CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    The total energy of a particle executing SHM is 80 J. What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position?

    A)  60 J                                       

    B)  10 J                       

    C)  40 J                       

    D)         45 J

    E)  zero

    Correct Answer: D

    Solution :

    \[E=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}=80\,J\]                             ?.. (1) \[\therefore \]  \[U=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{\left( \frac{3a}{4} \right)}^{2}}\]                 \[=\frac{9}{16}\times 80=45\,J\]


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