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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    For a particle executing simple harmonic motion, the kinetic energy \[k\] is given by\[k={{k}_{0}}{{\cos }^{2}}\omega t\]. The maximum value of potential energy is

    A) \[{{k}_{0}}\]                                      

    B)  zero

    C) \[{{k}_{0}}/2\]  

    D)         not obtainable

    Correct Answer: B

    Solution :

    \[K={{K}_{0}}=\]total energy As total energy remains conserved itself hence when U is maximum in\[SHM\],\[K=0\] i.e.,\[E\]is also equal to\[{{U}_{\max }}i.e.,\] \[{{U}_{\max }}=E={{K}_{0}}\]


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