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

  • question_answer
    When the displacement is half the amplitude, the ratio of potential energy to the total energy is [CPMT 1999; JIPMER 2000; Kerala PET 2002]

    A)            \[\frac{1}{2}\]                      

    B)            \[\frac{1}{4}\]

    C)            \[1\]                                        

    D)            \[\frac{1}{8}\]

    Correct Answer: B

    Solution :

               \[\frac{U}{E}=\frac{\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}}{\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}}=\frac{{{y}^{2}}}{{{a}^{2}}}=\frac{{{\left( \frac{a}{2} \right)}^{2}}}{a}=\frac{1}{4}\]


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