ 2\pi \,vt+\frac{\text{o }\!\!|\!\!\text{ }}{2} \right]\] \[\therefore \]Maximum value of \[y=2a\sin \frac{\text{o }\!\!|\!\!\text{ }}{2}\] Now, \[2a\sin \frac{\text{o }\!\!|\!\!\text{ }}{2}=a\sqrt{2}\] or \[\text{o }\!\!|\!\!\text{ =}\frac{\pi }{2}\]
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