question_answer

What is the ratio between the potential  energy and the total energy of a particle executing S.H.M., when its displacement is half of its amplitude?

Answer:

                \[\frac{\text{Potential energy}}{\text{Total energy}}=\frac{\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}}{\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}}\]                 \[\frac{{{y}^{2}}}{{{a}^{2}}}=\frac{{{(a/2)}^{2}}}{{{a}^{2}}}=\frac{1}{4}\mathbf{1:4}\]