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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    Two simple harmonic motions are represented by\[{{y}_{1}}=4\sin (4\pi t+\pi /2)\]and\[{{y}_{2}}=3\cos (4\pi t)\]. The resultant amplitude is

    A)  7                            

    B)         1

    C)  5                            

    D)         \[2+\sqrt{3}\]

    E)  \[2-\sqrt{3}\]

    Correct Answer: A

    Solution :

    Given \[{{y}_{1}}=4\sin \left( 4\pi t+\frac{\pi }{2} \right)\] or          \[{{y}_{1}}=4\cos 4\pi t\] and       \[{{y}_{2}}=3\cos 4\pi t\] Resultant   \[y={{y}_{1}}+{{y}_{2}}\]                 \[y=4\cos 4\pi t+3\cos 4\pi t\]                 \[y=7\cos 4\pi t\] \[\therefore \]Resultant amplitude \[a=7\]


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