CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    A particle starts SHM from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is\[3\frac{E}{4}\]. Its displacement at that instant is:

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

    B)  \[\frac{a}{2}\]

    C)  \[\frac{a}{\sqrt{\left( \frac{3}{2} \right)}}\]        

    D)         \[\frac{a}{\sqrt{3}}\]

    E)  \[a\]

    Correct Answer: B

    Solution :

    \[k=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\] \[\frac{3}{4}E=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\] \[\frac{3}{4}\left( \frac{1}{2}m{{\omega }^{2}}{{a}^{2}} \right)=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\] \[{{y}^{2}}={{a}^{2}}-\frac{3}{4}{{a}^{2}}\] \[=\frac{{{a}^{2}}}{4}\] \[\Rightarrow \]               \[y=\frac{a}{2}\]


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