JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Critical Thinking

  • question_answer
    The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0 sec, the tension in the string when the bob passes through O is                                                  [AMU 1995]

    A)            \[m\,(g+\pi \sqrt{2g\,h})\]    

    B)            \[m\,(g+\sqrt{{{\pi }^{2}}g\,h})\]

    C)            \[m\,\left( g+\sqrt{\frac{{{\pi }^{2}}}{2}g\,h} \right)\]        

    D)            \[m\,\left( g+\sqrt{\frac{{{\pi }^{2}}}{3}g\,h} \right)\]

    Correct Answer: A

    Solution :

                       Tension in the string when bob passes through lowest point    \[T=mg+\frac{m{{v}^{2}}}{r}=mg+mv\omega \]     (\[\because \] v = rw) putting \[v=\sqrt{2gh}\]and \[\omega ==\frac{2\pi }{T}=\frac{2\pi }{2}=\pi \]            we get \[T=m\ (g+\pi \sqrt{2gh})\]


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