A) \[\frac{5\pi }{6}\]
B) \[\frac{2\pi }{3}\]
C) \[\frac{\pi }{6}\]
D) \[\frac{2\pi }{5}\]
Correct Answer: C
Solution :
Given, \[x=A\sin \omega t+\delta \] At t = 0 (initially), \[x=\frac{A}{2}\] \[\Rightarrow \]\[\frac{A}{2}=A\sin \delta \Rightarrow \sin \delta =\frac{1}{2}\]\[\Rightarrow \]\[\delta =\frac{\pi }{6}or\frac{5\pi }{6}\]..(i) The velocity,\[\frac{dx}{dt}=\frac{d(A\sin \omega t)}{dt}\] \[\Rightarrow \]\[v=A\omega \cos (\omega t+\delta )\]At\[t=0,v=A\omega \cos \delta \] Now,\[\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\]and\[\cos \frac{5\pi }{6}=\frac{\sqrt{3}}{2}\] At t = 0, v is positive, so \[\delta \] must be equal to \[\frac{\pi }{6}.\]
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