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JCECE Medical JCECE Medical Solved Paper-2015

  • question_answer
    A particle of mass m is located in a unidimensional potential field, where the potential energy of the particle depends on the coordinates as \[U(x)={{U}_{0}}\]\[(1-\sin bx);\] where \[{{U}_{0}}\]and b are constant. Find the period of small oscillations that the particle performs about the equilibrium position.

    A) \[\frac{2\pi }{{{b}^{2}}}\sqrt{\frac{m}{{{U}_{0}}}}\]

    B)  \[\frac{\pi }{b}\sqrt{\frac{m}{{{U}_{0}}}}\]

    C)  \[\frac{{{\pi }^{2}}}{2b}\sqrt{\frac{m}{{{U}_{0}}}}\]

    D)  \[\frac{2\pi }{b}\sqrt{\frac{m}{{{U}_{0}}}}\]

    Correct Answer: D

    Solution :

     According to the question, given that \[U(x)={{U}_{0}}(1-\cos bx)=2{{U}_{0}}{{\sin }^{2}}\frac{bx}{2}\] For small oscillations, \[\sin \frac{bx}{2}\approx \frac{bx}{2}\] Thus, \[U(x)=2{{U}_{0}}{{\left( \frac{bx}{2} \right)}^{2}}=\frac{{{U}_{0}}{{b}^{2}}{{x}^{2}}}{2}\] Comparing with\[U=\frac{1}{2}k{{x}^{2}},\]we get \[k={{U}_{0}}{{b}^{2}}\] \[\therefore \]Time period, \[T=2\pi \sqrt{\frac{m}{k}}\] \[=2\pi \sqrt{\frac{m}{{{U}_{0}}{{b}^{2}}}}=\frac{2\pi }{b}\sqrt{\frac{m}{{{U}_{0}}}}\]


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