A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{6}\]
D) \[\pi \]
Correct Answer: B
Solution :
\[y=10\,\sin \left[ \frac{2\pi }{45}t+\alpha \right]\] \[y=5\,cm\,at\,t=0\] \[\therefore \] \[5=10\,(\sin \,\alpha )\] or \[\sin \alpha =\frac{1}{2}\] or \[\alpha =\frac{\pi }{6}\] Again if \[t=7.5\,s\] Then total phase \[=\frac{2\pi }{45}\times \frac{15}{2}+\frac{\pi }{6}\] \[=\frac{\pi }{3}+\frac{\pi }{6}\] \[=\frac{2\pi +\pi }{6}=\frac{3\pi }{6}=\frac{\pi }{2}\]
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