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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    The displacement of a particle is SHM varies according      to      the      relation \[x=4(cos\pi t+sin\pi t).\]The amplitude of the particle is

    A)  \[-4\]                                   

    B)  4

    C)  \[4\sqrt{2}\]                     

    D)  8

    Correct Answer: C

    Solution :

                     We have the displacement of the particle in SHM given by \[x=4(cos\pi t+\sin \pi t)\] \[x=4\cos \pi t+4sin\,\pi t\] From the given equation \[{{a}_{1}}=4\] and \[{{a}_{2}}=4\] \[\therefore \] amplitude \[a=\sqrt{a_{1}^{2}+a_{2}^{2}}\]                                 \[=\sqrt{{{4}^{2}}+{{4}^{2}}}\]                                 \[a=4\sqrt{2}\]


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