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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 44

  • question_answer
    The potential energy of a particle of mass 1 kg in motion along the x-axis is given by: \[U=4\left( 1-cos\,2x \right)J\], where x is in metres. The period of small oscillations (in sec) is -

    A) \[2\pi \]             

    B)        \[\pi \]    

    C) \[\frac{\pi }{2}\]   

    D)        \[\sqrt{2}\pi \]

    Correct Answer: C

    Solution :

    [c] \[U=4\left( 1-cos2x \right)J\] \[F=-\frac{dU}{dx}=-8\sin 2x\] for small oscillation \[sin2x\approx 2x\] \[F=-16x\] Comparing with \[F=-kx\] k=16 \[m{{\omega }^{2}}=16\] \[\omega =\sqrt{\frac{16}{m}}\] \[T=2\pi \sqrt{\frac{m}{16}}\] \[T=2\pi \sqrt{\frac{1}{16}}=\frac{\pi }{2}\sec \]


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