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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Energy of Simple Harmonic Motion

  • question_answer
    For a particle executing simple harmonic motion, the kinetic energy K is given by \[K={{K}_{o}}{{\cos }^{2}}\omega t\]. The maximum value of potential energy is                                                           [CPMT 1981]

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

    B)            Zero

    C)            \[\frac{{{K}_{0}}}{2}\]      

    D)            Not obtainable

    Correct Answer: A

    Solution :

                       Since maximum value of \[{{\cos }^{2}}\omega t\]is 1.                    \[\therefore {{K}_{\max }}={{K}_{o}}{{\cos }^{2}}\omega t={{K}_{o}}\]            Also \[{{K}_{\max }}=P{{E}_{\max }}={{K}_{o}}\]


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