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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Oscillations

  • question_answer
    A particle executing harmonic motion is having velocities \[{{v}_{1}}\] and \[{{v}_{2}}\] at distances is x, and x, from the equilibrium position. The amplitude of the motion is

    A) \[\sqrt{\frac{v_{1}^{2}{{x}_{2}}-v_{2}^{2}{{x}_{1}}}{v_{1}^{2}+v_{2}^{2}}}\]

    B) \[\sqrt{\frac{v_{1}^{2}x_{1}^{2}-v_{2}^{2}x_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}\]

    C) \[\sqrt{\frac{v_{1}^{2}x_{2}^{2}-v_{2}^{2}x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}\]

    D) \[\sqrt{\frac{v_{1}^{2}x_{2}^{2}+v_{2}^{2}x_{1}^{2}}{v_{1}^{2}+v_{2}^{2}}}\]

    Correct Answer: C

    Solution :

    [c] \[{{v}_{1}}=\omega \sqrt{{{a}^{2}}-x_{1}^{2}},\,\,{{v}_{2}}=\omega \sqrt{{{a}^{2}}-x_{2}^{2}}\] We get, \[a=\sqrt{\frac{v_{1}^{2}x_{2}^{2}-v_{2}^{2}x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}\]


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