Manipal Engineering Manipal Engineering Solved Paper-2012

  • 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: A

    Solution :

    \[{{K}_{\max }}={{K}_{0}}=\]total energy As total energy remains conserved in\[SHM\], 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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