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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    The amplitude of SHM \[y=2(\sin 5\pi t+\sqrt{2}\cos \pi t)\]is

    A)  2            

    B)                         \[2\sqrt{2}\]     

    C)         4                            

    D)         \[2\sqrt{3}\]

    E)  \[4\sqrt{2}\]

    Correct Answer: D

    Solution :

    The equation of SHM is \[y=A\sin (\omega t+\phi )\] Or           \[y=A(\sin \omega t\cos \phi +\cos \omega t\sin \phi )\]   ... (i) The given expression is \[y=2(\sin 5\pi t+\sqrt{2}\cos \pi t)\]          ...(ii) \[A\cos \phi =2\]and\[A\sin \phi =2\sqrt{2}\] Squaring and adding, we get                 \[A=2\sqrt{3}\]


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