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JEE Main & Advanced Sample Paper JEE Main Sample Paper-39

  • question_answer
    \[y=2(cm)\sin \left[ \frac{\pi t}{2}+\phi  \right]\] What is the maximum acceleration of the particle doing the SHM

    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, \[{{a}_{\max }}=\frac{{{\pi }^{2}}}{2}\]


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