WB JEE Medical WB JEE Medical Solved Paper-2007

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

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

    B)  \[\frac{a}{2}\]

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

    D)  \[\frac{a}{\sqrt{3}}\]

    Correct Answer: B

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner