CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    When the displacement is one-half the amplitude in SHM, the fraction of the total enersy that is potential is:

    A)  \[\frac{1}{2}\]  

    B)                         \[\frac{3}{4}\]                  

    C)  \[\frac{1}{4}\]                  

    D)         \[\frac{1}{8}\]

    E)  \[\frac{1}{6}\]

    Correct Answer: C

    Solution :

    \[E=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] \[U=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] \[\Rightarrow \]               \[U=\frac{1}{2}m{{\omega }^{2}}\times {{\left( \frac{1}{2}a \right)}^{2}}\] \[\therefore \]  \[U=\frac{1}{4}\times \frac{1}{2}m{{\omega }^{2}}{{a}^{2}}=\frac{1}{4}E\]


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