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NEET Sample Paper NEET Sample Test Paper-83

  • question_answer
    A particle is executing linear simple harmonic motion about origin\[\operatorname{x} = 0\]. Which of the group shown in figure represent the variation of the potential energy function U(x) versus x?

    A)  

    B)  

    C)             

    D)  

    Correct Answer: D

    Solution :

    In simple harmonic motion, the force acting on the particle (restoring force) is given by \[\operatorname{F} = - kx\] (where k is positive constant) Now,      \[F=-\frac{dU}{dx}\] \[\therefore \,\,\,\,-kx=-\frac{dU}{dx}\,\,\,\,\,or\,\,\,dU=kdx\] \[\therefore \,\,\,\,\,\,U(x)=\int_{0}^{x}{kdx}=\frac{1}{2}k{{x}^{2}}+C\]       (C is constant of integration) In SHM the potential energy of the oscillator is zero at mean position, i.e., \[\operatorname{U}\left( 0 \right) =0\], Hence, \[\operatorname{c}= 0\] \[\therefore \,\,\,\,\,\,\,U\text{(}x)=\frac{1}{2}k{{x}^{2}}.\]which is equation of upward opening parabola.


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