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

  • question_answer
    The total energy of a particle executing S.H.M. is 80 J. What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position  [Kerala (Engg.) 2001]

    A)            60 J                                           

    B)            10 J

    C)            40 J

    D)            45 J

    Correct Answer: D

    Solution :

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


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