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

  • question_answer
    The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement  (amplitude = a) is [MP PMT 1987; CPMT 1990; DPMT 1996;  MH CET 1997, 99; AFMC 1999; CPMT 2000]

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

    B)            \[a\sqrt{2}\]

    C)            \[\frac{a}{\sqrt{2}}\]        

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

    Correct Answer: C

    Solution :

                       Suppose at displacement y from mean position potential energy = kinetic energy Þ \[\frac{1}{2}m({{a}^{2}}-{{y}^{2}}){{\omega }^{2}}=\frac{1}{2}m{{\omega }^{\text{2}}}{{y}^{2}}\] Þ \[{{a}^{2}}=2{{y}^{2}}\] Þ \[y=\frac{a}{\sqrt{2}}\]


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