• # question_answer 58) 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)  $y=a$

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 $y=\frac{a}{2}$.