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

  • question_answer
    The angular velocity and the amplitude of a simple pendulum is \[\omega \] and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is          [CBSE PMT 1991]

    A)            \[{{X}^{2}}{{\omega }^{2}}/({{a}^{2}}-{{X}^{2}}{{\omega }^{2}})\] 

    B)            \[{{X}^{2}}/({{a}^{2}}-{{X}^{2}})\]

    C)            \[({{a}^{2}}-{{X}^{2}}{{\omega }^{2}})/{{X}^{2}}{{\omega }^{2}}\] 

    D)            \[({{a}^{2}}-{{X}^{2}})/{{X}^{2}}\]

    Correct Answer: D

    Solution :

                       Kinetic energy \[T=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{x}^{2}})\] and potential energy, \[V=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[\therefore \frac{T}{V}=\frac{{{a}^{2}}-{{x}^{2}}}{{{x}^{2}}}\]


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