JEE Main & Advanced Sample Paper JEE Main - Mock Test - 22

  • question_answer
    The kinetic energy of a particle executing SHM will be equal to (1/8)th of its potential energy when its displacement from the mean position is (where A is the amplitude)

    A) \[A\sqrt{2}\]                  

    B) A/2

    C) \[\frac{2\sqrt{2}}{3}A\] 

    D) \[A\sqrt{\frac{2}{3}}\]

    Correct Answer: C

    Solution :

    [c] : Kinetic energy when the displacement of the particle is x, is given by\[K=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\] Potential energy at this instant is given by \[U=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] According to the condition given in the question \[K=\frac{1}{8}U\]\[\Rightarrow \] \[\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})=\frac{1}{8}.\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] or \[({{A}^{2}}-{{x}^{2}})=\frac{{{x}^{2}}}{8}\] On rearranging, we get \[x=\frac{2\sqrt{2}}{3}A\]


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