NEET NEET SOLVED PAPER 2015 (C)

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

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

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

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

    Correct Answer: B

    Solution :

    Let A be the amplitude of oscillation then \[v_{1}^{2}={{\omega }^{2}}({{A}^{2}}-x_{1}^{2})\,\,\]                  ?(i) \[v_{2}^{2}={{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\,\,\,\]                 ?(ii) Subtracting Eq. (ii) from Eq. (i), we get \[v_{1}^{2}-v_{2}^{2}={{\omega }^{2}}(x_{2}^{2}-x_{1}^{2})\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\omega =\sqrt{\frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2}-x_{1}^{2}}}\] \[\Rightarrow \,\,\,\,\,\,\frac{2\pi }{T}=\sqrt{\frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2}-x_{1}^{2}}}\] \[\Rightarrow \,\,\,\,\,T=2\pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}\]


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