A) \[y=4\cos (0.6t)\]
B) \[y=2\sin \left( \frac{10}{3}t+\frac{\pi }{2} \right)\]
C) \[y=4\sin \left( \frac{10}{3}t+\frac{\pi }{2} \right)\]
D) \[y=2\cos \left( \frac{10}{3}t+\frac{\pi }{2} \right)\]
Correct Answer: B
Solution :
[b] From the figure, \[\frac{T}{2}=0.3\pi ,\] \[\therefore \,\,T=0.6\pi \,and\,{{\phi }_{0}}=-\frac{\pi }{2}\] Thus \[y=A\,\sin (\omega t+{{\phi }_{0}})=2\sin \left( \frac{2\pi }{0.6\pi }t+-\frac{\pi }{2} \right)\] \[=2\sin \left( \frac{10t}{3}-\frac{\pi }{2} \right)\].
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