NEET Sample Paper NEET Sample Test Paper-14

  • question_answer The potential energy of a particle of mass 0.1 kg moving along the \[x-\]axis is given by \[U=5x(x-4)\,J,\]where\[x\]is in metre. It can be concluded that:

    A)  the particle is acted upon by a constant force.

    B)  the speed of the particle is maximum at x = 2 m.

    C)  the particle cannot execute simple harmonic motion.

    D)  the period of oscillation of the particle is \[\frac{\pi }{20}s\]

    Correct Answer: B

    Solution :

    \[F=-\frac{dU}{dx}=-\frac{d}{dx}[5x(x-4)]=-(10x-20)\] \[\therefore \]\[a=\frac{F}{m}=-\left( \frac{10x-20}{0.1} \right)=-100(x-2)\] or \[a=-100X\] where \[X=x-2\] Hence, the motion is simple harmonic. In SHM, \[a=-{{\omega }^{2}}\times \]\[\therefore \]\[\omega =10\] or \[\frac{2\pi }{T}=10\,or\,T=\frac{\pi }{5}\,\sec .\] In the case of SHM. The speed is maximum at mean position. This position is given by \[X=0\]or \[x-2=0\]or \[x=2\,\,m.\] Hence the correction option is [b].


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