A) \[\frac{E}{2}\]
B) \[\frac{E}{4}\]
C) \[\frac{3E}{4}\]
D) \[\frac{\sqrt{3E}}{4}\]
Correct Answer: C
Solution :
Total energy in SHM \[E=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}},\] (where a = amplitude) Kinetic energy \[K=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\] \[=E-\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] when \[y=\frac{a}{2}\] \[\Rightarrow \] \[K=E-\frac{1}{2}m{{\omega }^{2}}\left( \frac{{{a}^{2}}}{4} \right)=E-\frac{E}{4}\] \[K=\frac{3E}{4}\]
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