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VMMC VMMC Medical Solved Paper-2010

  • question_answer
    A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic?

    A)  1 cm     

    B)                        \[\sqrt{2}\]cm

    C)  3 cm                     

    D)         \[2\sqrt{2}\]cm

    Correct Answer: D

    Solution :

    Let \[x\] be the point where Hence \[\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{x}^{2}})=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[2{{x}^{2}}={{a}^{2}}\] \[x=\frac{a}{\sqrt{2}}\] \[x=\frac{4}{\sqrt{2}}=2\sqrt{2}\,cm\]


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