A) \[\frac{\pi }{2}cm/{{s}^{2}}\]
B) \[\frac{{{\pi }^{2}}}{2}cm/{{s}^{2}}\]
C) \[\frac{{{\pi }^{2}}}{4}cm/{{s}^{2}}\]
D) \[\frac{\pi }{4}cm/{{s}^{2}}\]
Correct Answer: B
Solution :
\[y=2\sin \left( \frac{\pi t}{2}+\phi \right)\]velocity of particle\[\frac{dy}{dt}=2\times \frac{\pi }{2}\cos \left( \frac{\pi t}{2}+\phi \right)\] acceleration\[\frac{{{d}^{2}}y}{dt}=-\frac{{{\pi }^{2}}}{2}\sin \left( \frac{\pi t}{2}+\phi \right)\] Thus, \[{{a}_{\max }}=\frac{{{\pi }^{2}}}{2}\]You need to login to perform this action.
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