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  3. BHU PMT

BHU PMT BHU PMT (Screening) Solved Paper-2008

  • 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 :

                                    \[{{K}_{\max }}={{K}_{0}}=total\text{ }energy\]


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