CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    The KE and PE of a particle executing SHM of amplitude a will be equal when displacement is:

    A)  \[\frac{a}{2}\]                                  

    B)  \[a\sqrt{2}\]

    C)  \[2a\]                  

    D)         \[\frac{a\sqrt{2}}{3}\]

    E)  \[\frac{a}{\sqrt{2}}\]

    Correct Answer: E

    Solution :

    \[KE=PE\] \[\Rightarrow \]               \[\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] \[\Rightarrow \]               \[y=\frac{a}{\sqrt{2}}\]


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