A) 1/4
B) 1/2
C) 1/8
D) 1
Correct Answer: A
Solution :
Potential energy is given by \[=\frac{1}{2}m{{a}^{2}}{{\omega }^{2}}{{\sin }^{2}}\omega t\] ?(i) Total energy \[=\frac{1}{2}m{{a}^{2}}{{\omega }^{2}}\] ?(ii) Now from (i), (ii) and (iii), we get \[\frac{\text{Potential}\,\text{energy}}{\text{Total}\,\text{energy}}={{\sin }^{2}}\omega t=\frac{{{x}^{2}}}{{{a}^{2}}}\] Now for \[x=\frac{a}{2},\]we get \[\frac{\text{Potential}\,\text{energy}}{\text{Total}\,\text{energy}}={{\left( \frac{a}{2} \right)}^{2}}\times \frac{1}{{{a}^{2}}}=\frac{1}{4}\]
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