A) \[1/16\]
B) \[1/8\]
C) \[1/2\]
D) \[1/4\]
Correct Answer: A
Solution :
For SHM, potential energy \[=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] and total energy \[=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] Therefore, fraction of potential energy \[=\frac{\frac{1}{2}m\,{{\omega }^{2}}{{y}^{2}}}{\frac{1}{2}m\,{{\omega }^{2}}{{a}^{2}}}\] \[=\frac{{{y}^{2}}}{{{a}^{2}}}\] \[=\frac{{{\left( \frac{1}{4}\,a \right)}^{2}}}{{{a}^{2}}}\] \[\left[ As\,\,\,y=\frac{1}{4}a \right]\]
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