JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer A particle starts S.H.M. from the mean position. Its amplitude is a and total energy E. At one instant5 its kinetic energy is \[3\text{ }E/4,\] its displacement at this instant is

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

    B)  \[y=\frac{a}{2}\]

    C)  \[y=\frac{a}{\sqrt{3/2}}\]                           

    D)  \[y=a\]

    Correct Answer: B

    Solution :

     Total energy, \[E=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}};\] \[K.E.=\frac{3E}{4}=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}}).\] So, \[\frac{3}{4}=\frac{{{a}^{2}}-{{y}^{2}}}{{{a}^{2}}}\] or \[{{y}^{2}}=\frac{{{a}^{2}}}{4}\] or \[y=\frac{a}{2}\].


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