• # question_answer When the kinetic energy of a body executing S.H.M. is 1/3 of the potential energy. The displacement of the body is $x$ percent of the amplitude, where x is : A)  33                                          B)  87 C)  67                                          D)  50

The relation for kinetic energy of S.H.M. is given by                                          $=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})$          ??.(1) Potential energy is given by                                          $=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}$              .....(2) Now for the condition of question and from eqs. (1) and (2) $\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})=\frac{1}{3}\times \frac{1}{2}m{{\omega }^{2}}{{y}^{2}}$ or                                    $\frac{4}{6}m{{\omega }^{2}}{{y}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}$ or                                    ${{y}^{2}}=\frac{3}{4}{{a}^{2}}$ So,                                  $y=\frac{a}{2}\sqrt{3}=0.866a$                                          $\approx 87%$ of amplitude