A) \[\frac{E}{2}\]
B) \[\frac{E}{4}\]
C) \[\frac{3E}{4}\]
D) \[\frac{\sqrt{3}}{4}E\]
Correct Answer: C
Solution :
Total energy in SHM \[E=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\]; (where a = amplitude) Potential energy \[U=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})=E-\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] When \[y=\frac{a}{2}\]Þ \[U=E-\frac{1}{2}m{{\omega }^{2}}\left( \frac{{{a}^{2}}}{4} \right)=E-\frac{E}{4}=\frac{3E}{4}\]
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