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

  • question_answer
    A particle is executing SHM along a straight line. Its velocities at distances \[{{x}_{1}}\]and \[{{x}_{2}}\]from the mean position are \[{{V}_{1}}\]and\[{{V}_{2}}\], respectively. Its time period is

    A) \[2\pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{V_{1}^{2}-V_{2}^{2}}}\]

    B) \[2\pi \sqrt{\frac{V_{1}^{2}+V_{2}^{2}}{x_{1}^{2}+x_{2}^{2}}}\]

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

    D) \[2\pi \sqrt{\frac{x_{1}^{2}-x_{2}^{2}}{V_{1}^{2}-V_{2}^{2}}}\]

    Correct Answer: A

    Solution :

    [a] As we know, for particle undergoing SHM, \[V=\sqrt[\omega ]{{{A}^{2}}-{{X}^{2}}};\]                \[V_{1}^{2}={{\omega }^{2}}({{A}^{2}}-x_{1}^{2})\] \[V_{2}^{2}={{\omega }^{2}}({{A}^{2}}-x_{2}^{2})\] Substructing we get, \[\frac{V_{1}^{2}}{{{\omega }^{2}}}+x_{1}^{2}=\frac{V_{2}^{2}}{{{\omega }^{2}}}+x_{2}^{2}\] \[w=\sqrt{\frac{V_{1}^{2}-V_{2}^{2}}{x_{2}^{2}-x_{1}^{2}}}\Rightarrow T=2\pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{V_{1}^{2}-V_{2}^{2}}}\]


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