A) \[\frac{\sqrt{3}}{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{3}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
For a body executing SHM, the total energy \[{{\text{E}}_{\text{T}}}\] is given by \[{{E}_{T}}=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\]and \[{{E}_{K}}=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{y}^{2}})\] Given, \[y=\frac{A}{\sqrt{2}}\] \[{{E}_{K}}=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-\frac{{{A}^{2}}}{2} \right)=\frac{{{E}_{T}}}{2}\] Hence, \[\frac{{{E}_{K}}}{{{E}_{T}}}=\frac{1}{2}\] Note: In SHM kinetic energy is converted to potential energy and vice versa, but total energy remains same.
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