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KVPY Sample Paper KVPY Stream-SX Model Paper-17

  • 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:

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

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

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

    D) A

    Correct Answer: C

    Solution :

    \[PE=KE\] \[\Rightarrow \]   \[\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[\Rightarrow \]   \[x=\frac{A}{\sqrt{2}}\]


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