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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    A particle of mass m is located in a one-dimensional 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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