JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about\[x=0\]. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

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

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

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

    D)               A

    Correct Answer: B

    Solution :

    \[KE=PE\] \[\frac{1}{2}k({{A}^{2}}-{{x}^{2}})=\frac{1}{2}k{{x}^{2}}\] \[{{A}^{2}}\,=\,\,2{{x}^{2}}\] \[x=\frac{A}{\sqrt{2}}\]


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