AIIMS AIIMS Solved Paper-1999

  • question_answer
    For a particle executing simple harmonic motion, which of the following statements is not correct?

    A) restoring force is maximum at the extreme positions

    B) total energy of the particle always remains the same

    C) restoring force is always directed towards a fixed point

    D) acceleration of the particle is maximum at the equilibrium position.

    Correct Answer: D

    Solution :

                    The restoring force in a SHM is given by \[F=-Kx\]\[\Rightarrow \]if x is maximum F is also maximum. Hence the restoring force is maximum at the extreme positions. In SHM the total energy of the particle always remains constant and the restoring force as the name suggests, always acts towards a fixed point. We know that the displacement of a particle in SHM is given by    \[y=A\,\sin \omega t\] \[\Rightarrow \]  \[v=\frac{dy}{dt}=-A\omega \,\cos \,\omega t\] \[\therefore \]  \[a=\frac{dy}{dt}=-A{{\omega }^{2}}\,\,\sin \omega t\] \[\therefore \]  \[\frac{{{a}^{2}}}{{{A}^{2}}{{\omega }^{4}}}+\frac{{{v}^{2}}}{{{A}^{2}}{{\omega }^{2}}}=1\] \[\Rightarrow \]\[{{a}^{2}}={{\omega }^{2}}[{{A}^{2}}{{\omega }^{2}}-{{v}^{2}}]\] From this expression, it is clear that if v is maximum a is minimum and if v is minimum, a is maximum. As in the equilibrium position, the velocity is maximum, the acceleration is going to be minimum. Hence expression (d) is incorrect.


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