A) 1/16
B) 1/8
C) 1/2
D) ¼
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 \[=\frac{\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}}{\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}}\] \[=\frac{{{y}^{2}}}{{{a}^{2}}}\] \[=\frac{{{\left( \frac{1}{4}a \right)}^{2}}}{{{a}^{2}}}\] \[\left[ As\,y=\frac{1}{4}a \right]\] \[=\frac{1}{16}\]
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